Chapter 19.pdf

19 Temperature CHAPTER OUTLINE 19.1 19.2 19.3

19.4 19.5

Temperature and the Zeroth Law of Thermodynamics Thermometers and the Celsius Temperature Scale The Constant-Volume Gas Thermometer and the Absolute Temperature Scale Thermal Expansion of Solids and Liquids Macroscopic Description of an Ideal Gas

ANSWERS TO QUESTIONS Q19.1

Two objects in thermal equilibrium need not be in contact. Consider the two objects that are in thermal equilibrium in Figure 19.1(c). The act of separating them by a small distance does not affect how the molecules are moving inside either object, so they will still be in thermal equilibrium.

Q19.2

The copper’s temperature drops and the water temperature rises until both temperatures are the same. Then the metal and the water are in thermal equilibrium.

Q19.3

The astronaut is referring to the temperature of the lunar surface, specifically a 400°F difference. A thermometer would register the temperature of the thermometer liquid. Since there is no atmosphere in the moon, the thermometer will not read a realistic temperature unless it is placed into the lunar soil.

Q19.4

Rubber contracts when it is warmed.

Q19.5

Thermal expansion of the glass bulb occurs first, since the wall of the bulb is in direct contact with the hot water. Then the mercury heats up, and it expands.

Q19.6

If the amalgam had a larger coefficient of expansion than your tooth, it would expand more than the cavity in your tooth when you take a sip of your ever-beloved coffee, resulting in a broken or cracked tooth! As you ice down your now excruciatingly painful broken tooth, the amalgam would contract more than the cavity in your tooth and fall out, leaving the nerve roots exposed. Isn’t it nice that your dentist knows thermodynamics?

Q19.7

The measurements made with the heated steel tape will be too short—but only by a factor of 5 × 10 −5 of the measured length.

Q19.8

(a)

One mole of H 2 has a mass of 2.016 0 g.

(b)

One mole of He has a mass of 4.002 6 g.

(c)

One mole of CO has a mass of 28.010 g.

Q19.9

The ideal gas law, PV = nRT predicts zero volume at absolute zero. This is incorrect because the ideal gas law cannot work all the way down to or below the temperature at which gas turns to liquid, or in the case of CO 2 , a solid. 549

550

Temperature

Q19.10

Call the process isobaric cooling or isobaric contraction. The rubber wall is easy to stretch. The air inside is nearly at atmospheric pressure originally and stays at atmospheric pressure as the wall moves in, just maintaining equality of pressure outside and inside. The air is nearly an ideal gas to start with, but PV = nRT soon fails. Volume will drop by a larger factor than temperature as the water vapor liquefies and then freezes, as the carbon dioxide turns to snow, as the argon turns to slush, and as the oxygen liquefies. From the outside, you see contraction to a small fraction of the original volume.

Q19.11

Cylinder A must be at lower pressure. If the gas is thin, it will be at one-third the absolute pressure of B.

Q19.12

At high temperature and pressure, the steam inside exerts large forces on the pot and cover. Strong latches hold them together, but they would explode apart if you tried to open the hot cooker.

Q19.13

(a)

The water level in the cave rises by a smaller distance than the water outside, as the trapped air is compressed. Air can escape from the cave if the rock is not completely airtight, and also by dissolving in the water.

(b)

The ideal cave stays completely full of water at low tide. The water in the cave is supported by atmospheric pressure on the free water surface outside.

(a)

(b) FIG. Q19.13

Q19.14

Absolute zero is a natural choice for the zero of a temperature scale. If an alien race had bodies that were mostly liquid water—or if they just liked its taste or its cleaning properties—it is conceivable that they might place one hundred degrees between its freezing and boiling points. It is very unlikely, on the other hand, that these would be our familiar “normal” ice and steam points, because atmospheric pressure would surely be different where the aliens come from.

Q19.15

As the temperature increases, the brass expands. This would effectively increase the distance, d, from the pivot point to the center of mass of the pendulum, and also increase the moment of inertia of the pendulum. Since the moment of inertia is proportional to d 2 , and the period of a physical I pendulum is T = 2π , the period would increase, and the clock would run slow. mgd

Q19.16

As the water rises in temperature, it expands. The excess volume would spill out of the cooling system. Modern cooling systems have an overflow reservoir to take up excess volume when the coolant heats up and expands.

