Ch19 ISM

Chapter 19 The Second Law of Thermodynamics Conceptual Problems Modern automobile gasoline engines have efficiencies of about 25%. 1 • About what percentage of the heat of combustion is not used for work but released as heat? (a) 25%, (b) 50%, (c) 75%, (d ) 100%, (e) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work  done per cycle W to the heat absorbed from the high-temperature reservoir Qh. The percentage of the heat of combustion (heat absorbed from the hightemperature reservoir) is the ratio of Qc to Qh. We can use the relationship  between W, Qh, and Qc ( W  = Qh − Qc ) to find Qc/ Qh.

Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain:

ε 

Solving for Qc/ Qh yields:

Qc

=

Qh

Substitute for ε  to obtain:

Qc Qh

and

W  Qh

=

Qh

− Qc

Qh

= 1−

Qc Qh

= 1 − ε 

= 1 − 0.25 = 0.75

(c )

is correct.

If a heat engine does 100 kJ of work per cycle while releasing 400 kJ 2 • of heat, what is its efficiency? ( a) 20%, (b) 25%, (c) 80%, (d ) 400%, (e) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work  done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( W  = Qh − Qc ) to express the

efficiency of the heat engine in terms of Qc and W . Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain:

ε 

1867

=

W  Qh

=

W  W  + Qc

=

1 Q 1+ c W 

1868

Chapter 19

Substitute for Qc and W to obtain:

= and

1 400 kJ 1+ 100 kJ

(a )

= 0.2

is correct.

If the heat absorbed by a heat engine is 600 kJ per cycle, and it 3 • releases 480 kJ of heat each cycle, what is its efficiency? (a) 20%, (b) 80%, (c) 100%, (d ) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work  done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( W  = Qh − Qc ) to express the

efficiency of the heat engine in terms of Qc and Qh. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain: Substitute for Qc and Qh to obtain:

ε 

=

•

Qh

= 1− and

4

W 

=

Qh

− Qc

Qh

480 kJ 600 kJ

(a )

= 1−

Qc Qh

= 0.2

is correct.

Explain what distinguishes a refrigerator from a ″heat pump.″

Determine the Concept The job of a refrigerator is to move heat from its cold interior to the warmer kitchen environment. This process moves heat in a direction that is opposite its ″natural″ direction of flow, analogous to the use of a water pump to pump water out of a boat. The term heat pump is used to describe devices, such as air conditioners, that are used to cool living and working spaces in the summer and warm them in the winter.

An air conditioner’s COP is mathematically identical to that Q of a refrigerator, that is, COPAC = COPref =  c . However a heat pump’s COP is W  Q defined differently, as COPhp = h . Explain clearly why the two COPs are W  defined differently. Hint : Think of the end use of the three different devices. 5

•

[SSM]

Determine the Concept The COP is defined so as to be a measure of the effectiveness of the device. For a refrigerator or air conditioner, the important quantity is the heat drawn from the already colder interior, Qc. For a heat pump, the ideas is to focus on the heat drawn into the warm wa rm interior of the house, Qh.

1868

Chapter 19

Substitute for Qc and W to obtain:

= and

1 400 kJ 1+ 100 kJ

(a )

= 0.2

is correct.

If the heat absorbed by a heat engine is 600 kJ per cycle, and it 3 • releases 480 kJ of heat each cycle, what is its efficiency? (a) 20%, (b) 80%, (c) 100%, (d ) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work  done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( W  = Qh − Qc ) to express the

efficiency of the heat engine in terms of Qc and Qh. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain: Substitute for Qc and Qh to obtain:

ε 

=

•

Qh

= 1− and

4

W 

=

Qh

− Qc

Qh

480 kJ 600 kJ

(a )

= 1−

Qc Qh

= 0.2

is correct.

Explain what distinguishes a refrigerator from a ″heat pump.″

Determine the Concept The job of a refrigerator is to move heat from its cold interior to the warmer kitchen environment. This process moves heat in a direction that is opposite its ″natural″ direction of flow, analogous to the use of a water pump to pump water out of a boat. The term heat pump is used to describe devices, such as air conditioners, that are used to cool living and working spaces in the summer and warm them in the winter.

An air conditioner’s COP is mathematically identical to that Q of a refrigerator, that is, COPAC = COPref =  c . However a heat pump’s COP is W  Q defined differently, as COPhp = h . Explain clearly why the two COPs are W  defined differently. Hint : Think of the end use of the three different devices. 5

•

[SSM]

Determine the Concept The COP is defined so as to be a measure of the effectiveness of the device. For a refrigerator or air conditioner, the important quantity is the heat drawn from the already colder interior, Qc. For a heat pump, the ideas is to focus on the heat drawn into the warm wa rm interior of the house, Qh.

The Second Law of Thermodynamics

1869

Explain why you cannot cool your kitchen by leaving your refrigerator  6 • door open on a hot day. (Why does turning on a room air conditioner cool down the room, but opening a refrigerator door does not?) Determine the Concept As described by the second law of thermodynamics, more heat must be transmitted to the outside world than is removed by a refrigerator or air conditioner. The heating coils on a refrigerator are inside the room and so the refrigerator actually heats the room in which it is located. The heating coils on an air conditioner are outside one’s living space, so the waste heat is vented to the outside.

Why do steam-power-plant designers try to increase the temperature 7 • of the steam as much as possible? Determine the Concept Increasing the temperature of the steam increases the Carnot efficiency, and generally increases the efficiency of any heat engine.

To increase the efficiency of a Carnot engine, you should 8 • (a) decrease the temperature of the hot reservoir, (b) increase the temperature of  the cold reservoir, (c) increase the temperature of the hot reservoir, (d ) change the ratio of maximum volume to minimum volume. Determine the Concept Because the efficiency of a Carnot cycle engine is given T   by ε C = 1 − c , you should increase the temperature of the hot reservoir. ( c ) is T h

correct. 9 •• [SSM] Explain why the following statement is true: To increase the efficiency of a Carnot engine, you should make the difference between the two operating temperatures as large as possible; but to increase the efficiency of a Carnot cycle refrigerator , you should make the difference between the two operating temperatures as small as possible. Determine the Concept A Carnot-cycle refrigerator is more efficient when the temperatures are close together because it is easier to extract heat from an already cold interior if the temperature of the exterior is close to the temperature of the interior of the refrigerator. A Carnot-cycle heat engine is more efficient when the temperature difference is large because then more work is done by the engine for  each unit of heat absorbed from the hot reservoir.

A Carnot engine operates between a cold temperature reservoir of  10 •• o 27 C and a high temperature reservoir of 127°C. Its efficiency is ( a) 21%, (b) 25%, (c) 75%, (d ) 79%.

1870

Chapter 19

Determine the Concept The efficiency of a Carnot cycle engine is given by T c where T c and T h (in kelvins) are the temperatures of the cold and hot ε C = 1 − T h

reservoirs, respectively. Substituting numerical values for T c and T h yields:

C

= 1−

(b )

300 K  = 0.25 400 K 

is correct.

The Carnot engine in Problem 10 is run in reverse as a refrigerator. Its 11 •• COP is (a) 0.33, (b) 1.3, (c) 3.0 (d ) 4.7. Determine the Concept The coefficient of performance of a Carnot cycle engine Q run in reverse as refrigerator is given by COPref  = c . We can use the relationship W   between W, Qc, and Qh to eliminate W from this expression and then use the Q T  relationship, applicable only to a device operating in a Carnot cycle, c = c to Qh T h

express the refrigerator’s COP in terms of T c and T h. The coefficient of performance of a refrigerator is given by:

COPref  =

W  or, because W  = Qh

COPref  = Dividing the numerator and denominator of this fraction by Qc yields:

COPref  =

For a device operating in a Carnot cycle:

Qc

Substitute in the expression for  COPref  to obtain:

Qc

Qc Qh

− Qc

1 Qh Qc

Qh

=

−1

T c T h

COPref, C

=

1 T h T c

−1

− Qc ,

The Second Law of Thermodynamics Substitute numerical values and evaluate COPref, C:

COPref, C

(c )

=

1871

1 = 3.0 400 K  −1 300 K 

is correct.

On a humid day, water vapor condenses on a cold surface. During 12 •• condensation, the entropy of the water (a) increases, (b) remains constant, (c) decreases, (d ) may decrease or remain unchanged. Explain your answer. Determine the Concept When water vapor condenses, its entropy decreases (the liquid state is a more ordered state than is the vapor state) and the entropy of the

universe increases. ( a ) is correct. An ideal gas is taken reversibly from an initial state Pi, V i, T i to the 13 •• final state Pf , V f , T f . Two possible paths are (A) an isothermal expansion followed  by an adiabatic compression and (B) an adiabatic compression followed by an isothermal expansion. For these two paths, (a) Δ E int A > Δ E int B, (b) ΔS A > ΔS B, (c) ΔS A < ΔS B, (d ) None of the above. Determine the Concept The two paths are shown on the PV  diagram to the right. We can use the concept of a state function to choose from among the alternatives given as possible answers to the problem.

P

 B

 B

T i

f 

Pf  Pi

i

 A T f   A V 

V i

V f 

(a) Because E int is a state function and the initial and final states are the same for  the two paths, Δ E int, A = Δ E int, B . (b) and (c) S , like E int , is a state function and its change when the system moves from one state to another depends only on the system’s initial and final states. It is not dependent on the process by which the change occurs. Thus ΔS A = ΔS B . (d ) ( d ) is correct. Figure 19-12 shows a thermodynamic cycle for an ideal gas on an ST  14 •• diagram. Identify this cycle and sketch it on a PV diagram.

1872

Chapter 19

Determine the Concept The processes A→B and C→D are adiabatic and the  processes B→C and D→A are isothermal. Therefore, the cycle is the Carnot cycle shown in the adjacent PV  diagram.

P B

C

A

D V 

Figure 19-13 shows a thermodynamic cycle for an ideal gas on an SV  15 •• diagram. Identify the type of engine represented by this diagram. Determine the Concept  Note that A→B is an adiabatic expansion, B→C is a constant-volume process in which the entropy decreases, C→D is an adiabatic compression and D→A is a constant-volume process that returns the gas to its original state. The cycle is that of the Otto engine (see Figure 19-3). The points A, B, C, and D in Figure 19-13 correspond to points c, d, a, and b, respectively, in Figure 19-3.

Sketch an ST diagram of the Otto cycle. (The Otto cycle is discussed 16 •• in Section 19-1.) Determine the Concept The Otto cycle consists of four quasi-static steps. Refer  to Figure 19-3. There a→b is an adiabatic compression, b→c is a constant volume heating, c→d  is an adiabatic expansion and d →a is a constant-volume cooling. So, from a to b, S  is constant and T  increases, from b to c, heat is added to the system and both S and T  increase, from c→d S  is constant while T decreases, and from d to a both S and T decrease.

To determine how S depends on T  along b→c and d →a, consider the entropy change of the gas from point b to an arbitrary point on the path b→c where the entropy and temperature of the gas are S and T, respectively: Because W on = 0 for this constantvolume process: Substituting for Q yields:

ΔS  =

Q

T  where, because heat is entering the system, Q is positive.

Δ E int = Qin = Q = C V ΔT  = C V (T  − T b ) ΔS  =

C V (T  − T b ) T 

⎛  T   ⎞ = C V ⎜1 − b ⎟ ⎝  T  ⎠

The Second Law of Thermodynamics

On path b→c the entropy is given by:

The first and second derivatives, dS  dT  and d 2 S  dT 2 , give the slope and concavity of the path. Calculate these derivatives assuming C V is constant. (For an ideal gas C V is a  positive constant.):

S  = S b

dS 

⎛  T   ⎞ + ΔS  = S b + C V ⎜1 − b ⎟ ⎝  T  ⎠

= C V

dT  d 2 S  2

dT 

1873

T b T 2

= −2C V

T b T 3

These results tell us that, along path b→c, the slope of the path is positive and the slope decreases as T increases. The concavity of the path is negative for all T. Following the same procedure on  path d →a gives:

⎛  ⎝ 

S  = S d  + C V ⎜1 − dS 

= C V

dT  d 2 S  2

dT 

T d  ⎞

⎟

T  ⎠

T d  T 2

= −2C V

T d  T 3

These results tell us that, along path d →a, the slope of the path is positive and the slope decreases as T increases. The concavity of the path is negative for all T. An ST diagram for the Otto cycle is shown to the right.

S  d 

a

c

b

T 

17 ••

[SSM]

Sketch an SV diagram of the Carnot cycle for an ideal gas.

Determine the Concept Referring to Figure 19-8, process 1→2 is an isothermal expansion. In this process heat is added to the system and the entropy and volume increase. Process 2→3 is adiabatic, so S  is constant as V  increases. Process 3→4 is an isothermal compression in which S  decreases and V  also decreases. Finally,  process 4→1 is adiabatic, that is, isentropic, and S is constant while V decreases.

1874

Chapter 19

During the isothermal expansion (from  point 1 to point 2) the work done by the gas equals the heat added to the gas. The change in entropy of the gas from point 1 (where the temperature is T 1) to an arbitrary point on the curve is given by: For an isothermal expansion, the work done by the gas, and thus the heat added to the gas, are given by: Substituting for Q yields:

Since S  = S 1 + ΔS , we have:

The graph of S as a function of V for an isothermal expansion shown to the right was plotted using a spreadsheet  program. This graph establishes the curvature of the 1→2 and 3→4 paths for the SV graph.

ΔS  =

Q T 1

⎛ V  ⎞ ⎟⎟ V  ⎝  1 ⎠

Q = W  = nRT 1 ln⎜⎜

⎛ V  ⎞ ΔS  = nR ln⎜⎜ ⎟⎟ ⎝ V 1 ⎠ ⎛ V  ⎞ ⎟⎟ V  ⎝  1 ⎠

S  = S 1 + nR ln⎜⎜

       S

V 

An SV graph for the Carnot cycle (see Figure 19-8) is shown to the right.

2

1

3

4 V 

Sketch an SV diagram of the Otto cycle. (The Otto cycle is discussed in 18 •• Section 19-1.)

The Second Law of Thermodynamics

1875

Determine the Concept The Otto cycle is shown in Figure 19-3. Process a→b takes place adiabatically and so both Q = 0 and ΔS = 0 along this path. Process b→c takes place at constant volume. Qin, however, is positive and so, while ΔV = 0 along this path, Q > 0 and, therefore ΔS > 0. Process c→d also takes place adiabatically and so, again, both Q = 0 and ΔS = 0 along this path. Finally, process d →a is a constant-volume process, this time with heat leaving the system and ΔS < 0. A sketch of the SV diagram for the Otto cycle follows: S  c

d 

b

a

Figure 19-14 shows a thermodynamic cycle for an ideal gas on an SP 19 •• diagram. Make a sketch of this cycle on a PV diagram. Determine the Concept Process A→B is at constant entropy; that is, it is an adiabatic process in which the pressure increases. Process B→C is one in which P is constant and S  decreases; heat is exhausted from the system and the volume decreases. Process C→D is an adiabatic compression. Process D→A returns the system to its original state at constant pressure. The cycle is shown in the adjacent PV diagram.

P

C

B

D

A

V 

One afternoon, the mother of one of your friends walks into his room 20 •• and finds a mess. She asks your friend how the room came to be in such a state, and your friend replies, ″Well, it is the natural destiny of any closed system to degenerate toward greater and greater levels of entropy. That’s all, Mom.″ Her  reply is a sharp ″ Nevertheless, you’d better clean your room!″ Your friend retorts, ″But that can’t happen. It would violate the second law of thermodynamics.″ Critique your friend’s response. Is his mother correct to ground him for not cleaning his room, or is cleaning the room really impossible? Determine the Concept The son is out of line, here, but besides that, he’s also wrong. While it is true that systems tend to degenerate to greater levels of 

1876

Chapter 19

disorder, it is not true that order cannot be brought forth from disorder. What is required is an agent doing work – for example, your friend – on the system in order to reduce the level of chaos and bring about order. His cleanup efforts will  be rewarded with an orderly system after a sufficient time for him to complete the task. It is true that order will not come about from the disordered chaos of his room – unless he applies some elbow grease.

Estimation and Approximation Estimate the change in COP of your electric food freezer when it is 21 • removed from your kitchen to its new location in your basement, which is 8°C cooler than your kitchen. Picture the Problem We can use the definition of the coefficient of performance to express the ratio of the coefficient of performance in your basement to the coefficient of performance in the kitchen. If we further assume that the freezer  operates in a Carnot cycle, then we can use the proportion Qh Qc = T h T c to

express the ratio of the coefficients of performance in terms of the temperatures in the kitchen, basement, and freezer. The ratio of the coefficients of   performance in the basement and kitchen is given by:

Qc, basement

COP basement COPkit

=

W c, basement Qc,kit W c,kit

Because W  = Qh

− Qc for a heat

engine or refrigerator:

Qc, basement

COP basement COPkit

=

Qh, basement − Qc, basement Qc,kit Qh,kit

Divide the numerators and denominators by Qc,basement and Qc,kit and simplify to obtain:

− Qc,kit

1 Qh, basement

COP basement COPkit

=

Qc, basement

1 Qh,kit Qc,kit Qh,kit

=

−1

Qc,kit

−1 −1

Qh, basement Qc, basement

−1

The Second Law of Thermodynamics If we assume that the freezer unit operates in a Carnot cycle, then Qh T h = and our expression for the Qc T c

T h,kit

COP basement COPkit

=

1877

−1

T c,kit T h, basement T c, basement

−1

ratio of the COPs becomes: Assuming that the temperature in your kitchen is 20°C and that the temperature of the interior of your  freezer is −5°C, substitute numerical values and evaluate the ratio of the coefficients of performance:

COP basement COPkit

=

293 K  −1 268 K  = 1.47 285 K  −1 268 K 

or an increase of 47% in the  performance of the freezer!

Estimate the probability that all the molecules in your bedroom are 22 •• located in the (open) closet which accounts for about 10% of the total volume of  the room. Picture the Problem The probability that all the molecules in your bedroom are  N 

⎛ V   ⎞ located in the (open) closet is given by  p = ⎜⎜ 2 ⎟⎟ where N is the number of air  ⎝ V 1  ⎠ molecules in your bedroom and V 1 and V 2 are the volumes of your bedroom and closet, respectively. We can use the ideal-gas law to find the number of molecules 3  N . We’ll assume that the volume of your room is about 50 m and that the temperature of the air is 20°C.  N 

If the original volume of the air in your bedroom is V 1, the probability  p of finding the N molecules, normally in your bedroom, confined to your closet whose volume is V 2 is given by:

or, because V 2

Use the ideal-gas law to express N :

 N  =

Substitute numerical values and evaluate N :

⎛ V   ⎞  p = ⎜⎜ 2 ⎟⎟ ⎝ V 1  ⎠

= 101 V 1 ,

 N 

⎛  1  ⎞  p = ⎜ ⎟ ⎝ 10 ⎠ PV  kT 

(101.325 kPa )(50 m3 )  N  = (1.381×10−23 J/K )(293 K ) = 1.252 ×1027 molecules

(1)

1878

Chapter 19

Substitute for  N in equation (1) and evaluate p:

1.252×1027

⎛  1  ⎞ ⎟ ⎝ 10 ⎠

 p = ⎜

≈ 10−10

=

1 101.252×10

27

= 10 −1.252×10

27

27

23 •• [SSM] Estimate the maximum efficiency of an automobile engine that has a compression ratio of 8.0:1.0. Assume the engine operates according to the Otto cycle and assume γ  = 1.4. (The Otto cycle is discussed in Section 19-1. ) Picture the Problem The maximum efficiency of an automobile engine is given  by the efficiency of a Carnot engine operating between the same two temperatures. We can use the expression for the Carnot efficiency and the equation relating V  and T  for a quasi-static adiabatic expansion to express the Carnot efficiency of the engine in terms of its compression ratio.

