ch19.pdf

CHAPTER

19

Heat and the First Law of Thermodynamics

1* · Body A has twice the mass and twice the specific heat of body B. If they are supplied with equal amounts of heat, how do the subsequent changes in their temperatures compare? M A = 2M B; cA = 2cB; C = Mc; ∆T = Q/C CA = 4CB; ∆TA = ∆TB/4 2

· The temperature change of two blocks of masses M A and M B is the same when they absorb equal amounts of heat. It follows that the specific heats are related by (a) cA= (MA / MB)cB . (b) cA = (MB / MA)cB . (c) cA = cB . (d) none of the above. (b)

3

· The specific heat of aluminum is more than twice that of copper. Identical masses of copper and aluminum, both at 20°C, are dropped into a calorimeter containing water at 40°C. When thermal equilibrium is reached, (a) the aluminum is at a higher temperature than the copper. (b) the aluminum has absorbed less energy than the copper. (c) the aluminum has absorbed more energy than the copper. (d) both (a) and (c) are correct statements. (c)

4

· Sam the shepherd's partner, Bernard, who is a working dog, consumes 2500 kcal of food each day. (a) How many joules of energy does Bernard consume each day? (b) Sam and Bernard often find themselves sleeping out in the cold night. If the energy consumed by Bernard is dissipated as heat at a steady rate over 24 h, what is his power output in watts as a heater for Sam? (a) E = Q E = (4.184 × 2.5 × 106) J = 10.46 MJ (b) P = E/t P = (10.46 × 106/8.64 × 104) W = 121 W

5* · A solar home contains 105 kg of concrete (specific heat = 1.00 kJ/kg⋅K). How much heat is given off by the concrete when it cools from 25 to 20°C?

Chapter 19 Q = C∆T = mc∆T

Heat and the First Law of Thermodynamics Q = (105 × 103 × 5) J = 500 MJ

6

· How many calories must be supplied to 60 g of ice at -10°C to melt it and raise the temperature of the water to 40°C? Q = Cice∆Tice + mLf + Cwater∆Twater Q = 60(10 × 0.49 + 79.7 + 40) cal = 7.48 kcal

7

·· How much heat must be removed when 100 g of steam at 150°C is cooled and frozen into 100 g of ice at 0°C? (Take the specific heat of steam to be 2.01 kJ/kg⋅K.) Q = m(csteam ∆Tsteam + Lv + cw∆Tw + Lf) Q = 100(0.48 × 50 + 540 + 100 + 79.7) cal = 74.4 kcal

8

·· A 50-g piece of aluminum at 20°C is cooled to −196°C by placing it in a large container of liquid nitrogen at that temperature. How much nitrogen is vaporized? (Assume that the specific heat of aluminum is constant and is equal to 0.90 kJ/kg⋅K.) mNLvN = mAlcAl∆TAl; solve for mN mN = (50 × 0.9 × 216/199) g = 48.8 g If 500 g of molten lead at 327°C is poured into a cavity in a large block of ice at 0°C, how much of the ice melts? mw = [500(24.7 + 0.128 × 327)/333.5] g = 99.8 g − mPb(Lf,Pb + cpb∆T) + mwLf,w = 0; solve for mw