Q19.17

The coefficient of expansion of metal is larger than that of glass. When hot water is run over the jar, both the glass and the lid expand, but at different rates. Since all dimensions expand, there will be a certain temperature at which the inner diameter of the lid has expanded more than the top of the jar, and the lid will be easier to remove.

Chapter 19

Q19.18

The sphere expands when heated, so that it no longer fits through the ring. With the sphere still hot, you can separate the sphere and ring by heating the ring. This more surprising result occurs because the thermal expansion of the ring is not like the inflation of a blood-pressure cuff. Rather, it is like a photographic enlargement; every linear dimension, including the hole diameter, increases by the same factor. The reason for this is that the atoms everywhere, including those around the inner circumference, push away from each other. The only way that the atoms can accommodate the greater distances is for the circumference—and corresponding diameter—to grow. This property was once used to fit metal rims to wooden wagon and horse-buggy wheels. If the ring is heated and the sphere left at room temperature, the sphere would pass through the ring with more space to spare.

551

FIG. Q19.18

SOLUTIONS TO PROBLEMS Section 19.1

Temperature and the Zeroth Law of Thermodynamics

No problems in this section

Section 19.2

Thermometers and the Celsius Temperature Scale

Section 19.3

The Constant-Volume Gas Thermometer and the Absolute Temperature Scale

P19.1

Since we have a linear graph, the pressure is related to the temperature as P = A + BT , where A and B are constants. To find A and B, we use the data

a f 1.635 atm = A + a78.0° CfB

0.900 atm = A + −80.0° C B

(1) (2)

Solving (1) and (2) simultaneously, we find

A = 1.272 atm

and

B = 4.652 × 10 −3 atm ° C

Therefore,

P = 1.272 atm + 4.652 × 10 −3 atm ° C T

(a)

P = 0 = 1.272 atm + 4.652 × 10 −3 atm ° C T

At absolute zero which gives

e

j

e

j

T = −274° C .

(b)

At the freezing point of water P = 1.272 atm + 0 = 1.27 atm .

(c)

And at the boiling point P = 1.272 atm + 4.652 × 10 −3 atm ° C 100° C = 1.74 atm .

e

ja

f

552 P19.2

Temperature

P1V = nRT1 and P2 V = nRT2 imply that

P19.3

P19.4

P2 T2 = P1 T1

a

fa

f

(a)

P2 =

0.980 atm 273 K + 45.0 K P1T2 = = 1.06 atm 273 + 20.0 K T1

(b)

T3 =

T1 P3 P1

a f a293 K fa0.500 atmf = 149 K = = 0.980 atm

−124° C FIG. P19.2

a

f

9 9 TC + 32.0° F = −195.81 + 32.0 = −320° F 5 5

(a)

TF =

(b)

T = TC + 273.15 = −195.81 + 273.15 = 77.3 K

(a)

To convert from Fahrenheit to Celsius, we use

TC =

and the Kelvin temperature is found as

T = TC + 273 = 310 K

(b)

In a fashion identical to that used in (a), we find TC = −20.6° C T = 253 K

and P19.5

P19.6

b

FG 212° F − 32.0° F IJ = H 100° C − 0.00° C K

(a)

∆T = 450° C = 450° C

(b)

∆T = 450° C = 450 K

810° F

a f 100° C = aa60.0° Sf + b Subtracting, 100° C = aa75.0° Sf Require

0.00° C = a −15.0° S + b

a

a = 1.33 C° S° .

f

Then 0.00° C = 1.33 −15.0° S C°+ b b = 20.0° C .

b

g

So the conversion is TC = 1.33 C° S° TS + 20.0° C . P19.7

(a)

T = 1 064 + 273 = 1 337 K melting point T = 2 660 + 273 = 2 933 K boiling point

(b)

g a

f

5 5 TF − 32.0 = 98.6 − 32.0 = 37.0° C 9 9

∆T = 1 596° C = 1 596 K . The differences are the same.