Express the Carnot efficiency of an engine operating between the temperatures T c and T h: Relate the temperatures T c and T h to the volumes V c and V h for a quasistatic adiabatic compression from V c to V h: Substitute for 

T c T h

to obtain:

Express the compression ratio r :

Substituting for r yields:

Substitute numerical values for r and γ  (1.4 for diatomic gases) and evaluate ε C:

ε C

=1−

−1

γ  

T cV c

T c T h

−1

= T hV h ⇒ γ  

⎛ V h  ⎞ ε C = 1 − ⎜ ⎜ V  ⎟⎟ ⎝  c  ⎠ r  =

T c T h

−1

γ  

=

V h

−1

γ  

V c

⎛ V   ⎞ = ⎜⎜ h ⎟⎟ ⎝ V c  ⎠

−1

γ  

−1

γ  

V c V h

ε C

=1−

C

= 1−

1 γ  − r  1 1

(8.0)1.4−1

≈ 56%

You are working as an appliance salesperson during the summer. One 24 •• day, your physics professor comes into your store to buy a new refrigerator. Wanting to buy the most efficient refrigerator possible, she asks yo u about the efficiencies of the available models. She decides to return the next day to buy the most efficient refrigerator. To make the sale, you need to provide her with the following estimates: (a) the highest COP possible for a household refrigerator, and (b) and the highest rate possible for the heat to be released by the refrigerator  if the refrigerator uses 600 W of electrical power.

The Second Law of Thermodynamics

1879

Picture the Problem If we assume that the temperature on the inside of the refrigerator is 0 °C (273 K) and the room temperature to be about 30°C (303 K), then the refrigerator must be able to maintain a temperature difference of about 30 K. We can use the definition of the COP of a refrigerator and the relationship  between the temperatures of the hot and cold reservoir and Qh and Qc to find an upper limit on the COP of a household refrigerator. In (b) we can solve the definition of COP for  Qc and differentiate the resulting equation with respect to time to estimate the rate at which heat is being drawn from the refrigerator  compartment.

(a) Using its definition, express the COP of a household refrigerator:

COP =

Qc

Apply conservation of energy to the refrigerator to obtain:

W  + Qc

= Qh ⇒ W  = Qh − Qc

Substitute for W and simplify to obtain:

COP =

(1)

W 

Qc Qh

− Qc

=

1 Qh Qc

Assume, for the sake of finding the upper limit on the COP, that the refrigerator is a Carnot refrigerator  and relate the temperatures of the hot and cold reservoirs to Qh and Qc: Substitute for 

Qh Qc

to obtain:

Qh

=

Qc

−1

T h T c

COPmax

=

1 T h T c

Substitute numerical values and evaluate COPmax:

COPmax

(b) Solve equation (1) for Qc:

Qc

Differentiate equation (2) with respect to time to obtain:

dQc

Substitute numerical values and dQc evaluate : dt 

dQc

=

−1

1 = 9.1 303 K  −1 273 K 

= W (COP )

dt 

dt 

= (COP )

(2)

dW  dt 

= (9.1)(600 J/s ) = 5.5 kW

1880

Chapter 19

25 •• [SSM] The average temperature of the surface of the Sun is about 5400 K, the average temperature of the surface of Earth is about 290 K. The solar  constant (the intensity of sunlight reaching Earth’s atmosphere) is about 1.37 kW/m2. (a) Estimate the total power of the sunlight hitting Earth. (b) Estimate the net rate at which Earth’s entropy is increasing due to this solar  radiation. Picture the Problem We can use the definition of intensity to find the total power  of sunlight hitting the earth and the definition of the change in entropy to find the changes in the entropy of Earth and the Sun resulting from the radiation from the Sun.

(a) Using its definition, express the intensity of the Sun’s radiation on Earth in terms of the power  P delivered to Earth and Earth’s cross sectional area A:

 I  =

Solve for P and substitute for  A to obtain:

P = IA = I π  R where R is the radius of Earth.

Substitute numerical values and evaluate P:

P = π  1.37 kW/m 2 6.37 ×10 6 m

(b) Express the rate at which Earth’s entropy S Earth changes due to the flow of solar radiation:

dS Earth

Substitute numerical values and dS Earth evaluate : dt 

dS Earth

P  A

2

(

)(

)

2

= 1.746 ×1017 W = 1.75 ×1017 W

dt 

dt 

=

=

P T Earth

1.746 ×1017 W 290 K 

= 6.02 ×1014 J/K ⋅ s

A 1.0-L box contains N molecules of an ideal gas, and the positions of  26 •• the molecules are observed 100 times per second. Calculate the average time it should take before we observe all N molecules in the left half of the box if  N is equal to (a) 10, (b) 100, (c) 1000, and (d ) 1.0 mole. (e) The best vacuums that have been created to date have pressures of about 10 –12 torr. If a vacuum chamber  has the same volume as the box, how long will a physicist have to wait before all of the gas molecules in the vacuum chamber occupy only the left half of it? Compare that to the expected lifetime of the universe, which is about 1010 years. Picture the Problem If you had one molecule in a box, it would have a 50% chance of being on one side or the other. We don’t care which side the molecules are on as long as they all are on one side, so with one molecule you have a 100% chance of it being on one side or the other. With two molecules, there are four 

The Second Law of Thermodynamics

1881

 possible combinations (both on one side, both on the other, one on one side and one on the other, and the reverse), so there is a 25% (1 in 4) chance of them both  being on a particular side, or a 50% chance of them both being on either side. Extending this logic, the probability of  N molecules all being on one side of the  box is P = 2/2 N , which means that, if the molecules shuffle 100 times a second, the time it would take them to cover all the combinations and all get on one side 2 N  or the other is t  = . In (e) we can apply the ideal gas law to find the number  2(100) of molecules in 1.0 L of air at a pressure of 10−12 torr and an assumed temperature of 300 K. (a) Evaluate t for  N = 10 molecules:

(b) Evaluate t for  N = 100 molecules:

 t

 t

=

210 = 5.12 s ≈ 5 s 2(100 s −1 )

=

2100 2(100 s −1 )

= 6.34 ×1027 s ×

1y 3.156 × 107 s

≈ 2 ×1020 y (c) Evaluate t for  N = 1000 molecules:

To evaluate 21000 let 10 x = 21000 and take the logarithm of both sides of  the equation to obtain: Substitute to obtain:

 t

=

21000 2(100 s −1 )

(1000)ln 2 =  x ln10 ⇒ x = 301

 t

=

10301 2(100 s −1 )

= 0.5 ×10 299 s ×

1y 3.156 ×107 s

≈ 2 ×10291 y (d ) Evaluate t for  23  N = 1.0 mol =6.022 ×10 molecules: 23

To evaluate 26.022×10 let 23

10 x = 26.022×10 and take the logarithm of both sides of the equation to obtain:

23

t  =

26.022×10 2(100 s −1 )

6.022 ×10 23 )ln 2 =  x ln 10 ⇒ x ≈ 1023

1882

Chapter 19

Substituting for  x yields:

t  ≈

1010

23

1y ⎛   ⎞ ⎜ ⎟ 7 2(100 s ) ⎝ 3.156 × 10 s ⎠ −1

23

≈ 1010 y (e) Solve the ideal gas law for the number of molecules N in the gas: Assuming the gas to be at room temperature (300 K), substitute numerical values and evaluate N : Evaluate t for  N = 3.22×107 molecules: 7

To evaluate 23.22×10 let

 N  =

 N  =

PV  kT 

10−12 torr )(133.32 Pa/torr )(1.0 L ) (1.381×10−23 J/K )(300 K )

= 3.22 ×107 molecules  t

=

23.22×10

7

2(100 s −1 )

3.22 × 107 )ln 2 =  x ln10 ⇒ x ≈ 107

7

10 x = 23.22×10 and take the logarithm of both sides of the equation to obtain: Substituting for  x yields:

7

 t

=

1010 1y −1 × 2(100 s ) 3.156 × 107 s 7

≈ 1010 y Express the ratio of this waiting time to the lifetime of the universe t universe:

t  t universe

7

1010 y = 10 10 y

7

≈ 1010

or  7

t  ≈ 1010 t universe

Heat Engines and Refrigerators 27 • [SSM] A heat engine with 20.0% efficiency does 0.100 kJ of work  during each cycle. (a) How much heat is absorbed from the hot reservoir during each cycle? (b) How much heat is released to the cold reservoir during each cycle? Picture the Problem (a) The efficiency of the engine is defined to be = W  Qh where W is the work done per cycle and Qh is the heat absorbed from

the hot reservoir during each cycle. (b) Because, from conservation of energy, Qh = W  + Qc , we can express the efficiency of the engine in terms of the heat Qc

The Second Law of Thermodynamics

1883

released to the cold reservoir during each cycle. (a) Qh absorbed from the hot reservoir  during each cycle is given by: (b) Use Qh

= W  + Qc to obtain:

W 

=

100 J

Qh

=

Qc

= Qh − W  = 500 J − 100 J = 400 J

ε 

0.200

= 500 J

A heat engine absorbs 0.400 kJ of heat from the hot reservoir and does 28 • 0.120 kJ of work during each cycle. (a) What is its efficiency? (b) How much heat is released to the cold reservoir during each cycle? Picture the Problem (a) The efficiency of the engine is defined to be ε  = W  Qh where W is the work done per cycle and Qh is the heat absorbed from

the hot reservoir during each cycle. (b) We can apply conservation of energy to the engine to obtain Qh = W  + Qc and solve this equation for the heat Qc released to the cold reservoir during each cycle. (a) The efficiency of the heat engine is given by:

ε 

=

W  Qh

=

120 J 400 J

= 30%

(b) Apply conservation of energy to the engine to obtain:

Qh

= W  + Qc ⇒ Qc = Qh − W 

Substitute numerical values and evaluate Qc:

Qc

= 400 J − 120 J = 280 J

A heat engine absorbs 100 J of heat from the hot reservoir and releases 29 • 60 J of heat to the cold reservoir during each cycle. (a) What is its efficiency? (b) If each cycle takes 0.50 s, find the power output of this engine. Picture the Problem We can use its definition to find the efficiency of the engine and the definition of power to find its power output.

(a) The efficiency of the heat engine is given by:

ε 

=

Substitute numerical values and evaluate ε :

ε 

= 1−

W  Qh

=

Qh

60 J 100 J

− Qc

Qh

= 1−

= 40%

Qc

Qh

1884

Chapter 19

(b) The power output P of this engine is the rate at which it does work: Substitute numerical values and evaluate P:

P=

dW  dt 

=

d 

ε  Qh

dt 

= ε 

dQh dt 

⎛  100 J  ⎞ ⎟⎟ = 80 W 0.500 s ⎝   ⎠

P = (0.40)⎜⎜

A refrigerator absorbs 5.0 kJ of heat from a cold reservoir and releases 30 • 8.0 kJ to a hot reservoir. (a) Find the coefficient of performance of the refrigerator. (b) The refrigerator is reversible. If it is run backward as a heat engine between the same two reservoirs, what is its efficiency? Picture the Problem We can apply their definitions to find the COP of the refrigerator and the efficiency of the heat engine.

(a) The COP of a refrigerator is defined to be:

COP =

Qc

Apply conservation of energy to relate the work done per cycle to Qh and Qc:

W  = Qh

− Qc

Substitute for W to obtain:

COP =

Qc

Substitute numerical values and evaluate COP:

COP =

(b) The efficiency of a heat pump is defined to be:

ε 

=

Apply conservation of energy to the heat pump to obtain:

ε 

=

Substitute numerical values and evaluate ε  :

W 

Qh

− Qc

5.0 kJ 8.0 kJ − 5.0 kJ

= 1.7

W  Qh Qh

− Qc

Qh

= 1−

= 1−

5.0 kJ 8.0 kJ

Qc Qh

= 38%

31 •• [SSM] The working substance of an engine is 1.00 mol of a monatomic ideal gas. The cycle begins at P1 = 1.00 atm and V 1 = 24.6 L. The gas is heated at constant volume to P2 = 2.00 atm. It then expands at constant pressure until its volume is 49.2 L. The gas is then cooled at constant volume until its

The Second Law of Thermodynamics

1885

 pressure is again 1.00 atm. It is then compressed at constant pressure to its original state. All the steps are quasi-static and reversible. ( a) Show this cycle on a PV diagram. For each step of the cycle, find the work done by the gas, the heat absorbed by the gas, and the change in the internal energy of the gas. (b) Find the efficiency of the cycle. Picture the Problem To find the heat added during each step we need to find the temperatures in states 1, 2, 3, and 4. We can then find the work done on the gas during each process from the area under each straight-line segment and the heat that enters the system from Q = C V ΔT and Q = C P ΔT . We can use the 1st law of 

thermodynamics to find the change in internal energy for each step of the cycle. Finally, we can find the efficiency of the cycle from the work done each cycle and the heat that enters the system each cycle. (a) The cycle is shown to the right:

Apply the ideal-gas law to state 1 to find T 1: T 1 =

P1V 1 nR

=

(1.00 atm )(24.6 L) = 300 K  ⋅ L atm ⎛   ⎞ (1.00 mol)⎜ 8.206 ×10 −2 ⎟ mol ⋅ K  ⎠ ⎝ 

The pressure doubles while the volume remains constant between states 1 and 2. Hence:

T 2

= 2T 1 = 600 K 

The volume doubles while the  pressure remains constant between states 2 and 3. Hence:

T 3

= 2T 2 = 1200 K 

The pressure is halved while the volume remains constant  between states 3 and 4. Hence:

T 4

= 12 T 3 = 600 K 

1886

Chapter 19

For path 1

2: W 12

= PΔV 12 = 0

and Q12

J  ⎞ = C V ΔT 12 = 32 RΔT 12 = 32 ⎛  ⎜ 8.314 ⎟ (600 K − 300 K ) = 3.74 kJ mol ⋅ K  ⎠ ⎝ 

The change in the internal energy of  the system as it goes from state 1 to state 2 is given by the 1st law of  thermodynamics: Because W 12

= 0:

For path 2

3:

W on

Δ E int

= Qin + W on

Δ E int,12

= Q12 = 3.74 kJ

101.325 J ⎞ = −W 23 = − PΔV 23 = −(2.00 atm )(49.2 L − 24.6 L )⎛  ⎜ ⎟ = − 4.99 kJ ⎝  L ⋅ atm  ⎠

Q23

J  ⎞ = C P ΔT 23 = 52 RΔT 23 = 52 ⎛  ⎜ 8.314 ⎟ (1200 K − 600 K ) = 12.5 kJ mol K  ⋅ ⎝   ⎠

Apply Δ Eint

= Qin + W on to obtain:

For path 3

4: W 34

Δ E int, 23

= 12.5 kJ − 4.99 kJ = 7.5 kJ

= PΔV 34 = 0

and Q34

J  ⎞ = Δ E int,34 = C V ΔT 34 = 32 RΔT 34 = 32 ⎛  ⎜ 8.314 ⎟ (600 K − 1200 K ) = − 7.48 kJ ⋅ mol K  ⎝   ⎠

Apply Δ Eint

= Qin + W on to obtain:

For path 4

1:

W on

Δ E int, 34

= −7.48 kJ + 0 = − 7.48 kJ

101.325 J ⎞ = −W 41 = − PΔV 41 = −(1.00 atm )(24.6 L − 49.2 L )⎛  ⎜ ⎟ = 2.49 kJ ⎝  L ⋅ atm  ⎠

and Q41

J  ⎞ = C P ΔT 41 = 52 RΔT 41 = 52 ⎛  ⎜ 8.314 ⎟ (300 K − 600 K ) = − 6.24 kJ mol ⋅ K  ⎠ ⎝ 

The Second Law of Thermodynamics Apply Δ E int

= Qin + W on to obtain:

Δ E int, 41

1887

= −6.24 kJ + 2.49 kJ = − 3.75 kJ

For easy reference, the results of the preceding calculations are summarized in the following table: Process W on , kJ Qin , kJ Δ Eint (= Qin + W on ) , kJ 1→2 2→3 3→4 4→1

0 −4.99 0 2.49

3.74 12.5 −7.48 −6.24

(b) The efficiency of the cycle is given by: Substitute numerical values and evaluate ε :

3.74 7.5 −7.48 −3.75

ε 

=

=

W  by Qin

− W 23 + (− W 41 ) Q12 + Q23

=

4.99 kJ − 2.49 kJ 3.74 kJ + 12.5 kJ

≈ 15%

Remarks: Note that the work done per cycle is the area bounded by the rectangular path. Note also that, as expected because the system returns to its initial state, the sum of the changes in the internal energy for the cycle is zero.

The working substance of an engine is 1.00 mol of a diatomic ideal 32 •• gas. The engine operates in a cycle consisting of three steps: (1) an adiabatic expansion from an initial volume of 10.0 L to a pressure of 1.00 atm and a volume of 20.0 L, (2) a compression at constant pressure to its original volume of  10.0 L, and (3) heating at constant volume to its original pressure. Find the efficiency of this cycle. P

Picture the Problem The three steps in the process are shown on the PV  diagram. We can find the efficiency of  the cycle by finding the work done by the gas and the heat that enters the system per cycle.

(atm) 1

2.639

2

1

2

3

0 0

The pressures and volumes at the end  points of the adiabatic expansion are related according to:

γ  

P1V 1

20.0

10.0

= P2V 2

γ  

⎛ V   ⎞ ⇒ P1 = ⎜⎜ 2 ⎟⎟ ⎝ V 1  ⎠

γ  

P2

V (L)

1888

Chapter 19

Substitute numerical values and evaluate P1: Express the efficiency of the cycle:

1.4

⎛ 20.0 L ⎞ (1.00 atm ) = 2.639 atm P1 = ⎜ ⎟ ⎝ 10.0 L ⎠ ε 

=

W 

(1)

Qh

 No heat enters or leaves the system during the adiabatic expansion:

Q12

=0

Find the heat entering or leaving the system during the isobaric compression:

Q23

= C V ΔT 23 = 72 RΔT 23 = 72 PΔV 23 = 72 (1.00 atm )(10.0 L − 20.0 L ) = −35.0 atm ⋅ L

Find the heat entering or leaving the system during the constantvolume process:

Q31

= C V ΔT 31 = 52 RΔT 31 = 52 ΔPV 31 = 52 (2.639 atm − 1.00 atm )(10.0 L ) = 41.0 atm ⋅ L

Apply the 1st law of thermodynamics to the cycle ( Δ E int, cycle = 0 ) to obtain:

W on

= Δ E int − Qin = −Qin = Q12 + Q23 + Q31 = 0 − 35.0 atm ⋅ L + 41.0 atm ⋅ L = 6.0 atm ⋅ L

Substitute numerical values in equation (1) and evaluate ε  :

ε 

=

6.0 atm ⋅ L 41atm ⋅ L

= 15%

An engine using 1.00 mol of an ideal gas initially at a volume of  33 •• 24.6 L and a temperature of 400 K performs a cycle consisting of four steps: (1) an isothermal expansion at 400 K to twice its initial volume, (2) cooling at constant volume to a temperature of 300 K (3) an isothermal compression to its original volume, and (4) heating at constant volume to its original temperature of  400 K. Assume that C v = 21.0 J/K. Sketch the cycle on a PV diagram and find its efficiency. Picture the Problem We can find the efficiency of the cycle by finding the work  done by the gas and the heat that enters the system per cycle.

The Second Law of Thermodynamics The PV diagram of the cycle is shown to the right. A, B, C, and D identify the four states of the gas and the numerals 1, 2, 3, and 4 represent the four steps through which the gas is taken.