9* ··

10 ·· A 30-g lead bullet initially at 20°C comes to rest in the block of a ballistic pendulum. Assume that half the initial kinetic energy of the bullet is converted into thermal energy within the bullet. If the speed of the bullet was 420 m/s, what is the temperature of the bullet immediately after coming to rest in the block? 2 Q = 1/2(1/2mv2) = mcPb∆T; ∆T = v2/4c Pb; cPb in J/kg ∆T = (420 /4 × 128) K = 344.5 K 11 ·· A 1400-kg car traveling at 80 km/h is brought to rest by applying the brakes. If the specific heat of steel is 0.11 cal/g⋅K, what total mass of steel must be contained in the steel brake drums if the temperature of the brake drums is not to rise by more than 120°C? 1. Find Q = 1/2mv2 Q = 1/2 × 1400 × 22.22 J = 345.7 kJ = 82.6 kcal 2. M = Q/c∆T M = (82.6/0.11 × 120) kg = 6.26 kg 12 · A 200-g piece of lead is heated to 90°C and is then dropped into a calorimeter containing 500 g of water that is initially at 20°C. Neglecting the heat capacity of the container, find the final temperature of the lead and water. In the solution of Problems 12 to 23 (Calorimetry) the fundamental relationship Qout = Qin is used. mPbcPb(tPb − tf) = mwcw(tf − tw); solve for tf tf(500 + 200 × 0.128)=(500 × 20 + 200 × 0.128 × 90); tf = 23.4 oC 13* · The specific heat of a certain metal can be determined by measuring the temperature change that occurs when a piece of the metal is heated and then placed in an insulated container made of the same material and containing water. Suppose a piece of metal has a mass of 100 g and is initially at 100°C. The container has a mass of 200 g and contains 500 g of water at an initial temperature of 20.0°C. The final temperature is 21.4°C. What is the specific heat of the metal? m1c(t1i− tf) = m2c(tf − t2i) + mwcw(tf − t2i); find c 78.6c = 2.8c + 7; c = 0.093 cal/g.K

Chapter 19

Heat and the First Law of Thermodynamics

14 ·· A 25-g glass tumbler contains 200 mL of water at 24°C. If two 15-g ice cubes each at a temperature of −3°C are dropped into the tumbler, what is the final temperature of the drink? Neglect thermal conduction between the tumbler and the room. (mgcg + 200)(24 − tf) = 3micecice + mice (Lf + tf); find tf 205 × 24 − 205tf = 44.1 + 30(79.7 + tf); tf = 10.6 oC 15 ·· A 200-g piece of ice at 0°C is placed in 500 g of water at 20°C. The system is in a container of negligible heat capacity and is insulated from its surroundings. (a) What is the final equilibrium temperature of the system? (b) How much of the ice melts? (a) To melt the ice requires 200 × 79.7 cal ≅ 16 kcal; reducing the temperature of the water to 0oC releases 10 kcal. Therefore, not all the ice melts, and the final temperature is 0oC. (b) The mass of ice that melts is (10 kcal)/(79.7 cal/g) = 125 g. 16 ·· A 3.5-kg block of copper at a temperature of 80°C is dropped into a bucket containing a mixture of ice and water whose total mass is 1.2 kg. When thermal equilibrium is reached the temperature of the water is 8°C. How much ice was in the bucket before the copper block was placed in it? (Neglect the heat capacity of the bucket.) mCucCu∆tCu = mw∆tw + miceLf; solve for mice 3.5 × 0.0923 × 72 = 1.2 × 8 + 79.7mice; mice = 0.171 kg 17* ·· A well-insulated bucket contains 150 g of ice at 0°C. (a) If 20 g of steam at 100°C is injected into the bucket, what is the final equilibrium temperature of the system? (b) Is any ice left afterward? (a), (b) mstLv + mst(100 − tf) = mice(Lf + tf); tf = 4.97oC; (b) Since tf > 0oC, no ice is left. solve for tf 18 ·· A calorimeter of negligible mass contains 1 kg of water at 303 K and 50 g of ice at 273 K. Find the final temperature T. Solve the same problem if the mass of ice is 500 g. 1. miceLf + micetf = mw(30 − tf); solve for tf, mice = 50 g 50(79.7 + tf) = 1000(30 − tf); tf = 24.8oC 2. For mice = 500 g, only 376 g will melt. tf = 0oC 19 ·· A 200-g aluminum calorimeter contains 500 g of water at 20°C. A 100-g piece of ice cooled to −20°C is placed in the calorimeter. (a) Find the final temperature of the system, assuming no heat loss. (Assume that the specific heat of ice is 2.0 kJ/kg. ⋅K.) (b) A second 200-g piece of ice at −20°C is added. How much ice remains in the system after it reaches equilibrium? (c) Would you give a different answer for (b) if both pieces of ice were added at the same time? (a) (mAlcAl + mwcw)(20 − tf) = 20micecice + mice(Lf + tf) tf = 3.0oC (b) Find Q released to lower Al and H2O to 0oC Q = (200 × 0.215 + 600)3.0 = 1929 cal o Find Qice required to raise 200 g of ice to 0 C Qice = 200 × 0.478 × 20 = 1912 cal Find the amount of ice melted by 17 cal m = 17/79.7 g = 0.21 g; ice remaining = 199.8 g (c) The initial and final conditions are the same No 20 ·· The specific heat of a 100-g block of material is to be determined. The block is placed in a 25-g copper calorimeter that also holds 60 g of water. The system is initially at 20°C. Then 120 mL of water at 80°C are added to the calorimeter vessel. When thermal equilibrium is attained, the temperature of the water is 54°C. Determine the specific heat of the block.