Chapter 19

Section 19.4 P19.8

553

Thermal Expansion of Solids and Liquids

α = 1.10 × 10 −5 ° C −1 for steel

e

a

j

f

∆L = 518 m 1.10 × 10 −5 ° C −1 35.0° C − −20.0° C = 0.313 m P19.9

The wire is 35.0 m long when TC = −20.0° C .

b

∆L = Liα T − Ti

g

a f a f for Cu. ∆L = a35.0 mfe1.70 × 10 aC°f jc35.0° C − a −20.0° C fh = −1

α = α 20.0° C = 1.70 × 10 −5 C°

−1

−5

a

fe

ja

+3.27 cm

f

P19.10

∆L = Liα ∆T = 25.0 m 12.0 × 10 −6 C° 40.0° C = 1.20 cm

P19.11

For the dimensions to increase, ∆L = αLi ∆T

a

fa

1.00 × 10 −2 cm = 1.30 × 10 −4 ° C −1 2.20 cm T − 20.0° C

f

T = 55.0° C *P19.12 P19.13

*P19.14

ja

e

fa

f

∆L = αLi ∆T = 22 × 10 −6 C° 2.40 cm 30° C = 1.58 × 10 −3 cm

a

fa

f

a

fa

f

(a)

∆L = αLi ∆T = 9.00 × 10 −6 ° C −1 30.0 cm 65.0° C = 0.176 mm

(b)

∆L = αLi ∆T = 9.00 × 10 −6 ° C −1 1.50 cm 65.0° C = 8.78 × 10 −4 cm

(c)

∆V = 3αVi ∆T = 3 9.00 × 10 −6 ° C −1

e

F jGH 30.0aπ fa4 1.50f

Ia JK

2

f

cm3 65.0° C = 0.093 0 cm3

The horizontal section expands according to ∆L = αLi ∆T .

ja

e

fa

f

∆x = 17 × 10 −6 ° C −1 28.0 cm 46.5° C − 18.0° C = 1.36 × 10 −2 cm The vertical section expands similarly by

ja

e

FIG. P19.14

fa

f

∆y = 17 × 10 −6 ° C −1 134 cm 28.5° C = 6.49 × 10 −2 cm . The vector displacement of the pipe elbow has magnitude ∆r = ∆x 2 + ∆ y 2 =

a0.136 mmf + a0.649 mmf 2

2

= 0.663 mm

and is directed to the right below the horizontal at angle

θ = tan −1

FG ∆y IJ = tan FG 0.649 mm IJ = 78.2° H 0.136 mm K H ∆x K −1

∆r = 0.663 mm to the right at 78.2° below the horizontal

554 P19.15

Temperature

b

g

b

L Al 1 + α Al ∆T = LBrass 1 + α Brass ∆T L Al − LBrass ∆T = LBrassα Brass − L Alα Al

(a)

∆T =

g

a10.01 − 10.00f a10.00fe19.0 × 10 j − a10.01fe24.0 × 10 j −6

−6

∆T = −199° C so T = −179° C. This is attainable. ∆T =

(b)

a10.02 − 10.00f a10.00fe19.0 × 10 j − a10.02fe24.0 × 10 j −6

−6

∆T = −396° C so T = −376° C which is below 0 K so it cannot be reached. P19.16

g a50.0° Cf

jb

e

∆A = 2 17.0 × 10 −6 ° C −1 0.080 0 m

∆A = 2αAi ∆T :

(a)

2

∆A = 1.09 × 10 −5 m 2 = 0.109 cm 2 (b)

The length of each side of the hole has increased. Thus, this represents an increase in the area of the hole.

b

g

e

e

P19.17

∆V = β − 3α Vi ∆T = 5.81 × 10 −4 − 3 11.0 × 10 −6

P19.18

(a)

a

f

jjb50.0 galga20.0f =

0.548 gal

a

5.050 cm = 5.000 cm 1 + 24.0 × 10 −6 ° C −1 T − 20.0° C

L = Li 1 + α∆T :

f

T = 437° C (b)