2      ) 1.5    m     t    a      ( 1

A 4 1 D

   P

B 3

0.5 0

2 C

0

10

20

30

40

Because steps 2 and 4 are constantvolume processes, W 2 = W 4 = 0: Because the internal energy of the gas increases in step 4 while no work  is done, and because the internal energy does not change during step 1 while work is done by the gas, heat enters the system only during these  processes:

ε 

=

ε 

=

ε 

=

W  Qh W  Qh W  Qh

=

=

50

=

W 1 + W 2

+ W 3 + W 4 Qh,1 + Qh, 2 + Qh,3 + Qh, 4 W 1 + 0 + W 3 + 0 Qh,1 + Qh, 2

+ Qh,3 + Qh,4

W 1 + W 3 Qh,1 + Qh, 4

The work done during the isothermal expansion (1) is given by:

W 1

⎛ V   ⎞ = nRT ln⎜⎜ B ⎟⎟ ⎝ V A  ⎠

The work done during the isothermal compression (3) is given by:

W 3

⎛ V   ⎞ = nRT c ln⎜⎜ D ⎟⎟ ⎝ V C  ⎠

Q1

⎛ V   ⎞ = W 1 = nRT h ln⎜⎜ B ⎟⎟ ⎝ V A  ⎠

Q4

= C V ΔT  = C V (T h − T c )

Because there is no change in the internal energy of the system during step 1, the heat that enters the system during this isothermal expansion is given by: The heat that enters the system during the constant-volume step 4 is given by:

400 K 300 K

V (L)

Express the efficiency of the cycle:

1889

(1)

60

1890

Chapter 19

Substituting in equation (1) yields:

 Noting the

V B V A

= 2 and

V D V C

=

1 2

, substitute

⎛ V   ⎞ ⎛ V B  ⎞ ⎟⎟ + nRT c ln⎜⎜ D ⎟⎟ ⎝ V A  ⎠ ⎝ V C  ⎠ ε  = ⎛ V   ⎞ nRT h ln⎜⎜ B ⎟⎟ + C V (T h − T c ) ⎝ V A  ⎠ nRT h ln⎜⎜

and simplify to obtain:

⎛ 1 ⎞ ⎟ T h ln(2) − T c ln (2 ) 2 ⎠ ⎝  = = ε  = C V C V T h ln (2 ) + (T h − T c ) T h ln(2) + (T h − T c ) T h + T h ln (2) + T c ln⎜

nR

nR

T h − T c C V nR ln (2 )

(T h − T c )

Substitute numerical values and evaluate ε :

ε 

=

400 K − 300 K  = 13.1% J 21.0 K  (400 K − 300 K ) 400 K + J  ⎞ ⎛  (1.00 mol)⎜ 8.314 ⎟ ln (2) mol K  ⋅ ⎝   ⎠

Figure 19-15 shows the cycle followed by 1.00 mol of an ideal 34 •• monatomic gas initially at a volume of 25.0 L. All the processes are quasi-static. Determine (a) the temperature of each numbered state of the cycle, (b) the heat transfer for each part of the cycle, and (c) the efficiency of the cycle. Picture the Problem We can use the ideal-gas law to find the temperatures of  each state of the gas and the heat capacities at constant volume and constant  pressure to find the heat flow for the constant-volume and isobaric processes. Because the change in internal energy is zero for the isothermal process, we can use the expression for the work done on or by a gas during an isothermal process to find the heat flow during such a process. Finally, we can find the efficiency of  the cycle from its definition.

(a) Use the ideal-gas law to find the temperature at point 1:

T 1 =

P1V 1 nR

=

(100 kPa )(25.0 L ) (1.00 mol)⎛  ⎜ 8.314 ⎝ 

= 301K 

J  ⎞ ⎟ mol ⋅ K  ⎠

The Second Law of Thermodynamics Use the ideal-gas law to find the temperatures at points 2 and 3:

T 2

= T 3 = =

1891

P2V 2 nR

(200 kPa )(25.0 L ) J  ⎞ (1.00 mol)⎛  ⎜ 8.314 ⎟ mol ⋅ K  ⎠ ⎝ 

= 601 K  (b) Find the heat entering the system for the constant-volume process from 1 Q12

→ 2:

J  ⎞ = C V ΔT 12 = 32 RΔT 12 = 32 ⎛  ⎜ 8.314 ⎟ (601K − 301K ) = 3.74 kJ ⋅ mol K  ⎝   ⎠

Find the heat entering or leaving the system for the isothermal process from 2 → 3:

Q23

⎛ V   ⎞ ⎛ 50.0 L ⎞ J  ⎞ ⎟⎟ = 3.46 kJ = nRT 2 ln⎜⎜ 3 ⎟⎟ = (1.00 mol)⎛  ⎜ 8.314 ⎟ (601 K )ln⎜⎜ mol K  2 5 . 0 L V  ⋅ ⎝   ⎠ ⎝   ⎠ ⎝  2  ⎠

Find the heat leaving the system during the isobaric compression from 3 → 1: Q31

J  ⎞ = C P ΔT 31 = 52 RΔT 31 = 52 ⎛  ⎜ 8.314 ⎟ (301 K − 601K ) = − 6.24 kJ mol ⋅ K  ⎠ ⎝ 

(c) Express the efficiency of the cycle: Apply the 1st law of thermodynamics to the cycle:

Substitute numerical values in equation (1) and evaluate ε  :

ε 

=

W  Qin

W  =

=

W 

+ Q23

Q12

(1)

∑Q = Q

+ Q23 + Q31 = 3.74 kJ + 3.46 kJ − 6.24 kJ = 0.96 kJ  because, for the cycle, Δ Eint = 0 . ε 

=

12

0.96 kJ 3.74 kJ + 3.46 kJ

= 13%

An ideal diatomic gas follows the cycle shown in Figure 19-16. The 35 •• temperature of state 1 is 200 K. Determine (a) the temperatures of the other three numbered states of the cycle and (b) the efficiency of the cycle. Picture the Problem We can use the ideal-gas law to find the temperatures of  each state of the gas. We can find the efficiency of the cycle from its definition;

1892

Chapter 19

using the area enclosed by the cycle to find the work done per cycle and the heat entering the system between states 1 and 2 and 2 and 3 to determine Qin. (a) Use the ideal-gas law for a fixed amount of gas to find the temperature in state 2 to the temperature in state 1: Substitute numerical values and evaluate T 2: Apply the ideal-gas law for a fixed amount of gas to states 2 and 3 to obtain: Substitute numerical values and evaluate T 3: Apply the ideal-gas law for a fixed amount of gas to states 3 and 4 to obtain: Substitute numerical values and evaluate T 4: (b) The efficiency of the cycle is:

P1V 1 T 1

P2V 2

=

T 2

⇒ T 2 = T 1

= (200 K )

T 3

= T 2

T 3

= (600 K )

T 4

= T 3

T 4

= (1800 K )

ε 

=

P2V 2

P4V 4 P3V 3

P1V 1

(3.0 atm ) = (1.0 atm )

T 2

P3V 3

P2V 2

= T 2

P2 P1

600 K 

V 3 V 2

(300 L ) = (100 L ) = T 3

= T 1

1800 K 

P4 P3

(1.0 atm ) = (3.0 atm )

600 K 

W 

(1)

Qin

Use the area of the rectangle to find the work done each cycle:

W  = ΔPΔV 

Apply the ideal-gas law to state 1 to find the product of n and R:

nR =

= (300 L − 100 L )(3.0 atm − 1.0 atm ) = 400 atm ⋅ L P1V 1 T 1

=

(1.0 atm )(100 L) 200 K 

= 0.50 L ⋅ atm/K   Noting that heat enters the system  between states 1 and 2 and states 2 and 3, express Qin:

Qin

= Q12 + Q23 = C V ΔT 12 + C P ΔT 23 = 52 nRΔT 12 + 72 nRΔT 23 = (52 ΔT 12 + 72 ΔT 23 )nR

The Second Law of Thermodynamics

1893

Substitute numerical values and evaluate Qin: Qin

L ⋅ atm ⎞ = [ 52 (600 K − 200 K ) + 72 (1800 K − 600 K )]⎛  ⎜ 0.50 ⎟ = 2600 atm ⋅ L K   ⎠ ⎝ 

Substitute numerical values in equation (1) and evaluate ε  :

ε 

=

400 atm ⋅ L 2600 atm ⋅ L

= 15%

••• Recently, an old design for a heat engine, known as the Stirling engine 36 has been promoted as a means of producing power from solar energy. The cycle of a Stirling engine is as follows: (1) isothermal compression of the working gas (2) heating of the gas at constant volume, (3) an isothermal expansion of the gas, and (4) cooling of the gas at constant volume. (a) Sketch PV and ST diagrams for  the Stirling cycle. (b) Find the entropy change of the gas for each step of the cycle and show that the sum of these entropy changes is equal to zero. Picture the Problem (a) The PV and ST cycles are shown below. (b) We can show that the entropy change during one Stirling cycle is zero by adding up the entropy changes for the four processes. P

S 

2 Δ V = 0

(2)

3 T h

(3) T h

T c

1

(4)

(1)

4

T c

ΔV = 0

T 

(b) The change in entropy for one Stirling cycle is the sum of the entropy changes during the cycle: Express the entropy change for the isothermal process from state 1 to state 2:

ΔS cycle

ΔS 12

= ΔS 12 + ΔS 23 + ΔS 34 + ΔS 41 (1)

⎛ V   ⎞ = nR ln⎜⎜ 2 ⎟⎟ ⎝ V 1  ⎠

1894

Chapter 19

Similarly, the entropy change for the isothermal process from state 3 to state 4 is:

ΔS 34

or, because V 2 = V 3 and V 1 = V 4, ΔS 34

The change in entropy for a constantvolume process is given by:

⎛ V   ⎞ = nR ln⎜⎜ 4 ⎟⎟ ⎝ V 3  ⎠ ⎛ V   ⎞ ⎛ V   ⎞ = nR ln⎜⎜ 1 ⎟⎟ = −nR ln⎜⎜ 2 ⎟⎟ ⎝ V 2  ⎠ ⎝ V 1  ⎠

ΔS isochoric

=∫

dQ T 

T f 

=∫

nC V dT 

T i

T 

⎛ T   ⎞ = nC V ln⎜⎜ f  ⎟⎟ ⎝ T i  ⎠ For the constant-volume process from state 2 to state 3:

ΔS 23

⎛ T   ⎞ = C V ln⎜⎜ c ⎟⎟ ⎝ T h  ⎠

For the constant-volume process from state 4 to state 1:

ΔS 41

⎛ T   ⎞ ⎛ T   ⎞ = C V ln⎜⎜ h ⎟⎟ = −C V ln⎜⎜ c ⎟⎟ ⎝ T c  ⎠ ⎝ T h  ⎠

Substituting in equation (1) yields:

ΔS cycle

⎛ V   ⎞ ⎛ T   ⎞ ⎛ V   ⎞ ⎛ T   ⎞ = −nR ln⎜⎜ 2 ⎟⎟ + C V ln⎜⎜ c ⎟⎟ − nR ln⎜⎜ 2 ⎟⎟ − C V ln⎜⎜ c ⎟⎟ = 0 ⎝ V 1  ⎠ ⎝ T h  ⎠ ⎝ V 1  ⎠ ⎝ T h  ⎠

37 •• ″As far as we know, Nature has never evolved a heat engine″ —Steven Vogel, Life’s Devices, Princeton University Press (1988). (a) Calculate the efficiency of a heat engine operating between body temperature (98.6ºF) and a typical outdoor temperature (70ºF), and compare this to the human body’s efficiency for converting chemical energy into work (approximately 20%). Does this efficiency comparison contradict the second law of thermodynamics? (b) From the result of Part ( a), and a general knowledge of the conditions under  which most warm-blooded organisms exist, give a reason why no warm-blooded organisms have evolved heat engines to increase their internal energies. Picture the Problem We can use the efficiency of a Carnot engine operating  between reservoirs at body temperature and typical outdoor temperatures to find an upper limit on the efficiency of an engine operating between these temperatures.

(a) Express the maximum efficiency of an engine operating between body temperature and 70°F:

ε C

=1−

T c T h

The Second Law of Thermodynamics Use T  =

5 9

(t F − 32) + 273 to obtain:

Substitute numerical values and evaluate ε C :

T  body

C

1895

= 310 K and T room = 294 K 

= 1−

294 K  = 5.16% 310 K 

The fact that this efficiency is considerably less than the actual efficiency of a human body does not contradict the second law of thermodynamics. The application of the second law to chemical reactions such as the ones that supply the body with energy have not been discussed in the text but we can note that we don’t get our energy from heat swapping between our body and the environment. Rather, we eat food to get the energy that we need. (b) Most warm-blooded animals survive under roughly the same conditions as humans. To make a heat engine work with appreciable efficiency, internal body temperatures would have to be maintained at an unreasonably high level. 38 ••• The diesel cycle shown in Figure 19-17 approximates the behavior of a diesel engine. Process ab is an adiabatic compression, process bc is an expansion at constant pressure, process cd is an adiabatic expansion, and process da is cooling at constant volume. Find the efficiency of this cycle in terms of the volumes V a, V  b and V c. Picture the Problem The working fluid will be modeled as an ideal gas and the  process will be modeled as quasistatic. To find the efficiency of the diesel cycle we can find the heat that enters the system and the heat that leaves the system and use the expression that gives the efficiency in terms of these quantities. Note that no heat enters or leaves the system during the adiabatic processes ab and cd. Heat enters the system during the isobaric process bc and leaves the system during the isovolumetric process da.

Express the efficiency of the cycle in terms of Qc and Qh:

ε 

=

W  Qh

=

Qh

− Qc

Qh

= 1−

Express Q for the isobaric warming  process bc:

Qbc

= Qh = C P (T c − T b )

Because C V is independent of T, Qda (the constant-volume cooling  process) is given by:

Qda

= Qc = C V (T d  − T a )

Substitute for Qh and Qc and simplify using γ   = C P C V to obtain:

ε 

= 1−

C V (T d  − T a ) C P (T c − T b )

Qc Qh

= 1−

(T d  − T a ) γ  (T c − T b )

1896

Chapter 19

Using an equation for a quasistatic adiabatic process, relate the temperatures T a and T b to the volumes V a and V b:

−1

γ  

T aV a

Proceeding similarly, relate the temperatures T c and T d  to the volumes V c and V d: 

−1

γ  

T cV c

V bγ  −1

−1

= T bV b ⇒ T a = T b γ  

= T d V d  ⇒ T d  = T c

V c

−1

V d  − γ  

−1 ⎛  V c −1  ⎞ ⎜⎜ T c −1 − T b V b −1 ⎟⎟ V a  ⎠ ⎝  V d  ε  = 1 − γ  (T c − T b )

Because V a = V d :

⎛ ⎛ V   ⎞ ⎜⎜ c ⎟ ⎜ ⎜⎝ V a  ⎠⎟ ⎝  ε  = 1 −

 Noting that Pb = Pc , apply the idealgas law to relate T b and T c:

T b T c

(1)

γ  

−1

γ  

Use equations (1) and (2) to eliminate T a and T d: 

=

−1

γ  

V a

γ  

γ  

γ  

γ  

1

(2)

 ⎞ ⎜⎜ ⎟⎟ ⎟ T c ⎝ V a  ⎠ ⎟  ⎠ ⎛  T b  ⎞ γ  ⎜1 − ⎜ T  ⎟⎟ ⎝  c  ⎠ −1

γ  

−

T b ⎛ V b  ⎞

−1

γ  

V b V c

Substitute for the ratio of T b to T c and simplify to obtain:

⎛ V c  ⎞ ⎜⎜ ⎟⎟ ⎝ V a  ⎠ ε  = 1 −

−1

γ  

−

V b ⎛ V b  ⎞

⎜

⎟

V c ⎜⎝ V a  ⎠⎟

−1

γ  

⎛ V c  ⎞ ⎜⎜ ⎟⎟ V  = 1 − ⎝  a  ⎠

V c V  ⋅ a V c

−1

γ  

−

V b ⎛ V b  ⎞

⎜

−1

γ  

⎟

V c ⎜⎝ V a  ⎠⎟

⎛  V b  ⎞ ⎛ V c V b  ⎞ ⎜⎜1 − ⎟⎟ γ  ⎜ ⎜ V  − V  ⎟⎟ V  V  a ⎝  c  ⎠ ⎝  a a  ⎠ ⎛ V c  ⎞ ⎛ V b  ⎞ ⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ V  V  V c − V b = 1 − ⎝  a  ⎠ ⎝  a  ⎠ = 1 − −1 γ  V a (V c − V b ) ⎛ V c V b  ⎞ γ  ⎜ ⎜ V  − V  ⎟⎟ ⎝  a a  ⎠ γ  

γ  

γ  

γ  

γ  

γ  

Second Law of Thermodynamics 39 •• [SSM] A refrigerator absorbs 500 J of heat from a cold reservoir and releases 800 J to a hot reservoir. Assume that the heat-engine statement of the second law of thermodynamics is false, and show how a perfect engine working

The Second Law of Thermodynamics

1897

with this refrigerator can violate the refrigerator statement of the second law o f  thermodynamics. Determine the Concept The following diagram shows an ordinary refrigerator  that uses 300 J of work to remove 500 J of heat from a cold reservoir and releases 800 J of heat to a hot reservoir (see ( a) in the diagram). Suppose the heat-engine statement of the second law is false. Then a ″ perfect″ heat engine could remove energy from the hot reservoir and convert it completely into work with 100  percent efficiency. We could use this perfect heat engine to remove 300 J of  energy from the hot reservoir and do 300 J of work on the ordinary refrigerator  (see (b) in the diagram). Then, the combination of the perfect heat engine and the ordinary refrigerator would be a perfect refrigerator; transferring 500 J of heat from the cold reservoir to the hot reservoir without requiring any work (see ( c) in the diagram).This violates the refrigerator statement of the second law. Hot reservoir at temperature T h 300 J

800 J

    ⇓

 ⇓     ⇓   

Ordinary refrigerator 

300 J

300 J

500 J

⇓

    ⇓

Perfect heat engine

Perfect refrigerator 

    ⇓

    ⇓

500 J

500 J Cold reservoir at temperature T c

(a)

(b )

(c )

If two curves that represent quasi-static adiabatic processes could 40 •• intersect on a PV diagram, a cycle could be completed by an isothermal path  between the two adiabatic curves shown in Figure 19-18. Show that such a cycle violates the second law of thermodynamics. Determine the Concept The work done by the system is the area enclosed by the cycle, where we assume that we start with the isothermal expansion. It is only in this expansion that heat is extracted from a reservoir. There is no heat transfer in the adiabatic expansion or compression. Thus, we would completely convert heat to mechanical energy, without exhausting any heat to a cold reservoir, in violation of the second law of thermodynamics.

Carnot Cycles 41 • [SSM] A Carnot engine works between two heat reservoirs at temperatures T h = 300 K and T c = 200 K. (a) What is its efficiency? ( b) If it

1898

Chapter 19

absorbs 100 J of heat from the hot reservoir during each cycle, how much work  does it do each cycle? (c) How much heat does it release during each cycle? (d ) What is the COP of this engine when it works as a refrigerator between the same two reservoirs? Picture the Problem We can find the efficiency of the Carnot engine using ε  = 1 − T c / T h and the work done per cycle from ε  = W  / Qh . We can apply

conservation of energy to find the heat rejected each cycle from the heat absorbed and the work done each cycle. We can find the COP of the engine working as a refrigerator from its definition. (a) The efficiency of the Carnot engine depends on the temperatures of the hot and cold reservoirs:

ε C

= 1−

T c T h

(b) Using the definition of efficiency, relate the work done each cycle to the heat absorbed from the hot reservoir:

W  = ε C Qh

(c) Apply conservation of energy to relate the heat given off each cycle to the heat absorbed and the work  done:

Qc

(d ) Using its definition, express and evaluate the refrigerator’s coefficient of performance:

COP =

= 1−

200 K  300 K 

= 33.3%

= (0.333)(100 J ) = 33.3 J

= Qh − W  = 100 J − 33.3 J = 66.7 J = 67 J

Qc W 

=

66.7 J 33.3 J

= 2.0

An engine absorbs 250 J of heat per cycle from a reservoir at 300 K  42 • and releases 200 J of heat per cycle to a reservoir at 200 K. (a) What is its efficiency? (b) How much additional work per cycle could be done if the engine were reversible? Picture the Problem We can find the efficiency of the engine from its definition and the additional work done if the engine were reversible from W  = ε CQh , where ε C

is the Carnot efficiency.