Chapter 19

Heat and the First Law of Thermodynamics

(mBcB + mCucCu + mw1)∆tB = mw2∆t2; here ∆tB = 34 K, ∆t2 = 26 K, mw1 = 60 g, mw2 = 120 g. Solve for cB.

cB = 0.295 cal/g.K

21* ·· Between innings at his weekly softball game, Stan likes to have a sip or two of beer. He usually consumes about 6 cans, which he prefers at exactly 40°F. His wife Bernice puts a six-pack of 12-ounce aluminum cans of beer (1 ounce has a mass of 28.4 g) originally at 80°F in a well-insulated Styrofoam container and begins adding ice. How many 30g ice cubes must she add to the container so that the final temperature is 40°F? (Neglect heat losses through the container and the heat removed from the aluminum and assume that the beer is mostly water.) 1. Convert to Celsius degrees. 40 oF = 4.44oC; 80oF = 26.67oC 2. mBcB∆tB = n icemice(Lf + 4.44); solve for n ice n ice = (6 × 12 × 28.4 × 22.2/30 × 84.14) = 18 22 ·· A 100-g piece of copper is heated in a furnace to a temperature t. The copper is then inserted into a 150-g copper calorimeter containing 200 g of water. The initial temperature of the water and calorimeter is 16°C, and the final temperature after equilibrium is established is 38°C. When the calorimeter and its contents are weighed, 1.2 g of water are found to have evaporated. What was the temperature t? 1. 1.2Lv + (mw + mcalcCu)∆tcal = mCucCu∆tCu; find ∆tCu ∆tCu = [1.2 × 540 + (200 + 150 × 0.0923) × 22]/(100 × 0.0923) Co o o o 2. t = tf + ∆tCu ∆tCu = 578 C ; t = (578 + 38) C = 616 C 23 ·· A 200-g aluminum calorimeter contains 500 g of water at 20°C. Aluminum shot of mass 300 g is heated to 100°C and is then placed in the calorimeter. (a) Using the value of the specific heat of aluminum given in Table 19-1, find the final temperature of the system, assuming that no heat is lost to the surroundings. (b) The error due to heat transfer between the system and its surroundings can be minimized if the initial temperature of the water and calorimeter is chosen to be

1 2

∆ t w below room temperature, where ∆tw is the temperature change of the calorimeter and water