L Al = LBrass for some ∆T , or

We must get

b

g

b

Li , Al 1 + α Al ∆T = Li , Brass 1 + α Brass ∆T

e

g

j

e

j

5.000 cm 1 + 24.0 × 10 −6 ° C −1 ∆T = 5.050 cm 1 + 19.0 × 10 −6 ° C −1 ∆T Solving for ∆T , ∆T = 2 080° C , T = 3 000° C

so

This will not work because aluminum melts at 660° C . P19.19

b

a

g

f

(a)

V f = Vi 1 + β∆T = 100 1 + 1.50 × 10 −4 −15.0 = 99.8 mL

(b)

∆Vacetone = βVi ∆T

b

b

∆Vflask = βVi ∆T

g

g

acetone

Pyrex

b

= 3αVi ∆T

g

Pyrex

for same Vi , ∆T , ∆Vacetone β acetone 1.50 × 10 −4 1 = = = −6 ∆Vflask β flask × 6 . 40 10 −2 3 3.20 × 10

e

j

The volume change of flask is about 6% of the change in the acetone’s volume .

Chapter 19

P19.20

(a),(b)

555

The material would expand by ∆L = αLi ∆T , ∆L = α∆T , but instead feels stress Li F Y∆L = = Yα∆T = 7.00 × 10 9 N m 2 12.0 × 10 −6 C° A Li

e

a f a30.0° Cf

j

−1

= 2.52 × 10 6 N m 2 . This will not break concrete. P19.21

(a)

b

g

∆V = Vt β t ∆T − VAl β Al ∆T = β t − 3α Al Vi ∆T

e

j

e

ja

= 9.00 × 10 −4 − 0.720 × 10 −4 ° C −1 2 000 cm3 60.0° C

∆V = 99.4 cm3 (b)

f

overflows.

The whole new volume of turpentine is

ja

e

f

2 000 cm3 + 9.00 × 10 −4 ° C −1 2 000 cm3 60.0° C = 2 108 cm3

so the fraction lost is

99.4 cm3 = 4.71 × 10 −2 3 2 108 cm

and this fraction of the cylinder’s depth will be empty upon cooling:

a

f

4.71 × 10 −2 20.0 cm = 0.943 cm . *P19.22

The volume of the sphere is VPb =

a

f

4 3 4 π r = π 2 cm 3 3

3

= 33.5 cm3 .

The amount of mercury overflowing is

e

j

overflow = ∆VHg + ∆VPb − ∆Vglass = β Hg VHg + β PbVPb − β glassVglass ∆T where Vglass = VHg + VPb is the initial volume. Then

e j e j e j e 1 1 L O = Ma182 − 27f10 118 cm + a87 − 27f10 33.5 cm P 40° C = 0.812 cm C° C° N Q

j

overflow = β Hg − β glass VHg + β Pb − β glass VPb ∆T = β Hg − 3α glass VHg + 3α Pb − 3α glass VPb ∆T −6

P19.23

In

−6

3

3

F Y∆L = require ∆L = αLi ∆T A Li

F = Yα∆T A F 500 N = ∆T = −4 2 AYα 2.00 × 10 m 20.0 × 10 10 N m 2 11.0 × 10 −6 C°

e

∆T = 1.14° C

je

je

j

3

556 *P19.24

Temperature

Model the wire as contracting according to ∆L = αLi ∆T and then stretching according to ∆L Y F stress = = Y = αLi ∆T = Yα∆T . A Li Li

e

1 45° C = 396 N C°

j

(a)

F = YAα∆T = 20 × 10 10 N m 2 4 × 10 −6 m 2 11 × 10 −6

(b)

∆T =

3 × 10 8 N m 2 stress = = 136° C Yα 20 × 10 10 N m 2 11 × 10 −6 C°

e

j

To increase the stress the temperature must decrease to 35° C − 136° C = −101° C . (c) *P19.25

The original length divides out, so the answers would not change.

The area of the chip decreases according to

∆A = γA1 ∆T = A f − Ai

b

a

g

A f = Ai 1 + γ∆T = Ai 1 + 2α∆T

f

The star images are scattered uniformly, so the number N of stars that fit is proportional to the area.

a

f

ja

e

f

Then N f = N i 1 + 2α∆T = 5 342 1 + 2 4.68 × 10 −6 ° C −1 −100° C − 20° C = 5 336 star images .