(a) Express the efficiency of the engine in terms of the heat absorbed from the high-temperature reservoir  and the heat exhausted to the lowtemperature reservoir:

ε 

=

W  Qh

= 1−

=

Qh

200 J 250 J

− Qc

Qh

= 1−

= 20.0%

Qc Qh

The Second Law of Thermodynamics (b) Express the additional work done if the engine is reversible:

ΔW  = W Carnot − W Part (a )

Relate the work done by a reversible engine to its Carnot efficiency:

W  = ε CQh

Substitute numerical values and evaluate W :

W  = ⎜⎜1 −

Substitute numerical values in equation (1) and evaluate ΔW :

⎛  ⎝ 

ΔW 

1899

(1)

⎛  T   ⎞ = ⎜⎜1 − c ⎟⎟Qh ⎝  T h  ⎠

200 K  ⎞

⎟ (250 J ) = 83.3 J

300 K  ⎠⎟

= 83.3 J − 50 J = 33 J

A reversible engine working between two reservoirs at temperatures T h 43 •• and T c has an efficiency of 30%. Working as a heat engine, it releases 140 J per  cycle of heat to the cold reservoir. A second engine working between the same two reservoirs also releases 140 J per cycle to the cold reservoir. Show that if the second engine has an efficiency greater than 30%, the two engines working together would violate the heat-engine statement of the second law. Determine the Concept Let the first engine be run as a refrigerator. Then it will remove 140 J from the cold reservoir, deliver 200 J to the hot reservoir, and require 60 J of energy to operate. Now take the second engine and run it between the same reservoirs, and let it eject 140 J into the cold reservoir, thus replacing the heat removed by the refrigerator. If ε 2, the efficiency of this engine, is greater than 30%, then Qh2, the heat removed from the hot reservoir by this engine, is 140 J/(1 − ε 2) > 200 J, and the work done by this engine is W  = ε 2Qh2 > 60 J. The end result of all this is that the second engine can run the refrigerator, replacing the heat taken from the cold reservoir, and do additional mechanical work. The two systems working together then convert heat into mechanical energy without rejecting any heat to a cold reservoir, in violation of the second law.

A reversible engine working between two reservoirs at temperatures T h 44 •• and T c has an efficiency of 20%. Working as a heat engine, it does 100 J of work   per cycle. A second engine working between the same two reservoirs also does 100 J of work per cycle. Show that if the efficiency of the second engine is greater  than 20%, the two engines working together would violate the refrigerator  statement of the second law. Determine the Concept If the reversible engine is run as a refrigerator, it will require 100 J of mechanical energy to take 400 J of heat from the cold reservoir  and deliver 500 J to the hot reservoir. Now let the second engine, with ε 2 > 0.2, operate between the same two heat reservoirs and use it to drive the refrigerator.

1900

Chapter 19

Because ε 2 > 0.2, this engine will remove less than 500 J from the hot reservoir in the process of doing 100 J of work. The net result is then that no net work is done  by the two systems working together, but a finite amount of heat is transferred from the cold reservoir to the hot reservoir, in violation of the refrigerator  statement of the second law. A Carnot engine works between two heat reservoirs as a refrigerator. 45 •• During each cycle, 100 J of heat are absorbed and 150 J are released to the hot reservoir. (a) What is the efficiency of the Carnot engine when it works as a heat engine between the same two reservoirs? (b) Show that no other engine working as a refrigerator between the same two reservoirs can have a COP greater than 2.00. Picture the Problem We can use the definition of efficiency to find the efficiency of the Carnot engine operating between the two reservoirs.

(a) The efficiency of the Carnot engine is given by:

ε C

=

W  Qh

=

50 J 150 J

= 33%

(b) If the COP > 2, then 50 J of work will remove more than 100 J of heat from the cold reservoir and put more than 150 J of heat into the hot reservoir. So running engine described in Part (a) to operate the refrigerator with a COP > 2 will result in the transfer of heat from the cold to the hot reservoir without doing any net mechanical work in violation of the second law. A Carnot engine works between two heat reservoirs at temperatures 46 •• T h = 300 K and T c = 77.0 K. (a) What is its efficiency? (b) If it absorbs 100 J of  heat from the hot reservoir during each cycle, how much work does it do? (c) How much heat does it release to the low-temperature reservoir during each cycle? (d ) What is the coefficient of performance of this engine when it works as a refrigerator between these two reservoirs? Picture the Problem We can use the definitions of the efficiency of a Carnot engine and the coefficient of performance of a refrigerator to find these quantities. The work done each cycle by the Carnot engine is given by W  = ε CQh and we can

use the conservation of energy to find the heat rejected to the low-temperature reservoir. (a) The efficiency of a Carnot engine depends on the temperatures of the hot and cold reservoirs:

C

= 1−

T c T h

= 1−

77.0 K  = 74.3% 300 K 

The Second Law of Thermodynamics

(b) Express the work done each cycle in terms of the efficiency of  the engine and the heat absorbed from the high-temperature reservoir:

W  = ε C Qh

(c) Apply conservation of energy to obtain:

Qc

(d ) Using its definition, express and evaluate the refrigerator’s coefficient of performance:

COP =

1901

= (0.743)(100 J ) = 74.3 J

= Qh − W  = 100 J − 74.3 J = 26 J Qc W 

=

26 J 74.3 J

= 0.35

47 •• [SSM] In the cycle shown 1 in Figure 19-19, 1.00 mol of an ideal diatomic gas is initially at a pressure of 1.00 atm and a temperature of 0.0ºC. The gas is heated at constant volume to T 2 = 150ºC and is then expanded adiabatically until its pressure is again 1.00 atm. It is then compressed at constant pressure back  to its original state. Find ( a) the temperature after the adiabatic expansion, (b) the heat absorbed or released by the system during each step, (c) the efficiency of this cycle, and (d ) the efficiency of a Carnot cycle operating between the temperature extremes of this cycle. Picture the Problem We can use the ideal-gas law for a fixed amount of gas and the equations of state for an adiabatic process to find the temperatures, volumes, and pressures at the end points of each process in the given cycle. We can use Q = C V ΔT  and Q = C P ΔT  to find the heat entering and leaving during the

constant-volume and isobaric processes and the first law of thermodynamics to find the work done each cycle. Once we’ve calculated these quantities, we can use its definition to find the efficiency of the cycle and the definition of the Carnot efficiency to find the efficiency of a Carnot engine operating between the extreme temperatures. (a) Apply the ideal-gas law for a fixed amount of gas to relate the temperature at point 3 to the temperature at point 1:

Apply the ideal-gas law for a fixed amount of gas to relate the pressure at point 2 to the temperatures at  points 1 and 2 and the pressure at 1:

P1V 1 T 1

=

P3V 3 T 3

or, because P1 = P3, V  T 3 = T 1 3 V 1 P1V 1 T 1

=

P2V 2 T 2

⇒ P2 =

(1)

P1V 1T 2 V 2T 1

1902

Chapter 19

Because V 1 = V 2:

Apply an equation for an adiabatic  process to relate the pressures and volumes at points 2 and 3:  Noting that V 1 = 22.4 L, evaluate V 3:

Substitute numerical values in equation (1) and evaluate T 3 and t 3:

= P1

P2

γ  

P1V 1

T 2 T 1

= (1.00 atm )

= P3V 3

γ  

423 K  273 K 

⎛  P  ⎞ ⇒ V 3 = V 1 ⎜⎜ 1 ⎟⎟ ⎝ P3 ⎠

= 1.55 atm

1 γ  

1 1.4

⎛ 1.55 atm ⎞ ⎟⎟ = 30.6 L V 3 = (22.4 L )⎜⎜ 1 atm ⎝   ⎠

= (273 K )

T 3

30.6 L 22.4 L

= 373 K 

and

(b) Process 1→2 takes place at constant volume (note that γ  = 1.4 corresponds to a diatomic gas and that C P – C V = R):

t 3

= T 3 − 273 = 100°C

Q12

= CVΔT 12 = 52 RΔT 12 J  ⎞ = 52 ⎛  ⎜ 8.314 ⎟ (423 K − 273 K ) ⋅ mol K  ⎝   ⎠ = 3.12 kJ

Process 2→3 takes place adiabatically:

Q23

= 0

Process 3→1 is isobaric (note that C P = C V + R):

Q31

= C P ΔT 31 = 72 RΔT 12 J  ⎞ = 72 ⎛  ⎜ 8.314 ⎟ (273 K − 373 K ) mol ⋅ K  ⎠ ⎝  = − 2.91 kJ W 

(c) The efficiency of the cycle is given by:

ε 

Apply the first law of thermodynamics to the cycle:

= Qin + W on or, because Δ E int, cycle = 0 (the system

=

(2)

Qin

Δ E int

 begins and ends in the same state) and W on = −W  by the gas = Qin . Evaluating W yields:

W  =

∑Q = Q

+ Q23 + Q31 = 3.12 kJ + 0 − 2.91 kJ = 0.21 kJ 12

The Second Law of Thermodynamics Substitute numerical values in equation (2) and evaluate ε  : (d ) Express and evaluate the efficiency of a Carnot cycle operating between 423 K and 273 K:

=

C

0.21 kJ 3.12 kJ

= 1−

T c T h

1903

= 6.7%

= 1−

273 K  = 35.5% 423 K 

You are part of a team that is completing a mechanical-engineering 48 ••  project. Your team built a steam engine that takes in superheated steam at 270ºC and discharges condensed steam from its cylinder at 50.0ºC. Your team has measured its efficiency to be 30.0%. (a) How does this efficiency compare with the maximum possible efficiency for your engine? (b) If the useful power output of the engine is known to be 200 kW, how much heat does the engine release to its surroundings in 1.00 h? Picture the Problem We can find the maximum efficiency of the steam engine  by calculating the Carnot efficiency of an engine operating between the given temperatures. We can apply the definition of efficiency to find the heat discharged to the engine’s surroundings in 1.00 h.

(a) The efficiency of the steam engine as a percentage of the maximum possible efficiency is given by: The efficiency of a Carnot engine operating between temperatures F c and T h is: Substituting for ε max yields:

ε steam engine

=

0.300

ε max

ε max

ε max

= 1−

steam engine max

T c

= 1−

T h

=

323 K  543 K 

0.300 0.4052

= 40.52%

= 74.05%

or  steam engine

(b) Relate the heat Qc discharged to the engine’s surroundings to Qh and the efficiency of the engine:

=

W  Qh

=

= 0.740 Qh

− Qc

Qh

max

⇒ Qc = (1 − )Qh

1904

Chapter 19

Using its definition, relate the efficiency of the engine to the heat intake of the engine and the work it does each cycle: Substitute for Qh in the expression for  Qc and simplify to obtain:

W 

PΔt 

Qh

=

=

Qc

= (1 − ε )

ε 

ε 

PΔt  ε 

1  ⎞ = ⎛  ⎜ − 1⎟ PΔt  ⎝ ε   ⎠

Substitute numerical values and evaluate Qc (1.00 h ) :

⎛  1 − 1 ⎞ ⎛ 200 kJ ⎞ (3600 s ) = 1.68 GJ ⎟⎜ ⎟ s  ⎠ ⎝ 0.300  ⎠ ⎝ 

Qc (1.00 h ) = ⎜

*Heat Pumps 49 • [SSM] As an engineer, you are designing a heat pump that is capable of delivering heat at the rate of 20 kW to a house. The house is located where, in January, the average outside temperature is –10ºC. The temperature of the air in the air handler inside the house is to be 40ºC. (a) What is maximum possible COP for a heat pump operating between these temperatures? (b) What must be the minimum power of the electric motor driving the heat pump? (c) In reality, the COP of the heat pump will be only 60 percent of the ideal value. What is the minimum power of the engine when the COP is 60 percent of the ideal value? Picture the Problem We can use the definition of the COPHP and the Carnot efficiency of an engine to express the maximum efficiency of the refrigerator in terms of the reservoir temperatures. We can apply the definition of power to find the minimum power needed to run the heat pump.

(a) Express the COPHP in terms of T h and T c:

Substitute numerical values and evaluate COPHP:

COPHP

COPHP

=

Qh

=

1 Q 1− c Qh

=

313 K  = 6.26 313 K − 263 K 

W 

=

= 6.3

Qh Qh

=

− Qc 1 T  1− c T h

=

T h T h

− T c

The Second Law of Thermodynamics (b) The COPHP is also given by:

Substitute numerical values and evaluate Pmotor : (c) The minimum power of the engine is given by:

Pout

COPHP

=

 Pmotor  =

20 kW 6.26

⇒

Pmotor 

=

Pout

COPHP

= 3.2 kW

dQc Pmin

Pmotor  =

1905

dQc

dt 

dt 

=

ε HP

ε 

(COP

HP, max

)

where ε HP is the efficiency of the heat  pump. Substitute numerical values and evaluate Pmin:

Pmin

=

20 kW

(0.60)(6.26 )

= 5.3 kW

A refrigerator is rated at 370 W. ( a) What is the maximum amount of  50 • heat it can absorb from the food compartment in 1.00 min if the foodcompartment temperature of the refrigerator is 0.0ºC and it releases heat into a room at 20.0ºC? (b) If the COP of the refrigerator is 70% of that of a reversible refrigerator, how much heat can it absorb from the food compartment in 1.00 min under these conditions? Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. By expressing the COP in terms of  T c and T h we can write the amount of heat removed from the refrigerator as a function of T c, T h, P, and Δt .

= (COP )W  = (COP )PΔt 

(a) Express the amount of heat the refrigerator can remove in a given  period of time as a function of its COP:

Qc

Express the COP in terms of T h and T c and simplify to obtain:

COP =

=

Qc W 

=

1 − ε 

Qc ε  Qh

T c T h

Qh

= −1 =

ε 

=

1

=

− T c

ε 

− W 

ε  Qh

1 T  1− c T h

−1

1906

Chapter 19

Substituting for COP yields:

Qc

⎛  T   ⎞ = ⎜⎜ c ⎟⎟ PΔt  ⎝ T h − T c  ⎠

Substitute numerical values and evaluate Qc: Qc

273 K   ⎞ 60 s ⎞ ⎛  = ⎛  ⎜ ⎟ (370 W )⎜1.00 min × ⎟ = 303 kJ = 0.30 MJ min ⎠ ⎝ 293 K − 273 K  ⎠ ⎝ 

(b) If the COP is 70% of the efficiency of an ideal pump:

' 

Qc

= (0.70)(303 kJ ) = 0.21 MJ

A refrigerator is rated at 370 W. ( a) What is the maximum amount of  51 • heat it can absorb for the food compartment in 1.00 min if the temperature in the compartment is 0.0ºC and it releases heat into a room at 35ºC? ( b) If the COP of  the refrigerator is 70% of that of a reversible pump, how much heat can it absorb from the food compartment in 1.00 min? Is the COP for the refrigerator greater  when the temperature of the room is 35ºC or 20ºC? Explain. Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. By expressing the COP in terms of  T c and T h we can write the amount of heat removed from the refrigerator as a function of T c, T h, P, and Δt .

= (COP )W  = (COP )PΔt 

(a) Express the amount of heat the refrigerator can remove in a given  period of time as a function of its COP:

Qc

Express the COP in terms of T h and T c and simplify to obtain:

COP =

=

Qc W 

Qc

=

1 − ε 

ε Qh

Substituting for COP yields:

Qc

Qh

− W 

ε Qh

1

= −1

ε 

=

=

1 T  1− c T h

ε 

−1 =

T c T h

⎛  T   ⎞ = ⎜⎜ c ⎟⎟ PΔt  ⎝ T h − T c  ⎠

− T c

The Second Law of Thermodynamics

1907

Substitute numerical values and evaluate Qc: Qc

273 K   ⎞ 60 s ⎞ ⎛  = ⎛  ⎜ ⎟ (370 W )⎜1.00 min × ⎟ = 173 kJ = 0.17 MJ min ⎠ ⎝ 308 K − 273 K  ⎠ ⎝ 

(b) If the COP is 70% of the efficiency of an ideal pump:

Q'c

= (0.70 )(173 kJ ) = 0.12 MJ

Because the temperature difference increases when the room is warmer, the COP decreases. 52 ••• You are installing a heat pump, whose COP is half the COP of a reversible heat pump. You will use the pump on chilly winter nights to increase the air temperature in your bedroom. Your bedroom’s dimensions are 5.00 m × 3.50 m × 2.50 m. The air temperature should increase from 63°F to 68°F. The outside temperature is 35°F , and the temperature at the air handler in the room is 112°F. If the pump’s electric power consumption is 750 W, how long will you have to wait in order for the room’s air to warm (take the specific heat of air to be 1.005 kJ/(kg·°C)? Assume you have good window draperies and good wall insulation so that you can neglect the release of heat through windows, walls, ceilings and floors. Also assume that the heat capacity of the floor, ceiling, walls and furniture are negligible. Picture the Problem We can use the definition of the coefficient of performance of a heat pump and the relationship between the work done per cycle and the  pump’s power consumption to find your waiting time.

The coefficient of performance of the heat pump is defined as:

COPHP

=

Qh W 

=

Qh PΔt 

⇒ Δ t =

Qh

(COPHP ) P

where Qh is the heat required to raise the temperature of your bedroom, P is the power consumption of the heat  pump, and Δt is the time required to warm the bedroom. We’re given that the coefficient of   performance of the heat pump is half  the coefficient of performance of an ideal heat pump:

Substituting for  COPHP yields:

COPHP

=

Qh W 

= 12 COPmax

⎛  T   ⎞ = 12 ⎜⎜ h ⎟⎟ ⎝ T h − T c  ⎠ Δt 

=

2Qh

⎛  T h  ⎞ ⎜⎜ ⎟⎟ P T  − T  ⎝  h c  ⎠

1908

Chapter 19

The heat required to warm the room is related to the volume of the room, the density of air, and the desired increase in temperature: Substitute for Qh to obtain:

Qh

= mcΔT  = ρ VcΔT 

where  ρ  is the density of air and c is its specific heat capacity.