during the measurement. Then the final temperature is

1 2

∆ t w above room temperature. What should the initial

temperature of the water and container be if the room temperature is 20°C? (a) mshcAl(100 − tf) = (mcalcal + mw)(tf − 20); find tf 6450 − 64.5tf = 532.3tf − 10645; tf = 28.6oC (b) For the calorimeter, let ti = tr − t0 and tf = tr + t0, mshcAl(100 - 20 − t0) = (mcalcAl + mw)(2t0); o where tr = 20 C; write the calorimetry equation, 64.5 × 80 − 64.5t0 = 2 × 532.3t0; t0 = 4.57oC; solve for t0, ti. ti = 15.43oC. 24 · Joule's experiment establishing the mechanical equivalence of heat involved the conversion of mechanical energy into internal energy. Give some examples of the internal energy of a system being converted into mechanical energy. Steam turbine; internal combustion engine; a person performing mechanical work, e.g., climbing a hill. 25* · Can a system absorb heat with no change in its internal energy? Yes 26 · In the equation Q = ∆U + W (the formal statement of the first law of thermodynamics), the quantities Q and W represent (a) the heat supplied to the system and the work done by the system.

Chapter 19

Heat and the First Law of Thermodynamics

(b) the heat supplied to the system and the work done on the system. (c) the heat released by the system and the work done by the system. (d) the heat rele ased by the system and the work done on the system. (a) 27 · A diatomic gas does 300 J of work and also absorbs 600 cal of heat. What is the change in internal energy of the gas? From Equ. 19-10, ∆U = Q − W ∆U = (600 × 4.184 − 300) J = 2210 J 28 · If 400 kcal is added to a gas that expands and does 800 kJ of work, what is the change in the internal energy of the gas? From Equ. 19-10, ∆U = Q − W ∆U = (400 × 4.184 − 800) kJ = 874 kJ 29* · A lead bullet moving at 200 m/s is stopped in a block of wood. Assuming that all of the energy change goes into heating the bullet, find the final temperature of the bullet if its initial temperature is 20°C. Q = 1/2mv2 = mc∆t = mc(tf − ti); tf = ti + v2/2c tf = (20 + 2002/2 × 128)oC = 176oC 30 · (a) At Nia gara Falls, the water drops 50 m. If the change in potential energy goes into the internal energy of the water, compute the increase in its temperature. (b) Do the same for Yosemite Falls, where the water drops 740 m. (These temperature rises are not observed because the water cools by evaporation as it falls.) (a), (b) Here ∆U = mgh = mc∆t; ∆t = gh/c (a) ∆t = 9.81 × 50/4184 = 0.117 K; (b) ∆t = 1.74 K 31 · When 20 cal of heat are absorbed by a gas, the system performs 30 J of work. What is the change in the internal energy of the gas? From Equ. 19-10, ∆U = Q − W ∆U = (20 × 4.184 − 30) J = 53.7 J 32 ·· A lead bullet initially at 30°C just melts upon striking a target. Assuming that all of the initial kinetic energy of the bullet goes into the internal energy of the bullet to raise its temperature and melt it, calculate the speed of the bullet upon impact. Q = 1/2mv2 = mc(TMP − Ti) + mLf; solve for v v = 2[c( T MP − T i )+ L f ] = 354 m/s 33* ·· A piece of ice is dropped from a height H. (a) Find the minimum value of H such that the ice melts when it makes an inelastic collision with the ground. Assume that all the mechanical energy lost goes into melting the ice. (b) Is it reasonable to neglect the variation in the acceleration of gravity in doing this problem? (c) Comment on the reasonableness of neglecting air resistance. What effect would air resistance have on your answer? (a) To melt the ice (at t = 0oC), mgh = mLf; h = Lf/g h = 333.5/9.81 km = 34 km (b) Yes. Since h > Tc you would expect the equipartition theorem to hold and for T 2000 K, vibration of the atoms contributes to cp of H 2O and CO2 so that cp of these gases is 7.5R at high temperatures. Also, assume the gases do not dissociate. 106 ··· The combustion of benzene is represented by the chemical reaction 2(C6H6) + 15(O2) → 12(CO2) + 6(H2O). The amount of energy released in the combustion of two mol of benzene is 1516 kcal. One mol of benzene and 7.5 mol of oxygen at 300 K are confined in an insulated enclosure at a pressure of 1 atm. (a) Find the temperature and volume following combustion if the pressure is maintained at 1 atm. (b) If, following combustion, the thermal insulation about the container is removed and the system is cooled to 300 K, what is the final pressure? For Problems 106-109, the specific heat of the combustion products depends on the temperature. Although cp increases gradually from (9/2)R per mol to (15/2)R per mol at high temperatures, we shall make the assumption that cp = 4.5R below T = 2000K and cp = 7.5R above T = 2000 K. We shall use R = 2.0 cal/mol.K. (a) 1. Find Vi for 8.5 mol of gas at 300 K and 1 atm. 2. Find the total heat released. 3. Find Q needed to form the products at 100oC; there are 3 mol of H2O, 6 mol of CO2. 4. Find Q to heat 9 mol of gas to 2000 K 5. Find Q available to heat gases above 2000 K 6. Find final T of 9 mol of triatomic gases. 7. Find Vf from the ideal gas law. (b) Since Tf = Ti, Pf = Pi(n f /n i)(Vi/Vf)