Section 19.5 P19.26

P19.27

Macroscopic Description of an Ideal Gas

a

fe

n=

(b)

N = nN A

(a)

Initially, PV i i = n i RTi

a = a 2.99 molfe6.02 × 10

j

fa

f

j a1.00 atmfV = n Ra10.0 + 273.15f K P b0. 280V g = n Ra 40.0 + 273.15f K

23

molecules mol = 1.80 × 10 24 molecules i

Finally, Pf V f = n f RT f

f

i

i

i

0.280 Pf

giving

313.15 K = 1.00 atm 283.15 K Pf = 3.95 atm

or

Pf = 4.00 × 10 5 Pa abs. .

Dividing these equations,

(b)

je

9.00 atm 1.013 × 10 5 Pa atm 8.00 × 10 −3 m 3 PV = = 2.99 mol 8.314 N ⋅ mol K 293 K RT

(a)

a f P a1.02fb0.280V g = n Ra85.0 + 273.15f K

After being driven

d

i

i

Pd = 1.121Pf = 4.49 × 10 5 Pa P19.28

a fa f a fa f

3 150 0.100 3 PV = = 884 balloons 3 4π r P ′ 4π 0.150 3 1.20 If we have no special means for squeezing the last 100 L of helium out of the tank, the tank will be full of helium at 1.20 atm when the last balloon is inflated. The number of balloons is then reduced 0.100 m3 3 to to 884 − = 877 . 3 4π 0.15 m PV = NP ′V ′ =

e

a

4 3 π r NP ′ : 3

j f

N=

Chapter 19

P19.29

The equation of state of an ideal gas is PV = nRT so we need to solve for the number of moles to find N.

e

ja

fa ga

fa

f

1.01 × 10 5 N m 2 10.0 m 20.0 m 30.0 m PV = = 2. 49 × 10 5 mol n= RT 8.314 J mol ⋅ K 293 K

b

e

5

N = nN A = 2.49 × 10 mol 6.022 × 10 *P19.30

(a)

PV i i = n i RTi =

23

f

j

molecules mol = 1.50 × 10 29 molecules

mi RTi M

e

j

3

4.00 × 10 −3 kg 1.013 × 10 5 N 4π 6.37 × 10 6 m mole ⋅ K MPV i i = mi = RTi 8.314 Nm 50 K mole m2 3 = 1.06 × 10 21 kg (b)

Pf V f PV i i

=

n f RT f ni RTi

F 1.06 × 10 kg + 8.00 × 10 kg I T GH JK 50 K 1.06 × 10 kg F 1 IJ = 56.9 K T = 100 K G H 1.76 K nRT F 9.00 g I F 8.314 J I F 773 K I =G P= G J G J V H 18.0 g mol K H mol K K H 2.00 × 10 m JK = 21

2 ⋅1 =

20

f

21

f

P19.31

P19.32

P19.33

557

−3

a

1.61 MPa = 15.9 atm

fa f

(a)

T2 = T1

P2 = 300 K 3 = 900 K P1

(b)

T2 = T1

P2 V2 = 300 2 2 = 1 200 K P1V1

∑ Fy = 0 :

3

a fa f

b

g

ρ out gV − ρ in gV − 200 kg g = 0

bρ

out

ge

− ρ in 400 m

3

j = 200 kg

The density of the air outside is 1.25 kg m3 . n P = From PV = nRT , V RT The density is inversely proportional to the temperature, and the density of the hot air is

jFGH 283T K IJK e1.25 kg m jFGH 1 − 283T K IJK e400 m j = 200 kg e

ρ in = 1.25 kg m3

Then

in

3

3

in

283 K 1− = 0.400 Tin 283 K 0.600 = Tin = 472 K Tin

FIG. P19.33

558 *P19.34

P19.35

Temperature

Consider the air in the tank during one discharge process. We suppose that the process is slow enough that the temperature remains constant. Then as the pressure drops from 2.40 atm to 1.20 atm, the volume of the air doubles. During the first discharge, the air volume changes from 1 L to 2 L. Just 1 L of water is expelled and 3 L remains. In the second discharge, the air volume changes from 2 L to 4 L and 2 L of water is sprayed out. In the third discharge, only the last 1 L of water comes out. Were it not for male pattern dumbness, each person could more efficiently use his device by starting with the tank half full of water. (a)