Δt 

=

2 VcΔT 

⎛  T h  ⎞ ⎜⎜ ⎟⎟ P T  − T  ⎝  h c  ⎠

Substitute numerical values and evaluate Δt :

⎛  ⎝  Δ t =

2⎜1.293

⎛  kg  ⎞ J  ⎞ ⎛  5 C° ⎞ ( ) ⎜ ⎟ 5 . 00 m 3.50 m 2.50 m 1005 5 F × × ° × ⎟ ⎜ ⎟ ⎜ m 3 ⎠ kg ⋅ C° ⎠⎟ ⎝  9 F° ⎠ ⎝  ⎛  317 K   ⎞ (750 W ) ⎜ ⎟ ⎝ 317 K − 275 K  ⎠

= 56 s

Entropy Changes 53 • [SSM] You inadvertently leave a pan of water boiling away on the hot stove. You return just in time to see the last drop converted into steam. The  pan originally held 1.00 L of boiling water. What is the change in entropy of the water associated with its change of state from liquid to gas? Picture the Problem Because the water absorbed heat in the vaporization process Qabsorbed

its change in entropy is positive and given by ΔS H 2O

 by H 2 O

=

T 

. See Table 18-2

for the latent heat of vaporization of water. The change in entropy of the water is given by:

The heat absorbed by the water as it vaporizes is the product of its mass and latent heat of vaporization: Substituting for  Qabsorbed yields:  by H 2 O

Qabsorbed ΔS H O 2

=

Qabsorbed

 by H 2 O

T 

= mLv =

 by H 2 O

ΔS H O 2

=

ρ VLv

T 

VLv

The Second Law of Thermodynamics Substitute numerical values and evaluate ΔS H 2O :

ΔS H O 2

1909

kJ ⎞ ⎛ 1.00 kg ⎞ (1.00 L )⎛  ⎜⎜ 2257 ⎟⎟ ⎜ ⎟ L  ⎠ kg ⎠ ⎝  = ⎝  373 K 

= 6.05

kJ K 

What is the change in entropy of 1.00 mol of liquid water at 0.0ºC that 54 • freezes to ice at 0.0°C? Picture the Problem We can use the definition of entropy change to find the change in entropy of the liquid water as it freezes. Because heat is removed from liquid water when it freezes, the change in entropy of the liquid water is negative. See Appendix C for the molar mass of water and Table 18-2 for the latent heat of  fusion of water.

The change in entropy of the water is given by:: The heat removed from the water as it freezes is the product of its mass and latent heat of fusion:

Qremoved ΔS H O 2

=

Qremoved

= −mLf 

or, because m = nM H 2 O , from H 2 O

ΔS H O 2

T 

from H 2 O

Qremoved

Substitute numerical values and evaluate

from H 2 O

= −nM H O Lf  2

ΔS H O : 2

g  ⎞ ⎛  J ⎞ − (1.00 mol)⎛  ⎜18.015 ⎟ ⎜⎜ 333.5 ⎟⎟ mol ⎠ ⎝  g ⎠ J ⎝  = = − 22.0 273 K 

K 

Consider the freezing of 50.0 g of water once it is placed in the freezer  55 • compartment of a refrigerator. Assume the walls of the freezer are maintained at  –10ºC. The water, initially liquid at 0.0ºC, is frozen into ice and cooled to  –10ºC. Show that even though the entropy of the water decreases, the net entropy of the universe increases. Picture the Problem The change in the entropy of the universe resulting from the freezing of this water and the cooling of the ice formed is the sum of the entropy changes of the water-ice and the freezer. Note that, while the entropy of the water  decreases, the entropy of the freezer increases.

1910

Chapter 19

The change in entropy of the universe resulting from this freezing and cooling process is given by: Express

ΔS water  :

Express

ΔS freezing :

ΔS u = ΔS water  + ΔS freezer 

(1)

ΔS water  = ΔS freezing + ΔS cooling

(2)

ΔS freezing =

− Qfreezing

(3)

T freezing

where the minus sign is a consequence of the fact that heat is leaving the water  as it freezes. Relate Qfreezing to the latent heat of 

Qfreezing

= mLf 

fusion and the mass of the water: Substitute in equation (3) to obtain:

Express

ΔS cooling :

ΔS freezing =

− mLf  T freezing

⎛ T   ⎞ ΔS cooling = mC  p ln⎜⎜ f  ⎟⎟ ⎝ T i  ⎠

Substitute in equation (2) to obtain:

 Noting that the freezer gains heat (at 263 K) from the freezing water and cooling ice, express ΔS freezer : Substitute for  ΔS water  and

ΔS water  =

⎛ T   ⎞ + mC  p ln⎜⎜ f  ⎟⎟ T freezing ⎝ T i  ⎠

ΔS freezer  = =

− mLf 

ΔQice T freezer  mLf  T freezer 

+ +

ΔQcooling ice T freezer  mC  p ΔT  T freezer 

ΔS freezer in equation (1):

⎛ T   ⎞ mLf  mC  p ΔT  + mC  p ln⎜⎜ f  ⎟⎟ + + T freezing T  T  ⎝  i  ⎠ freezer  T freezer  ⎡ − Lf  ⎛ T   ⎞  L + C  ΔT ⎤ = m⎢ + C  p ln⎜⎜ f  ⎟⎟ + f   p ⎥ T freezer  ⎥⎦ ⎢⎣ T freezing ⎝ T i  ⎠

ΔS u =

− mLf 

The Second Law of Thermodynamics

1911

Substitute numerical values and evaluate ΔS u:

⎡ 333.5 ×103 J ⎢ J  ⎞ ⎛ 263 K  ⎞ kg ⎛  ⎟⎟ ⎟⎟ ln⎜⎜ + ⎜⎜ 2100 ΔS u = (0.0500 kg ) ⎢− 273 K  kg K  273 K  ⋅ ⎢ ⎝   ⎠ ⎝   ⎠ ⎢⎣ ⎤ J ⎛  J  ⎞ ⎟⎟ (273 K − 263 K )⎥ 333.5 ×10 3 + ⎜⎜ 2100 kg ⎝  kg ⋅ K  ⎠ ⎥ = 2.40 J/K  + ⎥ 263 K  ⎥ ⎦ and, because ΔS u > 0, the entropy of the universe increases. In this problem, 2.00 mol of an ideal gas at 400 K expand quasi56 • statically and isothermally from an initial volume of 40.0 L to a final volume of  80.0 L. (a) What is the entropy change of the gas? (b) What is the entropy change of the universe for this process? st

Picture the Problem We can use the definition of entropy change and the 1 law of thermodynamics to express ΔS  for the ideal gas as a function of its initial and final volumes.

(a) The entropy change of the gas is given by: Apply the first law of thermodynamics to the isothermal process to express Q in terms of W on:

The work done on the gas is given  by: Substitute for Q in equation (1) to obtain:

ΔS gas

=

Q T 

Q = Δ E int

− W on or, because Δ E int = 0 for an isothermal expansion of a gas, Q = −W on

W on

⎛ V  ⎞ ⎛ V  ⎞ = nRT ln⎜⎜ i ⎟⎟ ⇒ Q = −nRT ln⎜⎜ i ⎟⎟ ⎝ V f  ⎠ ⎝ V f  ⎠

ΔS gas

⎛ V  ⎞ = −nR ln⎜⎜ i ⎟⎟ ⎝ V f  ⎠

Substitute numerical values and evaluate ΔS :

ΔS gas

(1)

J  ⎞ ⎛ 40.0 L ⎞ J ⎟⎟ = 11.5 = −(2.00 mol)⎛  ⎜ 8.314 ⎟ ln⎜⎜ mol ⋅ K  ⎠ ⎝ 80.0 L ⎠ K  ⎝ 

1912

Chapter 19

(b) Because the process is reversible:

ΔS u = 0

Remarks: The entropy change of the environment of the gas is 11.5 J/K. 57 •• [SSM] A system completes a cycle consisting of six quasi-static steps, during which the total work done by the system is 100 J. During step 1 the system absorbs 300 J of heat from a reservoir at 300 K, during step 3 the system absorbs 200 J of heat from a reservoir at 400 K, and during step 5 it absorbs heat from a reservoir at temperature T 3. (During steps 2, 4 and 6 the system undergoes adiabatic processes in which the temperature of the system changes from one reservoir’s temperature to that of the next.) ( a) What is the entropy change of the system for the complete cycle? (b) If the cycle is reversible, what is the temperature T 3? Picture the Problem We can use the fact that the system returns to its original state to find the entropy change for the complete cycle. Because the entropy change for the complete cycle is the sum of the entropy changes for each process, we can find the temperature T 3 from the entropy changes during the 1st two  processes and the heat released during the third.

(a) Because S is a state function of  the system, and because the system’s final state is identical to its initial state:

ΔSsystem

(b) Relate the entropy changes for  each of the three heat reservoirs and the system for one complete cycle of  the system:

ΔS 1

= 0

1 complete cycle

or  Q1 T 1

+ ΔS 2 + ΔS 3 + ΔS system = 0

+

Q2 T 2

+

Q3 T 3

+0 = 0

Substitute numerical values. Heat is rejected by the two high-temperature reservoirs and absorbed by the cold reservoir:

− 300 J − 200 J 400 J + + =0

Solving for T 3 yields:

T 3

300 K 

400 K 

T 3

= 267 K 

In this problem, 2.00 mol of an ideal gas initially has a temperature of  58 •• 400 K and a volume of 40.0 L. The gas undergoes a free adiabatic expansion to twice its initial volume. What is ( a) the entropy change of the gas and (b) the entropy change of the universe?

The Second Law of Thermodynamics

1913

Picture the Problem The initial and final temperatures are the same for a free expansion of an ideal gas. Thus, the entropy change ΔS  for a free expansion from V i to V f f  is the same as ΔS for an isothermal process from V i to V f f.  We can use the definition of entropy change and the 1st law of thermodynamics to express ΔS  for  the ideal gas as a function of its initial and final volumes.

(a) The entropy change of the gas is given by: Apply the first law of thermodynamics to the isothermal process to express Q:

ΔS gas

=

Q

(1)

T 

Q = Δ E int

− W on or, because Δ E int int = 0 for a free expansion of a gas, Q = −W on

The work done on the gas is given  by: Substitute for Q in equation (1) to obtain:

W on

⎛ V  ⎞ ⎛ V  ⎞ = nRT ln⎜⎜ i ⎟⎟ ⇒ Q = −nRT ln⎜⎜ i ⎟⎟ ⎝ V f  ⎠ ⎝ V f  ⎠

ΔS gas

⎛ V  ⎞ = −nR ln⎜⎜ i ⎟⎟ ⎝ V f  ⎠

Substitute numerical values and evaluate ΔS :

ΔS gas

J  ⎞ ⎛ 40.0 L ⎞ J ⎟⎟ = 11.5 = −(2.00 mol)⎛  ⎜ 8.314 ⎟ ln⎜⎜ mol ⋅ K  ⎠ ⎝ 80.0 L ⎠ K  ⎝ 

(b) The change in entropy of the universe is the sum of the entropy changes of the gas and the surroundings: For the change in entropy of the surroundings we use the fact that, during the free expansion, the surroundings are unaffected: The change in entropy of the universe is the change in entropy of the gas:

ΔS u

= ΔS gas + ΔS surroundings

ΔS surroundings

ΔS u

=

= 11.5

Qrev T 

=

0 T 

=0

J K 

A 200-kg block of ice at 0.0ºC is placed in a large lake. The 59 •• temperature of the lake is just slightly higher than 0.0ºC, and the ice melts very

1914

Chapter 19

slowly. (a) What is the entropy change of the ice? (b) What is the entropy change of the lake? (c) What is the entropy change of o f the universe (the ice plus the lake)? lake )? Picture the Problem Because the ice gains heat as it melts, its entropy change is  positive and can be calculated from its definition. Because the temperature of the lake is just slightly greater than 0°C and the mass of water is so much greater than that of the block of ice, the absolute value of the entropy change of the lake will  be approximately equal to the entropy change of the ice as it melts.

(a) The entropy change of the ice is given by: Substitute numerical values and evaluate ΔS ice :

(b) Relate the entropy change of the lake to the entropy change of the ice: (c) The entropy change of the universe due to this melting process is the sum of the entropy changes of  the ice and the lake:

mLf 

ΔS ice

=

ΔS ice

⎛  kJ ⎞ (200 kg )⎜⎜ 333.5 ⎟⎟ kg ⎠ ⎝  = = 273 K 

ΔSlake

ΔS u

T 

≈ −ΔSice = − 244

244

kJ K 

kJ K 

= ΔSice + ΔSlake

Because the temperature of the lake is slightly greater than that of the ice, the magnitude of the entropy change of the lake is less than 244 kJ/K and the entropy change of the universe is greater than zero. The melting of the ice is an irreversible process and

ΔS u > 0 .

A 100-g piece of ice at 0.0ºC is placed in an insulated calorimeter with 60 •• negligible heat capacity containing 100 g of water at 100ºC. (a) What is the final temperature of the water once thermal equilibrium is established? (b) Find the entropy change of the universe for this process. Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a melting process and for constant-pressure processes to find the entropy change of  the universe (the entropy change of the piece of ice plus the entropy change of the water in the insulated container).

The Second Law of Thermodynamics

1915

∑Q = 0

(a) Apply conservation of energy to obtain:

i

i

or  Qmelting

+ Qwarming − Qcooling = 0

ice

water 

water 

Substitute to relate the masses of the ice and water to their temperatures, specific heats, and the final temperature of the water:

⎛  ⎝ 

(100 g )⎜⎜ 333.5

⎛  kJ ⎞ kJ  ⎞ ⎟⎟ + (100 g )⎜⎜ 4.18 ⎟⎟ t ⋅ ° kg ⎠ kg C ⎝   ⎠ ⎛  kJ  ⎞ ⎟⎟ (100°C − t ) = 0 − (100 g )⎜⎜ 4.18 kg C ⋅ ° ⎝   ⎠ = 10.1°C

Solving for t yields:

 t

(b) The entropy change of the universe is the sum of the entropy changes of the ice and the water:

ΔS u = ΔS ice + ΔS water 

Using the expression for the entropy change for a constant-pressure  process, express the entropy change of the melting ice and warming icewater:

ΔS ice = ΔS melting ice + ΔS warming water  =

mLf  T f 

⎛ T  ⎞ + mcP ln⎜⎜ f  ⎟⎟ ⎝ T i  ⎠

Substitute numerical values to obtain:

⎛  kJ ⎞ (0.100 kg )⎜⎜ 333.5 ⎟⎟ kg ⎠ ⎛  kJ  ⎞ ⎛ 283 K  ⎞ ⎝  ⎟⎟ = 137 ⎟⎟ ln⎜⎜ ΔSice = + (0.100 kg )⎜⎜ 4.18 273 K  kg K  273 K  ⋅ ⎝   ⎠ ⎝   ⎠ Find the entropy change of the cooling water:

ΔS water 

⎛  kJ  ⎞ ⎛ 283 K  ⎞ J ⎟⎟ = −115 ⎟⎟ ln⎜⎜ = (0.100 kg )⎜⎜ 4.18 kg ⋅ K  ⎠ ⎝ 373 K  ⎠ K  ⎝ 

Substitute for ΔS ice ice and ΔS water  water  and evaluate the entropy change of the universe:

ΔS u

= 137

J K 

− 115

J K 

= 22

J K 

J K 

1916

Chapter 19

Remarks: The result that Su > 0 tells us that this process is irreversible. 61 •• [SSM] A 1.00-kg block of copper at 100ºC is placed in an insulated calorimeter of negligible heat capacity containing 4.00 L of liquid water at 0.0ºC. Find the entropy change of (a) the copper block, (b) the water, and (c) the universe. Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a constant pressure process to find the entropy changes of the copper block, the water, and the universe.

(a) Use the equation for the entropy change during a constant-pressure  process to express the entropy change of the copper block:

⎛ T  ⎞ ΔS Cu = mCu cCu ln⎜⎜ f  ⎟⎟ ⎝ T i  ⎠

Apply conservation of energy to obtain:

∑Q = 0

(1)

i

i

or  Qcopper  block  + Qwarming water  = 0 Substitute to relate the masses of the block and water to their temperatures, specific heats, and the final temperature T f f  of the water:

⎛  kJ  ⎞ ⎟⎟ (T f  − 373 K ) (1.00 kg )⎜⎜ 0.386 ⋅ kg K  ⎝   ⎠ kg ⎞ ⎛  kJ  ⎞ ⎟ (T f  − 273 K ) = 0 + (4.00 L )⎛  ⎜1.00 ⎟ ⎜⎜ 4.18 L  ⎠ ⎝  kg ⋅ K  ⎠⎟ ⎝  Solve for T f f  to obtain:

T f 

= 275.26 K 

Substitute numerical values in equation (1) and evaluate ΔS Cu :

ΔS Cu

⎛  kJ  ⎞ ⎛ 275.26 K  ⎞ J ⎟⎟ = − 117 ⎟⎟ln⎜⎜ = (1.00 kg )⎜⎜ 0.386 kg ⋅ K  ⎠ ⎝  373 K   ⎠ K  ⎝ 

(b) The entropy change of the water is given by:

⎛ T   ⎞ ΔS water  = mwater c water  ln⎜⎜ f  ⎟⎟ ⎝ T i  ⎠

The Second Law of Thermodynamics

1917

Substitute numerical values and evaluate ΔS water  :

ΔS water 

⎛  kJ  ⎞ ⎛ 275.26 K  ⎞ J ⎟⎟ = 138 ⎟⎟ ln⎜⎜ = (4.00 kg )⎜⎜ 4.18 kg ⋅ K  ⎠ ⎝  273 K   ⎠ K  ⎝ 

(c) Substitute for  ΔS Cu and

ΔS water 

and evaluate the entropy change of  the universe:

ΔS u

= ΔSCu + ΔS water  = −117 = 20

J J + 138 K  K 

J K 

Remarks: The result that Su > 0 tells us that this process is irreversible.

If a 2.00-kg piece of lead at 100ºC is dropped into a lake at 10ºC, find 62 •• the entropy change of the universe. Picture the Problem Because the mass of the water in the lake is so much greater than the mass of the piece of lead, the temperature of the lake will increase only slightly and we can reasonably assume that its final temperature is 10°C. We can apply the equation for the entropy change during a constant  pressure process to find the entropy changes of the piece of lead, the water in the lake, and the universe.

Express the entropy change of the universe in terms of the entropy changes of the lead and the water in the lake:

ΔS u = ΔS Pb + ΔS w

(1)

Using the equation for the entropy change during a constant-pressure  process, express and evaluate the entropy change of the lead:

ΔS Pb

⎛ T   ⎞ ⎛  kJ  ⎞ ⎛ 283 K  ⎞ J ⎟⎟ = −70.69 ⎟⎟ ln⎜⎜ = mPb cPb ln⎜⎜ f  ⎟⎟ = (2.00 kg )⎜⎜ 0.128 kg ⋅ K  ⎠ ⎝ 373 K  ⎠ K  ⎝  ⎝ T i  ⎠

The entropy change of the water in the lake is given by:

ΔS w

=

Qw T w

=

QPb T w

=

mPb cPb ΔT Pb T w

1918

Chapter 19

Substitute numerical values and evaluate ΔS w:

ΔS w

⎛  kJ  ⎞ ⎟⎟ (90 K ) (2.00 kg )⎜⎜ 0.128 kg K  ⋅ ⎝   ⎠ = 283 K 

= 81.41J/K  Substitute numerical values in equation (1) and evaluate ΔS u:

ΔS u

= −70.69

J K 

+ 81.41

J K 

= 11

J K 

Entropy and Lost Work 63 •• [SSM] A a reservoir at 300 K absorbs 500 J of heat from a second reservoir at 400 K. (a) What is the change in entropy of the universe, and (b) how much work is lost during the process? Picture the Problem We can find the entropy change of the universe from the entropy changes of the high- and low-temperature reservoirs. The maximum amount of the 500 J of heat that could be converted into work can be found from the maximum efficiency of an engine operating between the two reservoirs.