Vi = (22.4 × 8.5 × 300/273) L = 209.2 L Q = (1516/2) kcal = 758 kcal Q to form steam = 3 × 18(73 + 540) cal = 33.10 kcal Q to heat CO2 = 6 × 4.5 × 2.0 × 73 cal = 3.94 kcal Q = 9 × 4.5 × 2.0 × 1627 cal = 131.79 kcal Q = (758 − 131.79 − 3.94 − 33.10) kcal = 589.2 kcal 3 ∆T = 589.2 × 10 /(9 × 7.5 × 2.0) K = 4364 K; Tf = 6364 K Vf = (9 × 8.314 × 6364/101) L = 4715 L = 4.715 m3 Pf = (1 × (9/8.5)(209.2/4715) atm = 0.047 atm

107 ··· Repeat Problem 106, parts (a) and (b), using as the combustible substance 1 mol of acetylene for which the combustion reaction is 2(C2H2) + 5(O2) → 4(CO2) + 2(H2O). The combustion of 1 mol of acetylene releases 300 kcal. (a) 1. Find Vi of 3.5 mol at 300 K and 1 atm. Vi = (22.4 × 3.5 × 300/273) L = 86.15 L

Chapter 19

Heat and the First Law of Thermodynamics

2. Find the total heat released. 3. Find Q needed to form products at 100oC; there are 1 mol of H2O and 2 mol of CO2. 4. Find Q to heat 3 mol of gas to 2000 K 5. Find Q available to heat gases above 2000 K 6. Find Tf of 3 mol of triatomic gases. 7. Find Vf using ideal gas law. (b) Pf = Pi(n f /n i)(Vi/Vf)

Q = 300 kcal Q to form steam = 18(73 + 540) cal = 11.03 kcal Q to heat CO2 = 2 × 4.5 × 2.0 × 73 cal = 1.31 kcal Q = 3 × 4.5 × 2.0 × 1627 cal = 43.93 kcal Q = (300 − 43.93 − 1.31 − 11.03) kcal = 243.7 kcal 3 ∆T = 243.7 × 10 /(3 × 7.5 × 2.0) K = 5416 K; Tf = 7418 K Vf = (3 × 8.314 × 7416/101) L = 1831 L = 1.831 m3 Pf = 1(3/3.5)(86.15/1831) atm = 0.0403 atm