PV = nRT

e je j b ga f m = nM = a 41.6 molfb 28.9 g molg =

1.013 × 10 5 Pa 1.00 m3 PV n= = = 41.6 mol RT 8.314 J mol ⋅ K 293 K (b)

1.20 kg , in agreement with the tabulated density of

3

1.20 kg m at 20.0°C. *P19.36

e

j

2

The void volume is 0.765Vtotal = 0.765π r 2 = 0.765π 1.27 × 10 −2 m 0.2 m = 7.75 × 10 −5 m3 . Now for the gas remaining PV = nRT n=

P19.37

(a)

b

n=

PV = nRT

e

j ga

−5 5 2 3 PV 12.5 1.013 × 10 N m 7.75 × 10 m = = 3.96 × 10 −2 mol RT 8.314 Nm mole K 273 + 25 K

f

PV RT

a

fe 3

−3 5 PVM 1.013 × 10 Pa 0.100 m 28.9 × 10 kg mol = m = nM = RT 8.314 J mol ⋅ K 300 K

b

f

ga

j

m = 1.17 × 10 −3 kg

P19.38

e

j

(b)

Fg = mg = 1.17 × 10 −3 kg 9.80 m s 2 = 11.5 mN

(c)

F = PA = 1.013 × 10 5 N m 2 0.100 m

(d)

The molecules must be moving very fast to hit the walls hard.

ja

e

At depth,

P = P0 + ρgh

At the surface,

P0 V f = nRT f :

Therefore

and

2

= 1.01 kN

PVi = nRTi P0 V f

bP + ρghgV

F T I FG P + ρgh IJ GH T JK H P K F 293 K IJ FG 1.013 × 10 = 1.00 cm G H 278 K K GH

V f = Vi Vf

f

f

0

i

=

Tf Ti

0

i

0

3

V f = 3.67 cm3

5

e

je

ja

f IJ JK

Pa + 1 025 kg m 3 9.80 m s 2 25.0 m 1.013 × 10 5 Pa

Chapter 19

P19.39

mf

PV = nRT :

mi

=

nf

m f = mi

so

Pf V f RTi Pf = RT f PV Pi i i

=

ni

FP I GH P JK f

i

∆m = mi − m f = mi P19.40

559

F P − P I = 12.0 kgFG 41.0 atm − 26.0 atm IJ = GH P JK H 41.0 atm K i

f

4.39 kg

i

My bedroom is 4 m long, 4 m wide, and 2.4 m high, enclosing air at 100 kPa and 20° C = 293 K . Think of the air as 80.0% N 2 and 20.0% O 2 . Avogadro’s number of molecules has mass

a0.800fb28.0 g molg + a0.200fb32.0 g molg = 0.028 8 kg mol F mI PV = nRT = G J RT H MK PVM e1.00 × 10 N m je38.4 m jb0.028 8 kg molg = = 45.4 kg m= RT b8.314 J mol ⋅ K ga293 K f

Then

5

gives *P19.41

2

3

~ 10 2 kg

The CO 2 is far from liquefaction, so after it comes out of solution it behaves as an ideal gas. Its molar mass is M = 12.0 g mol + 2 16.0 g mol = 44.0 g mol . The quantity of gas in the cylinder is m sample 6.50 g = = 0.148 mol n= M 44.0 g mol

b

Then

PV = nRT

gives

V=

g

b

ga

nRT 0.148 mol 8.314 J mol ⋅ K 273 K + 20 K = P 1.013 × 10 5 N m 2

e

je

je

j

f FG 1 N ⋅ m IJ F 10 L I = H 1 J K GH 1 m JK 3

3

3.55 L

P19.42

10 −9 Pa 1.00 m 3 6.02 × 10 23 molecules mol PVN A = = 2.41 × 10 11 molecules N= RT 8.314 J K ⋅ mol 300 K