(a) The entropy change of the universe is the sum of the entropy changes of the two reservoirs:

Substitute numerical values and evaluate ΔS u:

(b) Relate the heat that could have  been converted into work to the maximum efficiency of an engine operating between the two reservoirs: The maximum efficiency of an engine operating between the two reservoir temperatures is the efficiency of a Carnot device operating between the reservoir  temperatures:

ΔS u = ΔS h + ΔS c = −

Q T h

+

Q T c

⎛  1 1  ⎞ = −Q⎜⎜ − ⎟⎟ ⎝ T h T c  ⎠ ΔS u

⎛  1 1  ⎞ ⎟⎟ = (− 500 J ) ⎜⎜ − 400 K  300 K  ⎝   ⎠ = 0.42 J/K 

W  = ε maxQh

ε max

= ε C = 1 −

T c T h

The Second Law of Thermodynamics Substitute for ε max to obtain:

Substitute numerical values and evaluate W :

⎛  ⎝ 

T c  ⎞ ⎟Qh T h  ⎠⎟

⎛  ⎝ 

300 K  ⎞ ⎟ (500 J ) = 125 J 400 K  ⎠⎟

W  = ⎜⎜1 −

W  = ⎜⎜1 −

1919

In this problem, 1.00 mol of an ideal gas at 300 K undergoes a free 64 •• adiabatic expansion from V 1 = 12.3 L to V 2 = 24.6 L. It is then compressed isothermally and reversibly back to its original state. ( a) What is the entropy change of the universe for the complete cycle? (b) How much work is lost in this cycle? (c) Show that the work lost is T ΔS u. Picture the Problem Although no heat is lost by the gas in the adiabatic free expansion, the process is irreversible and the entropy of the gas increases. In the isothermal reversible process that returns the gas to its original state, the gas releases heat to the surroundings. However, because the process is reversible, the entropy change of the universe is zero. Consequently, the net entropy change is the negative of that of the gas in the isothermal compression.

(a) Relate the entropy change of the universe to the entropy changes of  the gas during 1 complete cycle:

ΔS u

= ΔS gas during

+ ΔS gas during

free expansion

or, because ΔS gas during

= ΔS gas during isothermal compresion

Substituting for  ΔS gas in the expression for  ΔS u yields:

= 0,

free expansion

ΔS u

The work done by the gas during its isothermal compression is given by:

isothermal compresion

=

−Q T 

W  by

⎛ V   ⎞ = −W on = −Q = −nRT ln⎜⎜ f  ⎟⎟ ⎝ V i  ⎠

ΔS u

⎛ V   ⎞ = −nR ln⎜⎜ f  ⎟⎟ ⎝ V i  ⎠

(1)

Substitute numerical values and evaluate ΔS u:

ΔS u

J  ⎞ ⎛ 12.3 L ⎞ J J ⎟⎟ = 5.763 = 5.76 = −(1.00 mol)⎛  ⎜ 8.314 ⎟ln⎜⎜ mol ⋅ K  ⎠ ⎝ 24.6 L ⎠ K  K  ⎝ 

(b) Use Equation 19-22 to find the amount of energy that becomes unavailable for doing work during this process:

W lost

J  ⎞ = T ΔS u = (300 K )⎛  ⎜ 5.763 ⎟ K  ⎠ ⎝  = 1.73 kJ

1920

Chapter 19

(c) Use equation (1) to express the  product of T and ΔS u:

⎛  ⎛ V   ⎞ ⎞ ⎛ V   ⎞ = T ⎜⎜ − nR ln⎜⎜ f  ⎟⎟ ⎟⎟ = −nRT ln⎜⎜ f  ⎟⎟ ⎝ V i  ⎠ ⎠ ⎝ V i  ⎠ ⎝  = W  by gas during its

T ΔS u

isothermal compression

General Problems A heat engine with an output of 200 W has an efficiency of 30%. It 65 • operates at 10.0 cycles/s. (a) How much work is done by the engine during each cycle? (b) How much heat is absorbed from the hot reservoir and how much is released to the cold reservoir during each cycle? Picture the Problem We can use the definition of power to find the work done each cycle and the definition of efficiency to find the heat that is absorbed each cycle. Application of the first law of thermodynamics will yield the heat given off  each cycle.

(a) Use the definition of power to relate the work done in each cycle to the frequency of each cycle: Substitute numerical values and evaluate W cycle: (b) Express the heat absorbed in each cycle in terms of the work done and the efficiency of the engine: Apply the 1st law of thermodynamics to find the heat given off in each cycle:

W cycle

= PΔt  =

P  f 

where f is the frequency of the engine. W cycle

=

200 W 10.0 s −1 W cycle

= 20.0 J

=

Qc,cycle

= Qh,cycle − W  = 67 J − 20 J

ε 

=

20.0 J

Qh,cycle

0.30

= 67 J

= 47 J

During each cycle, a heat engine operating between two heat 66 • reservoirs absorbs 150 J from the reservoir at 100ºC and releases 125 J to the reservoir at 20ºC. (a) What is the efficiency of this engine? ( b) What is the ratio of its efficiency to that of a Carnot engine working between the same reservoirs? (This ratio is called the second law efficiency.) Picture the Problem We can use their definitions to find the efficiency of the engine and that of a Carnot engine operating between the same reservoirs.

The Second Law of Thermodynamics (a) The efficiency of the engine is given by: Substitute numerical values and evaluate ε : (b) Find the efficiency of a Carnot engine operating between the same reservoirs: Express the ratio of the two efficiencies:

ε 

=

W  Qh

= 1−

C

Qh

− Qc

Qh

125 J 150 J

= 1−

= C

=

T c T h

= 1−

1921

Qc Qh

= 16.67% = 16.7%

= 1−

16.67% 21.45%

293 K  = 21.45% 373 K 

= 0.777

67 • [SSM] An engine absorbs 200 kJ of heat per cycle from a reservoir  at 500 K and releases heat to a reservoir at 200 K. Its efficiency is 85 percent of  that of a Carnot engine working between the same reservoirs. (a) What is the efficiency of this engine? (b) How much work is done in each cycle? (c) How much heat is released to the low-temperature reservoir during each cycle? Picture the Problem We can use the definition of efficiency to find the work  done by the engine during each cycle and the first law of thermodynamics to find the heat released to the low-temperature reservoir during each cycle.

(a) Express the efficiency of the engine in terms of the efficiency of a Carnot engine working between the same reservoirs: Substitute numerical values and evaluate ε  :

= 0.85

C

⎛  T   ⎞ = 0.85⎜⎜1 − c ⎟⎟ ⎝  T h  ⎠

⎛  200 K  ⎞ ⎟⎟ = 0.510 = 51% = 0.85⎜⎜1 − 500 K  ⎝   ⎠

(b) Use the definition of efficiency to find the work done in each cycle:

W  = ε  Qh

(c) Apply the first law of  thermodynamics to the cycle to obtain:

Qc,cycle

= (0.510)(200 kJ ) = 102 kJ

= 0.10 MJ = Qh,cycle − W  = 200 kJ − 102 kJ = 98 kJ

Estimate the change in entropy of the universe associated with an 68 • Olympic diver diving into the water from the 10-m platform.

1922

Chapter 19

Picture the Problem Assume that the mass of the diver is 75 kg and that the temperature of the water in the pool is 25 °C. The energy added to the water in the  pool is the change in the gravitational potential energy of the diver during the dive.

The change in entropy of the universe associated with a dive is given by:

ΔS u

= ΔS water  =

Qadded to water  T water 

where Qadded to water is the energy entering the water as a result of the kinetic energy of the diver as he enters the water.

The energy added to the water is the change in the gravitational potential energy of the diver: Substitute numerical values and evaluate ΔS u :

mgh

ΔS u

=

ΔS u

(75 kg )(9.81 m/s 2 )(10 m ) = (25 + 273)K 

T water 

≈ 25

J K 

To maintain the temperature inside a house at 20ºC, the electric power  69 • consumption of the electric baseboard heaters is 30.0 kW on a day when the outside temperature is –7ºC. At what rate does this house contribute to the increase in the entropy of the universe? Picture the Problem The change in entropy of the universe is the change in entropy of the house plus the change in entropy of the environment. We can find the change in entropy of the house by exploiting the given information that the temperature inside the house is maintained at a constant temperature. We can find the change in entropy of the surrounding by dividing the heat added by the temperature.

Entropy is a state function, and the state of the house does not change. Therefore the entropy of the house does not change: Heat is absorbed by the surroundings at the same rate R that energy is delivered to the house:

= ΔShouse + ΔSsurroundings or, because ΔS house = 0 , ΔS u = ΔSsurroundings ΔS u

ΔS surroundings

=

Q T surroundings

=

 RΔt  T surroundings

The Second Law of Thermodynamics Substitute for ΔS surroundings yields:

Substitute numerical values and evaluate ΔS u/Δt :

ΔS u

ΔS u Δt 

 RΔt 

=

T surroundings

=

30.0 kW 266 K 

⇒

ΔS u

=

Δt 

= 113

1923

 R T surroundings

W K 

Calvin Cliffs Nuclear Power Plant, located on the Hobbes River, 70 •• generates 1.00 GW of power. In this plant, liquid sodium circulates between the reactor core and a heat exchanger located in the superheated steam that drives the turbine. Heat is absorbed by the liquid sodium in the core, and released by the liquid sodium (and into the superheated steam) in the heat exchanger. The temperature of the superheated steam is 500 K. Heat is released into the river, and the water in the river flows by at a temperature of 25ºC. (a) What is the highest efficiency that this plant can have? (b) How much heat is released into the river  every second? (c) How much heat must be released by the core to supply 1.00 GW of electrical power? (d ) Assume that new environmental laws have been  passed to preserve the unique wildlife of the river. Because of these laws, the  plant is not allowed to heat the river by more than 0.50ºC. What is the minimum flow rate that the water in the Hobbes River must have? Picture the Problem We can use the expression for the Carnot efficiency of the  plant to find the highest efficiency this plant can have. We can then use this efficiency to find the power that must be supplied to the plant to generate 1.00 GW of power and, from this value, the power that is wasted. The rate at which heat is being released to the river is related to the requisite flow rate of the river by dQ dt  = cΔT  dV  dt .

(a) The Carnot efficiency of a plant operating between temperatures T c and T h is given by: Substitute numerical values and evaluate ε C: (c) The power that must be supplied, at 40.4% efficiency, to produce an output of 1.00 GW is given by:

(b) Relate the wasted power to the  power generated and the power  supplied:

ε max

max

= ε C = 1 −

= 1−

Psupplied

=

T c T h

298 K  = 0.404 500 K  Poutput

=

1.00 GW

ε max

0.404

= 2.48 GW Pwasted

= Psupplied − Pgenerated

1924

Chapter 19

Substitute numerical values and evaluate Pwasted :

Pwasted

(d ) Express the rate at which heat is  being dumped into the river:

dQ

= 2.48 GW − 1.00 GW = 1.48 GW

dt 

= cΔT 

dm

dt  dV 

= cΔT  ρ  Solve for the flow rate dV /dt of the river: Substitute numerical values (see Table 19-1 for the specific heat of  water) and evaluate dV /dt :

dV  dt 

dV  dt 

=

d 

= cΔT  ( ρ V ) dt 

dt 

dQ dt  cΔT  ρ 

1.48 × 10 9

=

J s

⎛  J  ⎞ kg  ⎞ ⎜⎜ 4180 ⎟⎟ (0.50 K )⎛  ⎜10 3 3 ⎟ kg ⎠ ⎝  m  ⎠ ⎝  = 7.1×10 5 L/s

An inventor comes to you to explain his new invention. It is a novel 71 •• heat engine using water vapor as the working substance. He claims that the water  vapor absorbs heat at 100°C, does work at the rate of 125 W, and releases heat to the air at the rate of only 25.0 W, when the air temperature is 25°C. (a) Explain to him why he cannot be correct. (b) After careful analysis of the data in his  prospectus folder, you decide he has made an error in the measurement of his exhausted-heat value. What is the minimum rate of exhausting heat that would make you consider believing him? Picture the Problem We can use the inventor’s data to calculate the thermal efficiency of his steam engine and then compare this value to the efficiency of a Carnot engine operating between the same temperatures.

(a) The Carnot efficiency of an engine operating between these temperatures is: The thermal efficiency of the inventor’s device, in terms of the rate at which it expels heat to the air and does work is:

ε C

= 1−

T c T h

dW  ε 

=

dt  dQh dt 

=

= 1−

298 K  373 K 

= 20.1%

dW 

=

dt  dW  dQc dt 

+

dt 

125 W 125 W + 25.0 W

= 83.3%

The Second Law of Thermodynamics

1925

You should explain to him that, because the efficiency he claims for his invention is greater than the efficiency of a Carnot engine operating between the same two temperatures, his data is not consistent with what is known about the thermodynamics of engines. He must have made a mistake in his analysis of his data−or he is a con man looking for people to swindle. (b) The maximum efficiency of a steam engine that has ever been achieved is about 50% of the Carnot efficiency of an engine operating between the same temperatures. Setting the efficiency of his steam engine equal to half the Carnot efficiency of the engine yields:

dW  1 2

ε C

=

dt  dQh

dW 

=

dt 

Solve for dQc/dt to obtain:

dQc dt 

Assuming that the inventor has measured the work done per cycle  by his invention correctly:

Ignoring his claim that 25.0 W of  work are done per cycle, let’s assume that his device does take in 150 W of energy each cycle and find how much work would do with an efficiency half that of a Carnot engine: Substituting numerical values yields:

dQc dt 

dQc dt 

=

dQh dt 

−

dW 

,a

dt  reasonable value for dQc/dt is:

dt 

+

dt 

⎛  2  ⎞ dW  = ⎜⎜ − 1⎟⎟ ⎝ ε C  ⎠ dt  2 = ⎛  − 1 ⎞⎟ (125 W ) ≈ 1100 W ⎜ ⎝ 0.201  ⎠

a value totally inconsistent with the inventor’s claims for his engine. dW  1 2

ε C

=

dt  dQh

⇒

dW  dt 

= 12 ε C

dQh dt 

dt 

dW  dt 

Because

dt  dW  dQc

dQc dt 

= 12 (0.201)(150 W ) ≈ 15 W = 150 W − 15 W = 135 W

The cycle represented in Figure 19-12 (next to Problem 19-14) is for  72 •• 1.00 mol of an ideal monatomic gas. The temperatures at points A and B are 300 and 750 K, respectively. What is the efficiency of the cyclic process ABCDA?

1926

Chapter 19

Picture the Problem Because the cycle represented in Figure 19-12 is a Carnot cycle, its efficiency is that of a Carnot engine operating between the temperatures of its isotherms.

The Carnot efficiency of the cycle is given by: Substitute numerical values and evaluate ε C:

ε C

= 1−

C

= 1−

T c T h

300 K  = 60.0% 750 K 

73 •• [SSM] (a) Which of these two processes is more wasteful? (1) A  block moving with 500 J of kinetic energy being slowed to rest by sliding (kinetic) friction when the temperature of the environment is 3 00 K, or (2) A reservoir at 400 K releasing 1.00 kJ of heat to a reservoir at 300 K? Explain your  choice. Hint: How much of the 1.00 kJ of heat could be converted into work by an ideal cyclic process? (b) What is the change in entropy of the universe for each  process? Picture the Problem All 500 J of mechanical energy are lost, i.e., transformed into heat in process (1). For process (2), we can find the heat that would be converted to work by a Carnot engine operating between the given temperatures and subtract that amount of work from 1.00 kJ to find the energy that is lost. In Part (b) we can use its definition to find the change in entropy for each process.

(a) For process (2): The efficiency of a Carnot engine operating between temperatures T h and T c is given by:

Substitute for  ε C to obtain:

W 2,max

ε C

= W recovered = ε CQin

= 1−

T c T h

and hence W recovered

⎛  T   ⎞ = ⎜⎜1 − c ⎟⎟Qin ⎝  T h  ⎠

W recovered

300 K  ⎞ = ⎛  ⎜1 − ⎟(1.00 kJ ) = 250 J ⎝  400 K  ⎠

or  750 J are lost. Process (2) is more wasteful of available work. (b) Find the change in entropy of the universe for process (1):

ΔS 1

=

ΔQ

T 

=

500 J 300 K 

= 1.67 J/K 

The Second Law of Thermodynamics

Express the change in entropy of  the universe for process (2):

ΔS 2 = ΔS h + ΔS c = −

1927

ΔQ ΔQ + T h

T c

⎛  1 1  ⎞ = ΔQ ⎜⎜ − ⎟⎟ ⎝ T c T h  ⎠ Substitute numerical values and evaluate ΔS 2:

ΔS 2

⎛  1 1  ⎞ ⎟⎟ = (1.00 kJ )⎜⎜ − 300 K  400 K  ⎝   ⎠ = 0.833 J/K 

Helium, a monatomic gas, is initially at a pressure of 16 atm, a volume 74 •• of 1.0 L, and a temperature of 600 K. It is quasi-statically expanded at constant temperature until its volume is 4.0 L and is then quasi-statically compressed at constant pressure until its volume and temperature are such that a quasi-static adiabatic compression will return the gas to its original state. (a) Sketch this cycle on a PV diagram. (b) Find the volume and temperature after the compression at constant pressure. (c) Find the work done during each step of the cycle. (d ) Find the efficiency of the cycle. Picture the Problem Denote the three states of the gas as 1, 2, and 3 with 1 being the initial state. We can use the ideal-gas law and the equation of state for an adiabatic process to find the temperatures, volumes, and pressures at points 1, 2, and 3. To find the work done during each cycle, we can use the equations for the work done during isothermal, isobaric, and adiabatic processes. Finally, we find the efficiency of the cycle from the work done each cycle and the heat that enters the system during the isothermal expansion.

(a) The PV diagram of the cycle is shown to the right.

(b) Apply the ideal-gas law to the isothermal expansion 1→2 to find P2:

P2

= P1

V 1 V 2

⎛ 1.0 L ⎞ ⎟⎟ = 4.0 atm = (16 atm )⎜⎜ 4.0 L ⎝   ⎠

1928

Chapter 19

Apply an equation for an adiabatic  process to relate the pressures and volumes at 1 and 3:

1 γ  

γ  

P1V 1

= P3V 3

γ  

⎛  P  ⎞ ⇒ V 3 = V 1 ⎜⎜ 1 ⎟⎟ ⎝ P3 ⎠ 1 1.67

Substitute numerical values and evaluate V 3:

⎛ 16 atm  ⎞ ⎟⎟ V 3 = (1.0 L )⎜⎜ 4.0 atm ⎝   ⎠ = 2.3 L

Apply an equation for an adiabatic  process (γ  =1.67) to relate the temperatures and volumes at 1 and 3:

⎛ V  ⎞ − − T 3V 3 1 = T 1V 1 1 ⇒ T 3 = T 1 ⎜⎜ 1 ⎟⎟ ⎝ V 3 ⎠

Substitute numerical values and evaluate T 3:

⎛  1.0 L  ⎞ ⎟⎟ T 3 = (600 K )⎜⎜ 2.294 L ⎝   ⎠ = 3.4 ×10 2 K 

(c) Express the work done each cycle:

W  = W 12

For the process 1→2:

For the process 2→3:

γ  

= 2.294 L

−1

γ  

γ  

1.67 −1

W 12

W 23

+ W 23 + W 31

= 344 K 

(1)

⎛ V   ⎞ ⎛ V   ⎞ = nRT 1 ln⎜⎜ 2 ⎟⎟ = P1V 1 ln⎜⎜ 2 ⎟⎟ ⎝ V 1  ⎠ ⎝ V 1  ⎠ ⎛ 4.0 L ⎞ ⎟⎟ = (16 atm )(1.0 L )ln⎜⎜ 1.0 L ⎝   ⎠ = 22.18 atm ⋅ L = P2 ΔV 23 = (4.0 atm )(2.294 L − 4.00 L ) = −6.824 atm ⋅ L

For the process 3→1:

= −C V ΔT 31 = − 32 nR(T 1 − T 3 ) = − 32 (P1V 1 − P3V 3 ) = − 32 [(16 atm )(1.0 L ) − (4.0 atm )(2.294 L )] = −10.24 atm ⋅ L W  = 22.18 atm ⋅ L − 6.824 atm ⋅ L Substitute numerical values in equation (1) and evaluate W : − 10.24 atm ⋅ L W 31

= 5.116 atm ⋅ L = 5 atm ⋅ L

The Second Law of Thermodynamics

(d ) Use its definition to express the efficiency of the cycle:

ε 

Substitute numerical values and evaluate ε :

=

=

W  Qin

=

W  Q12

=

1929

W  W 12

5.116 atm ⋅ L 22.18 atm ⋅ L

≈ 20%

75 •• [SSM] A heat engine that does the work of blowing up a balloon at a  pressure of 1.00 atm absorbs 4.00 kJ from a reservoir at 120ºC. The volume of the  balloon increases by 4.00 L, and heat is released to a reservoir at a temperature T c, where T c < 120ºC. If the efficiency of the heat engine is 50% of the efficiency of a Carnot engine working between the same two reservoirs, find the temperature T c. Picture the Problem We can express the temperature of the cold reservoir as a function of the Carnot efficiency of an ideal engine and, given that the efficiency of the heat engine is half that of a Carnot engine, relate T c to the work done by and the heat input to the real heat engine.