108 ··· Carbon monoxide and oxygen combine to form carbon dioxide with an energy release of 280 kJ/mol of CO according to the reaction 2(CO) + O2 → 2(CO2). Two mol of CO and one mol of O2 at 300 K are confined in an 80-L container; the combustion reaction is initiated with a spark. (a) What is the pressure in the container prior to the reaction? (b) If the reaction proceeds adiabatically, what are the final temperature and pressure? (c) If the resulting CO2 gas is cooled to 0°C, what is the pressure in the container? (a) 1. Find Pi of 3 mol at 300 K in 80 L Pi = (3 × 8.314 × 300/80) kPa = 93.53 kPa (b) 1. Find Cv of combustion product below Cv = 2(7/2) × 8.314 = 58.2 J/K 2000 K. 2. Find Q to raise 2 mol of CO2 to 2000 K Q = 58.2 × 1700 J = 98.94 kJ 3. Find Q to raise CO2 above 2000 K Q = (560 − 98.94) kJ = 461.1 kJ 3 4. Find Tf ∆T = (461.1 × 10 /2 × 6.5 × 8.314) K = 4266 K; Tf = 6266 K 5. Find Pf = Pi(n f /n i)(Tf /Ti) Pf = 93.53(2/3)(6266/300) kPa = 1.30 MPa (c) Pf = Pi(n f /n i)(Tf /Ti) Pf = 93.53(2/3)(273/300) kPa = 54.7 kPa 109*···Suppose that instead of pure oxygen, just enough air is mixed with the two mol of CO in the container of Problem 108 to permit complete combustion. Air is 80% N2 and 20% O2 by weight, and the nitrogen does not participate in the reaction. What then are the answers to parts (a), (b), and (c) of Problem 108? Note that for N2, cv at temperatures above 2000 K is (5/2)R + R since there is only one vibrational mode that contributes to the specific heat. (a) 1. Write the reaction for 2 mol of CO 2(CO) + O2 + 4N2 → 2(CO2) + 4N 2 2. Find Pi of 7 mol at 300 K in 80 L Pi = (7 × 8.314 × 300/80) kPa = 218.2 kPa (b) 1. Find Cv of product gases for T < 2000 K Cv = [2 × (7/2) + 4 × (5/2)]R = 141.3 J/K 2. Find Q to heat gases to 2000 K Q = 1700 × 141.3 J = 240.2 kJ 3. Find Q available to raise gases above 2000 K Q = (560 − 240.2) kJ = 319.8 kJ 3 4. Find Tf; note that Cv = 2 × 6.5 ∆T = (319.8 × 10 /27 × 8.314) K = 1425 K; Tf = 3425 K 5. Find Pf = Pi(n f /n i)(Tf /Ti) Pf = 218.2(6/7)(3425/300) kPa = 2.135 MPa (c) Pf = Pi(n f /n i)(Tf /Ti) Pf = 2135(273/3425) kPa = 170.2 kPa 110 ··· Use the expression given in Problem 103 for the internal energy per mole of a solid according to the Einstein model

Chapter 19

Heat and the First Law of Thermodynamics

to show that the molar heat capacity at constant volume is given by 2

/T  TE  eTE c v ′ = 3R   T E / T 2  T  (e − 1)

U=

3 RT E . The specific heat at constant volume is dU/dT and performing the indicated operation one eT E / T − 1 2

T E eT E /T 3 R   obtains cv = 2 .  T  ( eT E /T − 1 ) 111 ··· (a) Use the results of Problem 110 to show that the Dulong–Petit law, c v ′≈ 3R , holds for the Einstein model when T > TE. (b) For diamond, TE is approximately 1060 K. Numerically integrate ∆U = ∫ c v′ dT to find the increase in the internal energy if 1 mol of diamond is heated from 300 to 600 K. Compare your result to that obtained in Problem 103. With TE = 1060 K, cv has the values shown below T, K cv, J/mol.K 300 9.65 400 14.33 500 17.38 600 19.35 The specific heat as a function of temperature is shown in the Figure. Integrating numerically one obtains ∆U = 100[1/2(9.65) + 14.33 + 17.38) + 1/2(19.35)] = 4621 J in good agreement with the result of Problem 103.

112 ··· A refinement of the Einstein model by Debye resulted in the following expression for the specific heat: 3

T  ′ cv = 9R   T D 

y

4

x

x e ∫0 (e x − 1)2 dx

where TD is called the Debye temperature and y = TD / T. (a) Show that when T >> TD, the above expression reduces to the Dulong–Petit result cv = 3R. (Hint: When T >> TD, y