P19.43

P0 V = n1 RT1 =

b

FG m IJ RT H MK F m IJ RT P V = n RT = G HMK P VM F 1 1I − J m −m = G R HT T K 0

2

1

2

1

1

2

2

2

0

1

2

ga

f

560 P19.44

Temperature

(a)

Initially the air in the bell satisfies P0 Vbell = nRTi or

a

f

P0 2.50 m A = nRTi

(1)

When the bell is lowered, the air in the bell satisfies

a

f

Pbell 2.50 m − x A = nRT f

(2)

where x is the height the water rises in the bell. Also, the pressure in the bell, once it is lowered, is equal to the sea water pressure at the depth of the water level in the bell.

a

a

f

f

Pbell = P0 + ρg 82.3 m − x ≈ P0 + ρg 82.3 m

(3)

The approximation is good, as x < 2.50 m. Substituting (3) into (2) and substituting nR from (1) into (2),

a

fa

f

P0 + ρg 82.3 m 2.50 m − x A = P0 Vbell

Tf Ti

.

Using P0 = 1 atm = 1.013 × 10 5 Pa and ρ = 1.025 × 10 3 kg m3

L O fMM TT FGH1 + ρga82P.3 mf IJK PP N Q LM 277.15 K F e1.025 × 10 kg m je9.80 m s ja82.3 mf I G1 + JJ = a 2.50 mfM1 − 293.15 K G 1.013 × 10 N m K H MN a

x = 2.50 m 1 −

−1

f

0

0

3

3

−1

2

2

5

OP PP Q

x = 2.24 m (b)

If the water in the bell is to be expelled, the air pressure in the bell must be raised to the water pressure at the bottom of the bell. That is,

a

f

Pbell = P0 + ρg 82.3 m 5

e

ja

je

f

= 1.013 × 10 Pa + 1.025 × 10 3 kg m 3 9.80 m s 2 82.3 m 5

Pbell = 9.28 × 10 Pa = 9.16 atm

Additional Problems P19.45

The excess expansion of the brass is

b

a f a f ∆a ∆Lf = 2.66 × 10 m

∆ ∆L = 19.0 − 11.0 × 10 −6 −4

(a)

The rod contracts more than tape to a length reading 0.950 0 m − 0.000 266 m = 0.949 7 m

(b)

g a° Cf a0.950 mfa35.0° Cf

∆Lrod − ∆Ltape = α brass − α steel Li ∆T

0.950 0 m + 0.000 266 m = 0.950 3 m

−1

Chapter 19

P19.46

At 0°C, 10.0 gallons of gasoline has mass,

ρ=

from

m V

jb

e

m = ρV = 730 kg m3 10.0 gal

80 m I = 27.7 kg gFGH 0.003 1.00 gal JK 3

The gasoline will expand in volume by

b

ga

f

∆V = βVi ∆T = 9.60 × 10 −4 ° C −1 10.0 gal 20.0° C − 0.0° C = 0.192 gal At 20.0°C,

10.192 gal = 27.7 kg 10.0 gal = 27.7 kg

F 10.0 gal I = 27.2 kg GH 10.192 gal JK

The extra mass contained in 10.0 gallons at 0.0°C is 27.7 kg − 27.2 kg = 0.523 kg . P19.47

Neglecting the expansion of the glass, ∆h = ∆h =

V β∆T A

b0.250 cm 2g e1.82 × 10 π e 2.00 × 10 cmj 3

4 3π

−3

2

−4

ja

f

° C −1 30.0° C = 3.55 cm

FIG. P19.47 P19.48

(a)

The volume of the liquid increases as ∆V = Vi β∆T . The volume of the flask increases as ∆Vg = 3αVi ∆T . Therefore, the overflow in the capillary is Vc = Vi ∆T β − 3α ; and in the

b

capillary Vc = A∆h . Therefore, ∆h = (b)

Vi β − 3α ∆T . A

b

g

b g

For a mercury thermometer

β Hg = 1.82 × 10 −4 ° C −1

and for glass,

3α = 3 × 3.20 × 10 −6 ° C −1

Thus

β − 3α ≈ β

or

α