Using its definition, relate the efficiency of a Carnot engine working between the same reservoirs to the temperature of the cold reservoir: Relate the efficiency of the heat engine to that of a Carnot engine working between the same temperatures: Substitute for  ε C to obtain:

The work done by the gas in expanding the balloon is:

ε C

ε 

T c

= 1−

=

W  Qin

T c T h

⇒ T c = T h (1 − ε C )

= 12 ε C ⇒

ε C

=

2W  Qin

⎛  2W  ⎞ ⎟⎟ = T h ⎜⎜1 − Q ⎝  in  ⎠

W  = PΔV  = (1.00 atm )(4.00 L )

= 4.00 atm ⋅ L

1930

Chapter 19

Substitute numerical values and evaluate T c:

⎛  101.325 J ⎞ ⎞ ⎜ 2 ⎛  ⎜ 4.00 atm ⋅ L × ⎟⎟ atm ⋅ L  ⎠ ⎟ ⎝  ⎜ T c = (393 K ) 1 − = 313 K  ⎜ ⎟ 4.00 kJ ⎜ ⎟ ⎝   ⎠ Show that the coefficient of performance of a Carnot engine run as a 78 •• refrigerator is related to the efficiency of a Carnot engine operating between the same two temperatures by ε C × COPC = Tc T h . Picture the Problem We can use the definitions of the COP and ε C to show that their relationship is ε C × COPC = T C T h .

Using the definition of the COP, relate the heat removed from the cold reservoir to the work done each cycle:

COP =

Apply energy conservation to relate Qc, Qh, and W :

Qc

Substitute for Qc to obtain:

COP =

Divide the numerator and denominator by Qh and simplify to obtain:

Qc W 

= Qh − W 

COP =

Qh

− W 

W 

Qh

− W 

W 

1−

=

W 

Qh W 

Qh

Because

ε C

=

W  Qh

= 1−

T c T h

: COPc

=

1 − ε C

=

ε C

⎛  ⎝ 

T c  ⎞

1 − ⎜⎜1 −

⎟⎟ T c T h  ⎠ T h =

ε C

ε C

and ε C

× COPc =

T c T h

77 •• A freezer has a temperature T c = –23ºC. The air in the kitchen has a temperature T h = 27ºC. The freezer is not perfectly insulated and some heat 1

The Second Law of Thermodynamics

1931

leaks through the walls of the freezer at a rate of 50 W. Find the power of the motor that is needed to maintain the temperature in the freezer. Picture the Problem We can use the definition of the COP to express the work  the motor must do to maintain the temperature of the freezer in terms of the rate at which heat flows into the freezer. Differentiation of this expression with respect to time will yield an expression for the power of the motor that is needed to maintain the temperature in the freezer.

Using the definition of the COP, relate the heat that must be removed from the freezer to the work done by the motor: Differentiate this expression with respect to time to express the power  of the motor:

COP =

P

=

dt 

COPmax

Substitute for COPmax to obtain:

P=

 P

W 

dW 

Express the maximum COP of the motor:

Substitute numerical values and evaluate P:

Qc

dQc

=

=

⇒ W  =

Qc

COP

dQc dt 

COP

T c

ΔT 

ΔT 

dt  T c

⎛  50 K  ⎞ ⎟⎟ = 10 W = (50 W )⎜⎜ 250 K  ⎝   ⎠

In a heat engine, 2.00 mol of a diatomic gas are taken through the 78 •• cycle ABCA as shown in Figure 19-20. (The PV diagram is not drawn to scale.) At A the pressure and temperature are 5.00 atm and 600 K. The volume at B is twice the volume at A. The segment BC is an adiabatic expansion and the segment CA is an isothermal compression. (a) What is the volume of the gas at A? (b) What are the volume and temperature of the gas at B? (c) What is the temperature of the gas at C? (d ) What is the volume of the gas at C? (e) How much work is done by the gas in each of the three segments of the cycle?  (f ) How much heat is absorbed or released by the gas in each segment of this cycle? Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points A, B, and C and then find the work  done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the constant pressure, adiabatic, and isothermal processes of the cycle.

1932

Chapter 19

(a) Apply the ideal-gas law to find the volume of the gas at A:

V A

nRT A

=

PA

J  ⎞ (2.00 mol)⎛  ⎜ 8.314 ⎟ (600 K ) mol ⋅ K  ⎠ ⎝  = 101.325 kPa 5.00 atm ×

= 1.969 ×10 − 2 m 3 ×

atm 1L

10 −3 m 3

= 19.69 L

= 19.7 L (b) We’re given that V B

= 2V A .

V B

= 2(19.69 L ) = 39.38 L = 39.4 L

T B

= T A

Hence: Apply the ideal-gas law to this constant-pressure process to obtain:

(c) Because the process C→A is isothermal: (d ) Apply an equation for an adiabatic process (γ  = 1.4) to find the volume of the gas at C:

Substitute numerical values and evaluate V C:

V B V A

= (600 K )

2V A V A

= 1200 K  T C

= T A = 600 K 

−1

γ  

T BV B

V C

−1

= T CV C ⇒ V C γ  

⎛ T   ⎞ = V B ⎜⎜ B ⎟⎟ ⎝ T C  ⎠

⎛ 1200 K  ⎞ ⎟⎟ = (39.38 L )⎜⎜ 600 K  ⎝   ⎠ = 223 L

1

−1

γ  

1 1.4−1

= 222.77 L

(e) The work done by the gas during the constant-pressure  process AB is given by:

W AB

= PA (V B − V A ) = PA (2V A − V A ) = PAV A

Substitute numerical values and evaluate W AB:

W AB

= (5.00 atm )(19.69 L ) = 98.45 atm ⋅ L ×

101.325 J atm ⋅ L

= 9.9754 ×103 J = 9.98 kJ

The Second Law of Thermodynamics Apply the first law of thermodynamics to express the work done on the gas during the adiabatic expansion BC:

W BC

1933

= Δ E int, BC − Qin, BC = Δ E int, BC − 0 = Δ E int, BC = −ncV ΔT BC = − 52 nRΔT BC

Substitute numerical values and evaluate W BC: W BC

J  ⎞ = − 52 (2.00 mol)⎛  ⎜ 8.314 ⎟ (600 K − 1200 K ) = 2.494 ×10 4 J mol ⋅ K  ⎠ ⎝  = 24.9 kJ

The work done by the gas during the isothermal compression CA is: W CA

⎛ V   ⎞ ⎛  19.69 L  ⎞ J  ⎞ ⎟⎟ = nRT C ln⎜⎜ A ⎟⎟ = (2.00 mol)⎛  ⎜ 8.314 ⎟ (600 K )ln⎜⎜ mol K  222.77 L V  ⋅ ⎝   ⎠ ⎝   ⎠ ⎝  C  ⎠ = −24.20 kJ = − 24.2 kJ

( f ) The heat absorbed during the constant-pressure expansion AB is: QAB

J  ⎞ = ncP ΔT A − B = 72 nRΔT A − B = 72 (2.00 mol)⎛  ⎜ 8.314 ⎟ (1200 K − 600 K ) mol K  ⋅ ⎝   ⎠ = 34.92 kJ = 34.9 kJ

The heat absorbed during the adiabatic expansion BC is:

QBC

= 0

Use the first law of thermodynamics to find the heat absorbed during the isothermal compression CA:

QCA

= W CA + Δ E int, CA = W CA

= − 24.2 kJ  because Δ E int, CA = 0 for an isothermal  process.

79 •• [SSM] In a heat engine, 2.00 mol of a diatomic gas are carried through the cycle ABCDA shown in Figure 19-21. (The PV diagram is not drawn to scale.) The segment AB represents an isothermal expansion, the segment BC an adiabatic expansion. The pressure and temperature at A are 5.00 atm and 600 K. The volume at B is twice the volume at A. The pressure at D is 1.00 atm. (a) What is the pressure at B? ( b) What is the temperature at C? ( c) Find the total work done by the gas in one cycle.

1934

Chapter 19

Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points B, C, and D. We can then find the work done by the gas and the efficiency of the cycle by using the expressions for  the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle.

(a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB to find the  pressure at B:

PB

= PA

V A V B

= (5.00 atm )

= 2.50 atm ×

V A

2V A

101.325 kPa 1atm

= 253.3 kPa

= 253 kPa (b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC to express the temperature at C: Use the ideal-gas law to find the volume of the gas at B:

T C

= T B

V B

=

PCV C

(1)

PBV B

nRT B PB

J  ⎞ (2.00 mol)⎛  ⎜ 8.314 ⎟ (600 K ) mol ⋅ K  ⎠ ⎝  = 253.3 kPa

= 39.39 L Use the equation of state for an adiabatic process and γ  = 1.4 to find the volume occupied by the gas at C: Substitute numerical values in equation (1) and evaluate T C:

1 γ  

V C

T C

1 1.4

⎛ P  ⎞ ⎛ 2.50 atm ⎞ ⎟⎟ = V B ⎜⎜ B ⎟⎟ = (39.39 L )⎜⎜ 1.00 atm P ⎝   ⎠ ⎝  C  ⎠ = 75.78 L = (600 K )

(1.00 atm )(75.78 L ) (2.50 atm )(39.39 L )

= 462 K  (c) The work done by the gas in one cycle is given by:

W  = W AB + W BC

+ W CD + W DA

The Second Law of Thermodynamics

1935

The work done during the isothermal expansion AB is:

⎛ V   ⎞ ⎛ 2V  ⎞ J  ⎞ = nRT A ln⎜⎜ B ⎟⎟ = (2.00 mol)⎛  ⎜ 8.314 ⎟ (600 K )ln⎜⎜ A ⎟⎟ = 6.915 kJ mol ⋅ K  ⎠ ⎝  ⎝ V A  ⎠ ⎝  VA  ⎠

W AB

The work done during the adiabatic expansion BC is: W BC

J  ⎞ = −C V ΔT BC = − 52 nRΔT BC = − 52 (2.00 mol)⎛  ⎜ 8.314 ⎟ (462 K − 600 K ) mol ⋅ K  ⎠ ⎝  = 5.737 kJ

The work done during the isobaric compression CD is: W CD

= PC (V D − V C ) = (1.00 atm )(19.7 L − 75.78 L ) = −56.09 atm ⋅ L ×

101.325 J atm ⋅ L

= −5.680 kJ =0

Express and evaluate the work  done during the constant-volume  process DA:

W DA

Substitute numerical values and evaluate W :

W  = 6.915 kJ + 5.737 kJ − 5.680 kJ + 0

= 6.972 kJ = 6.97 kJ

In a heat engine, 2.00 mol of a monatomic gas are taken through the 80 •• cycle ABCA as shown in Figure 19-20. (The PV diagram is not drawn to scale.) At A the pressure and temperature are 5.00 atm and 600 K. The volume at B is twice the volume at A. The segment BC is an adiabatic expansion and the segment CA is an isothermal compression. (a) What is the volume of the gas at A? (b) What are the volume and temperature of the gas at B? (c) What is the temperature of the gas at C? (d ) What is the volume of the gas at C? (e) How much work is done by the gas in each of the three segments of the cycle?  (f ) How much heat is absorbed by the gas in each segment of the cycle? Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points A, B, and C and then find the work  done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the isobaric, adiabatic, and isothermal processes of the cycle.

1936

Chapter 19

(a) Apply the ideal-gas law to find the volume of the gas at A:

V A

nRT A

=

PA

J  ⎞ (2.00 mol)⎛  ⎜ 8.314 ⎟ (600 K ) mol ⋅ K  ⎠ ⎝  = 101.325 kPa 5.00 atm × atm

= 19.69 L = 19.7 L (b) We’re given that:

V B

= 2V A = 2(19.69 L ) = 39.39 L = 39.4 L

Apply the ideal-gas law to this isobaric process to find the temperature at B:

T B

= T A

V B V A

= (600 K )

2V A V A

= 1200 K  = T A = 600 K 

(c) Because the process CA is isothermal:

T C

(d ) Apply an equation for an adiabatic process (γ  = 5/3) to express the volume of the gas at C:

⎛ T B ⎞ − −1 −1 T BV B = T CV C ⇒ V C = V B ⎜⎜ ⎟⎟ ⎝ T C  ⎠

Substitute numerical values and evaluate V C:

1

γ  

γ  

V C

1

γ  

⎛ 1200 K  ⎞ ⎟⎟ = (39.39 L )⎜⎜ 600 K  ⎝   ⎠ = 111.4 L = 111 L

3 2

(e) The work done by the gas during the isobaric process AB is given by:

W AB

= PA (V B − V A ) = PA (2V A − V A ) = PAV A

Substitute numerical values and evaluate W AB:

W AB

= (5.00 atm )(19.69 L ) = 98.45 atm ⋅ L ×

101.325 J atm ⋅ L

= 9.975 kJ = 9.98 kJ

The Second Law of Thermodynamics Apply the first law of thermodynamics to express the work done by the gas during the adiabatic expansion BC:

W BC

1937

= Δ E int, BC − Qin, BC = Δ E int, BC − 0 = Δ E int, BC = −(ncV ΔT BC ) = − 32 nRΔT BC

Substitute numerical values and evaluate W BC: W BC

J  ⎞ = − 32 (2.00 mol)⎛  ⎜ 8.314 ⎟ (600 K − 1200 K ) = 14.97 kJ ⋅ mol K  ⎝   ⎠ = 15.0 kJ

The work done by the gas during the isothermal compression CA is:

W CA

⎛ V   ⎞ ⎛ 19.69 L ⎞ J  ⎞ ⎟⎟ = nRT C ln⎜⎜ A ⎟⎟ = (2.00 mol)⎛  ⎜ 8.314 ⎟ (600 K )ln⎜⎜ mol K  1 11 . 4 L V  ⋅ ⎝   ⎠ ⎝   ⎠ ⎝  C  ⎠ = −17.29 kJ − 17.3 kJ

( f ) The heat absorbed during the isobaric expansion AB is: Qin, AB

J  ⎞ = ncP ΔT AB = 52 nRΔT AB = 52 (2.00 mol)⎛  ⎜ 8.314 ⎟ (1200 K − 600 K ) mol ⋅ K  ⎠ ⎝  = 24.9 kJ

Express and evaluate the heat absorbed during the adiabatic expansion BC:

QBC

= 0

Use the first law of thermodynamics to express and evaluate the heat absorbed during the isothermal compression CA:

QCA

= W CA + Δ E int, CA = W CA = − 17.3 kJ

 because Δ E int = 0 for an isothermal  process.

In a heat engine, 2.00 mol of a monatomic gas are carried through the 81 •• cycle ABCDA shown in Figure 19-21. (The PV diagram is not drawn to scale.) The segment AB represents an isothermal expansion, the segment BC an adiabatic expansion. The pressure and temperature at A are 5.00 atm and 600 K. The volume at B is twice the volume at A. The pressure at D is 1.00 atm.

1938

Chapter 19

(a) What is the pressure at B? ( b) What is the temperature at C? ( c) Find the total work done by the gas in one cycle. Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points B, C, and D and then find the work  done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle.

(a) Apply the ideal-gas law for a fixed amount of gas to the isothermal  process AB:

PB

= PA

V A V B

= (5.00 atm )

= 2.50 atm ×

V A

2V A

101.325 kPa 1atm

= 253.3 kPa = 253 kPa (b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic  process BC: Use the ideal-gas law to find the volume at B:

T C

= T B

V B

=

PCV C

(1)

PBV B

nRT B PB

J  ⎞ (2.00 mol)⎛  ⎜ 8.314 ⎟ (600 K ) ⋅ mol K  ⎝   ⎠ = 253.3 kPa

= 39.39 L Use the equation of state for an adiabatic process and γ  = 5/3 to find the volume occupied by the gas at C: Substitute numerical values in equation (1) and evaluate T C:

(c) The work done by the gas in one cycle is given by:

1 γ  

V C

T C

⎛ P  ⎞ ⎛ 2.50 atm ⎞ ⎟⎟ = V B ⎜⎜ B ⎟⎟ = (39.39 L )⎜⎜ 1.00 atm P ⎝   ⎠ ⎝  C  ⎠ = 68.26 L

(1.00 atm )(68.26 L ) (2.50 atm )(39.39 L ) = 415.9 K = 416 K  = (600 K )

W  = W AB + W BC + W CD

+ W DA

(2)

35

The Second Law of Thermodynamics

1939

The work done during the isothermal expansion AB is:

W AB

⎛ V   ⎞ ⎛ 2V   ⎞ J  ⎞ = nRT A ln⎜⎜ B ⎟⎟ = (2.00 mol)⎛  ⎜ 8.314 ⎟ (600 K )ln⎜⎜ A ⎟⎟ = 6.915 kJ mol ⋅ K  ⎠ ⎝  ⎝ V A  ⎠ ⎝  V A  ⎠

The work done during the adiabatic expansion BC is: W BC

= −C V ΔT BC = − 32 nRΔT BC J  ⎞ = − 32 (2.00 mol)⎛  ⎜ 8.314 ⎟ (415.9 K − 600 K ) mol ⋅ K  ⎠ ⎝  = 4.592 kJ

The work done during the isobaric compression CD is: W CD

= PC (V D − V C ) = (1.00 atm )(19.7 L − 68.26 L ) = −48.56 atm ⋅ L ×

101.325 J atm ⋅ L

= −4.920 kJ =0

The work done during the constantvolume process DA is:

W DA

Substitute numerical values in equation (2) to obtain:

W  = 6.915 kJ + 4.592 kJ − 4.920 kJ + 0

= 6.59 kJ

Compare the efficiency of the Otto cycle to the efficiency of the 82 •• Carnot cycle operating between the same maximum and minimum temperatures. (The Otto cycle is discussed in Section 19-1.) Picture the Problem We can express the efficiency of the Otto cycle using the result from Example 19-2. We can apply the relation TV γ  −1 = constant to the adiabatic processes of the Otto cycle to relate the end-point temperatures to the volumes occupied by the gas at these points and eliminate the temperatures at c and d . We can use the ideal-gas law to find the highest temperature of the gas during its cycle and use this temperature to express the efficiency of a Carnot engine. Finally, we can compare the efficiencies by examining their ratio.

1940

Chapter 19

The efficiency of the Otto engine is given in Example 19-2:

ε Otto

= 1−

T d  − T a

(1)

T c − T b

where the subscripts refer to the various  points of the cycle as shown in Figure 19-3. −1

Apply the relation TV γ  −1 = constant to the adiabatic process a→b to obtain:

⎛ V   ⎞ T b = T a ⎜⎜ a ⎟⎟ ⎝ V b  ⎠

Apply the relation TV γ  −1 = constant to the adiabatic process c→d to obtain:

⎛ V   ⎞ T c = T d ⎜⎜ d  ⎟⎟ ⎝ V c  ⎠

Subtract the first of these equations from the second to obtain:

⎛ V   ⎞ ⎛ V   ⎞ T c − T b = T d ⎜⎜ d  ⎟⎟ − T a ⎜⎜ a ⎟⎟ ⎝ V c  ⎠ ⎝ V b  ⎠

In the Otto cycle, V a = V d and V c = V b. Substitute to obtain:

⎛ V   ⎞ ⎛ V   ⎞ T c − T b = T d ⎜⎜ a ⎟⎟ − T a ⎜⎜ a ⎟⎟ ⎝ V b  ⎠ ⎝ V b  ⎠ −1 ⎛ V a  ⎞ = (T d  − T a )⎜⎜ ⎟⎟ ⎝ V b  ⎠

γ  

−1

γ  

−1

γ  

−1

γ  

γ  

γ  

−1

−1

γ  

Substitute in equation (1) and simplify to obtain:

ε Otto

= 1−

T d  − T a −1

γ  

⎛ V   ⎞ (T d  − T a )⎜⎜ a ⎟⎟ ⎝ V b  ⎠ −1 ⎛ V b  ⎞ T  = 1 − ⎜⎜ ⎟⎟ = 1 − a T b ⎝ V a  ⎠ γ  

 Note that, while T a is the lowest temperature of the cycle, T b is not the highest temperature. Apply the ideal-gas law to c and b to obtain an expression for the cycle’s highest temperature T c:

Pc T c

=

Pb T b

⇒ T c = T b

Pc Pb

> T b

The Second Law of Thermodynamics

The efficiency of a Carnot engine operating between the maximum and minimum temperatures of the Otto cycle is given by: Express the ratio of the efficiency of  a Carnot engine to the efficiency of  an Otto engine operating between the same temperatures:

ε Carnot

= 1−

T a T c

1− ε Carnot

=

ε Otto

1−

Hence,

1941

T a T c T a

> 1 because T c > T b.

T b

>

Carnot

Otto

83 ••• [SSM] A common practical cycle, often used in refrigeration, is the  Brayton cycle, which involves (1) an adiabatic compression, (2) an isobaric (constant pressure) expansion,(3) an adiabatic expansion, and (4) an isobaric compression back to the original state. Assume the system begins the adiabatic compression at temperature T 1, and transitions to temperatures T 2, T 3 and T 4 after  each leg of the cycle. (a) Sketch this cycle on a PV diagram. (b) Show that the T  − T 1 efficiency of the overall cycle is given by ε  = 1 − 4 . (c) Show that this T 3 − T 2

( (

) )

1− γ   γ   efficiency, can be written as ε  = 1 − r ( ) , where r is the pressure ratio Phigh/Plow of the maximum and minimum pressures in the cycle.

Picture the Problem The efficiency of the cycle is the ratio of the work done to the heat that flows into the engine. Because the adiabatic transitions in the cycle do not have heat flow associated with them, all we must do is consider the heat flow in and out of the engine during the isobaric transitions.

(a) The  Brayton heat engine cycle is shown to the right. The paths 1→2 and 3→4 are adiabatic. Heat Qh enters the gas during the isobaric transition from state 2 to state 3 and heat Qc leaves the gas during the isobaric transition from state 4 to state 1.

Qh

P

2

⇓

3

1

⇓

4

Qc

(b) The efficiency of a heat engine is given by:

ε 

=

W  Qin

=

Qh

− Qc

Qin

V 

(1)

1942

Chapter 19

During the constant-pressure expansion from state 1 to state 2 heat enters the system:

Q23

= Qh = nC P ΔT  = nC P (T 3 − T 2 )

During the constant-pressure compression from state 3 to state 4 heat enters the system:

Q41

= −Qc = − nC P ΔT  = − nC P (T 1 − T 4 )

Substituting in equation (1) and simplifying yields:

ε 

=

nC P (T 3 − T 2 ) − (− nC P (T 1 − T 4 )) nC P (T 3 − T 2 )

(T 3 − T 2 ) + (T 1 − T 4 ) (T 3 − T 2 ) (T  − T  ) = 1− 4 1 (T 3 − T 2 ) =

(c) Given that, for an adiabatic transition, TV  −1 = constant , use the ideal-gas law to eliminate V and obtain: Let the pressure for the transition from state 1 to state 2 be Plow and the  pressure for the transition from state 3 to state 4 be Phigh. Then for the adiabatic transition from state 1 to state 2: Similarly, for the adiabatic transition from state 3 to state 4:

Subtract T 1 from T 4 and simplify to obtain:

γ  

T  P

−1

γ  

= constant

γ  

T 1

Plow−1 γ  

T 4

⎛  Plow  ⎞ ⎜ ⎟ = ⇒ T  1 −1 ⎜ ⎟ Phigh ⎝ Phigh  ⎠ γ  

=

T 2

γ  

⎛  P  ⎞ = ⎜⎜ low ⎟⎟ ⎝ Phigh  ⎠

−1

γ  

γ  

T 2

−1

γ  

γ  

T 3

⎛  P  ⎞ T 4 − T 1 = ⎜ low ⎟ ⎜ Phigh ⎟ ⎝   ⎠ ⎛  P  ⎞ = ⎜⎜ low ⎟⎟ ⎝ Phigh  ⎠

−1

γ  

γ  

⎛  P  ⎞ T 3 − ⎜ low ⎟ ⎜ Phigh ⎟ ⎝   ⎠

−1

γ  

γ  

(T 3 − T 2 )

−1

γ  

γ  

T 2

The Second Law of Thermodynamics Dividing both sides of the equation  by T 3 − T 2 yields:

Substitute in the result of Part ( b) and simplify to obtain:

− T 1 T 3 − T 2 T 4

⎛  P  ⎞ = ⎜⎜ low ⎟⎟ ⎝ Phigh  ⎠

⎛  Plow  ⎞ ⎟ ε  = 1 − ⎜ ⎜ Phigh ⎟ ⎝   ⎠

−1

γ  

γ  

−1

γ  

γ  

1943

⎛ P  ⎞ = 1 − ⎜⎜ high ⎟⎟ ⎝  Plow  ⎠

1− γ   γ  

1− γ  

= 1 − (r ) where r  =

γ  

Phigh Plow

84 ••• Suppose the Brayton cycle engine (see Problem 83) is run in reverse as a refrigerator in your kitchen. In this case, the cycle begins at temperature T 1 and expands at constant pressure until its temperature T 4. Then the gas is adiabatically compressed, until its temperature is T 3. And then it is compressed at constant  pressure until its temperature T 2. Finally, it adiabatically expands until it returns to its initial state at temperature T 1. (a) Sketch this cycle on a PV diagram. (b) T 4 − T 1 Show that the coefficient of performance isCOPB = . T 3 − T 2 − T 4 + T 1

(

(

)

)

(c) Suppose your ″Brayton cycle refrigerator ″ is run as follows. The cylinder  containing the refrigerant (a monatomic gas) has an initial volume and pressure of  60 mL and 1.0 atm. After the expansion at constant pressure, the volume and temperature are 75 mL and –25°C. The pressure ratio r = Phigh/Plow for the cycle is 5.0. What is the coefficient of performance for your refrigerator? (d ) To absorb heat from the food compartment at the rate of 120 W, what is the rate at which electrical energy must be supplied to the motor of this refrigerator? ( e) Assuming the refrigerator motor is actually running for only 4.0 h each day, how much does it add to your monthly electric bill. Assume 15 cents per kWh of electric energy and thirty days in a month. Picture the Problem The efficiency of the Brayton refrigerator cycle is the ratio of the heat that enters the system to the work done to operate the refrigerator. Because the adiabatic transitions in the cycle do not have heat flow associated with them, all we must do is consider the heat flow in and out of the refrigerator  during the isobaric transitions.

1944

Chapter 19

(a) The Brayton refrigerator cycle is shown to the right. The paths 1→2 and 3→4 are adiabatic. Heat Qc enters the gas during the constant pressure transition from state 1 to state 4 and heat Qh leaves the gas during the constant-pressure transition from state 3 to state 2.

Qh

P

2

⇓

3

1

⇓

4

Qc

(b) The coefficient of performance of  the Brayton cycle refrigerator is given by:

COPB

=

Qh W 

=

Qh Qh

− Qc

V 

(1)

During the constant-pressure compression from state 3 to state 2 heat leaves the system:

Q32

= −Qh = −nC P ΔT  = −nC P (T 2 − T 3 )

During the constant-pressure expansion from state 1 to state 4 heat enters the system:

Q14

= Qc = nC P ΔT  = nC P (T 4 − T 1 )

Substituting in equation (1) and simplifying yields:

COPB

− T 1 ) − nC P (T 4 − T 1 ) − nC P (T 2 − T 3 ) (T 4 − T 1 ) = − (T 4 − T 1 ) − (T 2 − T 3 ) T 4 − T 1 = T 3 − T 2 − T 4 + T 1 =

nC P (T 4

(c) The COPB requires the temperatures corresponding to states 1, 2, 3, and 4. We’re given that the temperature in state 4 is:

T 4

= −25°C + 273 K = 248 K 

For the constant-pressure transition from state 1 to state 4, the quotient T /V is constant:

T 1 V 1

=

Substitute numerical values and evaluate T 1:

T 1 = ⎜

T 4 V 4

⎛ V  ⎞ ⇒ T 1 = ⎜⎜ 1 ⎟⎟T 4 ⎝ V 4  ⎠

⎛ 60 mL ⎞ (248 K ) = 198 K  ⎟ ⎝ 75 mL ⎠

The Second Law of Thermodynamics Given that, for an adiabatic transition, TV  −1 = constant , use the ideal-gas law to eliminate V and obtain: For the adiabatic transition from state 4 to state 3:

Substitute numerical values and evaluate T 3: Similarly, for the adiabatic transition from state 2 to state 1:

Substitute numerical values in the expression derived in Part (a) and evaluate COPB:

(d ) From the definition of COPB:

The rate at which energy must be supplied to this refrigerator is given  by:

γ  

T  P

−1

γ  

= constant

γ  

T 3

−1

γ  

P3

⎛ P3  ⎞ ⎜⎜ ⎟⎟ T  ⇒ = 3 − P4 1 ⎝ P4  ⎠ γ  

=

T 4

−1

γ  

γ  

γ  

T 4

1.67 −1

T 3 = (5) 1.67 (248 K ) = 473 K  −1

γ  

⎛ P  ⎞ = ⎜⎜ 2 ⎟⎟ ⎝  P1  ⎠ = 378 K 

γ  

T 2

COPB

=

1.67 −1

T 1 = (5) 1.67 (198 K )

248 K − 198 K  473 K − 378 K − 248 K + 198 K 

= 1.1 W  =

dW  dt 

Qc

COPB

=

1

dQc

COPB dt 

or, if the frequency of the AC power  input is f,  fQc dW 

=

dt  Qc

Substituting for Qc yields:

dW  dt  n=

COPB

= nC P ΔT  = nC P (T 4 − T 1 )

Express the heat Qc that is drawn from the cold reservoir:

Use the ideal-gas law to express the number of moles of the gas:

1945

=

 fnC P (T 4

P4V 4  RT 4

− T 1 )

COPB

1946

Chapter 19

Because the gas is monatomic, C P = 52 R . Substitute for n and C P to obtain:

dW  dt 

5 2

 f 

= =

P4V 4  RT 4

 R(T 4

− T 1 )

COPB 5 2

 fP4V 4 (T 4

− T 1 )

(COPB )T 4

Substitute numerical values and evaluate dW /dt :

⎛  10 −3 m 3  ⎞ ⎟⎟(248 K − 198 K ) (60 s )(101.325 kPa )⎜⎜ 75 mL × L dW  ⎝   ⎠ = = 207 W dt  (1.11)(248 K ) −1

5 2

= 0.21 kW (e) The monthly cost of operation is given by Monthly Cost

= Cost Per Unit of  Power × Power Consumption = rate × daily consumption × number of  days per month

Substitute numerical values and evaluate the monthly cost of operation: Monthly Cost 85

•••

Using

=

$0.15 4.0 h × 207 W × × 30 d ≈ $4 kWh d

ΔS = C v ln (T2

T1 ) − nR ln (V2 V1 )  (Equation 19-16) for the

entropy change of an ideal gas, show explicitly that the entropy change is zero for  a quasi-static adiabatic expansion from state (V 1, T 1) to state (V 2, T 2). Picture the Problem We can use nR = C P

− C V ,

γ  

= C P

C V , and

TV γ  −1 = a constant to show that the entropy change for a quasi-static adiabatic expansion that proceeds from state (V 1,T 1) to state (V 2,T 2) is zero.

Express the entropy change for a general process that proceeds from state 1 to state 2:

⎛ T   ⎞ ⎛ V   ⎞ ΔS  = C V ln⎜⎜ 2 ⎟⎟ + nR ln⎜⎜ 2 ⎟⎟ ⎝ T 1  ⎠ ⎝ V 1  ⎠

For an adiabatic process:

T 2 T 1

⎛ V   ⎞ = ⎜⎜ 1 ⎟⎟ ⎝ V 2  ⎠

−1

γ  

The Second Law of Thermodynamics

Substitute for 

T 2 T 1

1947

and simplify to obtain:

⎡ ⎛ V  ⎞ ⎢ C V ln⎜⎜ 1 ⎟⎟ −1 ⎛ V   ⎞ ⎛ V   ⎞ ⎝ V 2  ⎠ + nR ln⎜⎜ 2 ⎟⎟ = ln⎜⎜ 2 ⎟⎟ ⎢⎢nR + V  ⎝ V 1  ⎠ ⎝ V 1  ⎠ ⎢ ln 2 V 1 ⎢⎣

−1

γ  

⎛ V  ⎞ ΔS  = C V ln⎜⎜ 1 ⎟⎟ ⎝ V 2 ⎠

γ  

⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦

⎡ ⎛ V 1  ⎞ ⎤ ⎜⎜ ⎟⎟ ⎥ 1 ln γ   C  ( ) − V ⎛ V 2 ⎞ ⎢⎢  ⎞ ⎝ V 2 ⎠ ⎥ = ln⎛  ⎜⎜ V 2 ⎟⎟ [nR − (γ   − 1)C V ] = ln⎜⎜ ⎟⎟ nR + V  ⎥ ⎝ V 1  ⎠ ⎢⎢ ⎝ V 1  ⎠ − ln 1 ⎥ V 2 ⎢⎣ ⎥⎦ Use the relationship between C P and C V to obtain:

nR = C P

− C V

Substituting for nR and γ  and simplifying yields:

⎛ C   ⎞ ⎤ ⎛ V   ⎞ ⎡ ΔS  = ln⎜⎜ 2 ⎟⎟ ⎢C P − C V − ⎜⎜  p − 1⎟⎟C V ⎥ ⎝ V 1  ⎠ ⎣ ⎝ C V  ⎠ ⎦ = 0

86 ••• (a) Show that if the refrigerator statement of the second law of  thermodynamics were not true, then the entropy of the universe could decrease. (b) Show that if the heat-engine statement of the second law were not true, then the entropy of the universe could decrease. (c) A third statement of the second law is that the entropy of the universe cannot decrease. Have you just proved that this statement is equivalent to the refrigerator and heat-engine statements? Picture the Problem (a) Suppose the refrigerator statement of the second law is violated in the sense that heat Qc is taken from the cold reservoir and an equal mount of heat is transferred to the hot reservoir and W  = 0. The entropy change of the universe is then ΔS u = Qc/T h − Qc/T c. Because T h > T c, S u < 0, i.e., the entropy of the universe would decrease.

(b) In this case, is heat Qh is taken from the hot reservoir and no heat is rejected to the cold reservoir; that is, Qc = 0, then the entropy change of the universe is ΔS u = −Qh/T h + 0, which is negative. Again, the entropy of the universe would decrease.

1948

Chapter 19

(c) The heat-engine and refrigerator statements of the second law only state that some heat must be rejected to a cold reservoir and some work must be done to transfer heat from the cold to the hot reservoir, but these statements do not specify the minimum amount of heat rejected or work that must be done. The statement ΔS u ≥ 0 is more restrictive. The heat-engine and refrigerator statements in conjunction with the Carnot efficiency are equivalent to ΔS u ≥ 0. 87 ••• Suppose that two heat engines are connected in series, such that the heat released by the first engine is used as the heat absorbed by the second engine as shown in Figure 19-22. The efficiencies of the engines are ε 1 and ε 2, respectively. Show that the net efficiency of the combination is given by ε net = ε 1 + ε 2 − ε 1ε 2 . Picture the Problem We can express the net efficiency of the two engines in terms of W 1, W 2, and Qh and then use ε 1 = W 1/Qh and ε 2 = W 2/Qm to eliminate W 1, W 2, Qh, and Qm.

Express the net efficiency of the two heat engines connected in series:

ε net

Express the efficiencies of engines 1 and 2:

ε 1

Solve for W 1 and W 2 and substitute to obtain:

ε net

Express the efficiency of engine 1 in terms of Qm and Qh:

ε 1

Substitute for Qm/Qh and simplify to obtain:

ε net

=

=

W 1 + W 2 Qh

W 1 Qh

=

and ε 2

ε 1Qh

= 1−

=

W 2 Qm

+ ε 2Qm Qh

Qm Qh

⇒

Qm Qh

= ε 1 +

Qm Qh

ε 2

= 1 − ε 1

= ε 1 + (1 − ε 1 )ε 2 =

ε 1

+ ε 2 − ε 1ε 2

88 ••• Suppose that two heat engines are connected in series, such that the heat released by the first engine is used as the heat absorbed by the second engine, as shown in Figure 19-22. Suppose that each engine is an ideal reversible heat engine. Engine 1 operates between temperatures T h and Tm and Engine 2 operates between T m and T c, where T h > T m > T c. Show that the net efficiency of  T  the combination is given by ε net = 1 − c . (Note that this result means that two T h

The Second Law of Thermodynamics

1949

reversible heat engines operating ″in series″ are equivalent to one reversible heat engine operating between the hottest and coldest reservoirs.) Picture the Problem We can express the net efficiency of the two engines in terms of W 1, W 2, and Qh and then use ε 1 = W 1/Qh and ε 2 = W 2/Qm to eliminate W 1, W 2, Qh, and Qm. Finally, we can substitute the expressions for the efficiencies of  the ideal reversible engines to obtain ε net = 1 − T c T h .

Express the efficiencies of ideal reversible engines 1 and 2:

ε 1

= 1−

T m

(1)

T h

and ε 2

= 1−

The net efficiency of the two engines connected in series is given by:

ε net

Express the efficiencies of engines 1 and 2:

ε 1

Solve for W 1 and W 2 and substitute in equation (3) to obtain:

ε net

Express the efficiency of engine 1 in terms of Qm and Qh:

ε 1

Substitute for 

Qm Qh

to obtain:

Substitute for ε 1 and ε 2 and simplify to obtain:

=

=

(2)

T m

W 1 + W 2

(3)

Qh

W 1 Qh

=

T c

and

ε 1Qh

= 1−

ε 2

=

W 2 Qm

+ ε 2Qm

= ε 1 +

Qh Qm Qh

⇒

Qm Qh

= ε 1 + (1 − ε 1 )ε 2

ε net

= 1−

T m

= 1−

T m T h

Qh

ε 2

= 1 − ε 1

ε net

T h

Qm

⎛ T   ⎞ ⎛  T   ⎞ + ⎜⎜ m ⎟⎟ ⎜⎜1 − c ⎟⎟ ⎝ T h  ⎠ ⎝  T m  ⎠ +

T m T h

−

T c T h

= 1−

T c T h

89 ••• [SSM] The English mathematician and philosopher Bertrand Russell (1872-1970) once said that if a million monkeys were given a million typewriters and typed away at random for a million years, they would produce all of  Shakespeare’s works. Let us limit ourselves to the following fragment of  Shakespeare ( Julius Caesar III:ii):