Cap20 Prob

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Chapter 20

Thermal Properties and Processes

Conceptual Problems

1 • Why does the mercury level of a thermometer first decrease slightlywhen the thermometer is first placed in warm water?

Determine the Concept The glass bulb warms and expands first, before the

mercury warms and expands.

2 • A large sheet of metal has a hole cut in the middle of it. When thesheet is heated, the area of the hole will (a) not change, (b) always increase,(c) always decrease, (d ) increase if the hole is not in the exact center of the sheet,(e) decrease only if the hole is in the exact center of the sheet.

Determine the Concept The heating of the sheet causes the average separation of

its molecules to increase. The consequence of this increased separation is that thearea of the hole always increases. )(b is correct.

3 • [SSM] Why is it a bad idea to place a tightly sealed glass bottle thatis completely full of water, into your kitchen freezer in order to make ice?

Determine the Concept Water expands greatly as it freezes. If a sealed glass bottle full of water is placed in a freezer, as the water freezes there will be noroom for the expansion to take place. The bottle will be broken.

4 • The windows of your physics laboratory are left open on a night whenthe temperature of the outside dropped well below freezing. A steel ruler and awooden ruler were left on the window sill, and when you arrive in the morningthey are both very cold. The coefficient of linear expansion of wood is about

5 × 10−6 K −1. Which ruler should you use to make the most accurate lengthmeasurements? Explain your answer.

Determine the Concept You should use the wooden ruler. Because thecoefficient of expansion for wood is about half that for metal, the metal ruler willhave shrunk considerably more than will have the wooden ruler.

5 • Bimetallic strips are used both for thermostats and for electrical circuit breakers. A bimetallic strip consists of a pair of thin strips of metal that havedifferent coefficients of linear expansion and are bonded together to form onedoubly thick strip. Suppose a bimetallic strip is constructed out of one steel stripand one copper strip, and suppose the bimetallic strip is curled in the shape of acircular arc with the steel strip on the outside. If the temperature of the strip isdecreased, will the strip straighten out or curl more tightly?

1863

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Chapter 201864

Determine the Concept The strip will curl more tightly. Because the coefficient

of linear expansion for copper (17 × 10−6K −

1) is greater than the coefficient of

linear expansion for steel (11 × 10−6 K −1), the length of the copper strip will

decrease more than the length of the steel strip−resulting in a tighter curl.

6 • Metal A has a coefficient of linear expansion that is three times the

coefficient of linear expansion of metal B. How do their coefficients of volumeexpansion β compare? (a)

β

A = β

B, (b) β

A = 3 β

B, (c) β

A = 6 β

B, (d ) β

A = 9 β

B,

(e) You cannot tell from the data given.

Determine the Concept We know that the coefficient of volume expansion isthree times the coefficient of linear expansion and so can use this fact to express

the ratio of A β to B β .

Express the coefficient of volumeexpansion of metal A in terms of its

coefficient of linear expansion:

AA 3α β =

Express the coefficient of volumeexpansion of metal B in terms of itscoefficient of linear expansion:

BB 3α β =

Dividing the first of these equations by the second yields:

B

A

B

A

B

A

3

3

α

α

α

α

β

β ==

Because BA 3 = :3

3

B

B

B

A ==α

α

β

β ⇒ ( )b is correct.

7 • The summit of Mount Rainier is 14 410 ft above sea level.Mountaineers say that you cannot hard boil an egg at the summit. This statementis true because at the summit of Mount Rainier (a) the air temperature is too lowto boil water, (b) the air pressure is too low for alcohol fuel to burn, (c) thetemperature of boiling water is not hot enough to hard boil the egg, (d ) the oxygencontent of the air is too low to support combustion, (e) eggs always break in

climbers′ backpacks.

Determine the Concept Actually, an egg can be hard boiled, but it takes quite a

bit longer than at sea level. )(c is the best response.

8 • Which gases in Table 20-3 cannot be condensed by applying pressureat 20ºC? Explain your answer.

Determine the Concept Gases that cannot be liquefied by applying pressure at

20°C are those for which T c < 293 K. These are He, Ar, Ne, H2, O2, NO.

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Thermal Properties and Processes 1865

9 •• [SSM] The phase diagram in Figure 20-15 can be interpreted toyield information on how the boiling and melting points of water change withaltitude. (a) Explain how this information can be obtained. (b) How might thisinformation affect cooking procedures in the mountains?

Determine the Concept

(a) With increasing altitude, decreases; from curve OC, the temperature of the

liquid-gas interface decreases as the pressure decreases, so the boiling

temperature decreases. Likewise, from curve OB, the melting temperature

increases with increasing altitude.

(b) Boiling at a lower temperature means that the cooking time will have to be

increased.

10 •• Sketch a phase diagram for carbon dioxide using information fromSection 20-3.

Determine the Concept The following phase diagram for carbon dioxide wasconstructed using information in Section 20-3.

atm, P

K ,T

5.1

216.6 304.2

Gas

Gas

SolidVapor

Liquid

pointCritical

11 •• Explain why the carbon dioxide on Mars is found in the solid state inthe polar regions even though the atmospheric pressure at the surface of Mars isonly about 1 percent of the atmospheric pressure at the surface of Earth.

Determine the Concept At very low pressures and temperatures, carbon dioxide

can exist only as a solid or gas (or vapor above the gas). The atmosphere of Marsis 95 percent carbon dioxide. Mars, on average, is warm enough so that theatmosphere is mostly gaseous carbon dioxide. The polar regions are cold enoughto enable solid carbon dioxide (dry ice) to exist, even at the low pressure.

12 •• Explain why decreasing the temperature of your house at night inwinter can save money on heating costs. Why doesn’t the cost of the fuelconsumed to heat the house back to the daytime temperature in the morning equal

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Chapter 201866

the savings realized by cooling it down in the evening and keeping it coolthroughout the night?

Determine the Concept The amount of heat lost by the house is proportional to

the difference between the temperature inside the house and that of the outside air.

Hence, the rate at which the house loses heat (that must be replaced by the

furnace) is greater at night when the temperature of the house is kept high than

when it is allowed to cool down.

13 •• [SSM] Two solid cylinders made of materials A and B have thesame lengths; their diameters are related by d A = 2d B. When the same temperaturedifference is maintained between the ends of the cylinders, they conduct heat atthe same rate. Their thermal conductivities are therefore related by which of thefollowing equations? (a) k A = k B/4, (b) k A = k B/2, (c) k A = k B, (d ) k A = 2k B,(e) k A = 4k BB

Picture the Problem The rate at which heat is conducted through a cylinder isgiven by xT kAdt dQ I ΔΔ== // (see Equation 20-7) where A is the cross-

sectional area of the cylinder.

The heat current in cylinder A is the

same as the heat current in cylinder

B:

BA I I =

Substituting for the heat currents

yields: L

T Ak

L

T Ak

Δ=

ΔBBAA ⇒

A

BBA

A

Ak k =

Because d A = 2d B:⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

B

BBA

4 A

Ak k ⇒ B4

1A k k =

)(a is correct.

14 •• Two solid cylinders made of materials A and B have the samediameter; their lengths are related by LA = 2 LB. When the same temperaturedifference is maintained between the ends of the cylinders, they conduct heat atthe same rate. Their thermal conductivities are therefore related by which of thefollowing equations? (a) k A = k B/4, (b) k A = k B/2, (c) k A = k B, (d ) k A = 2k B,(e) k A = 4k B. B

Determine the Concept We can use the expression for the heat current in aconductor, Equation 20-7, to relate the heat current in each cylinder to its thermalconductivity, cross-sectional area, temperature difference, and length.

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The heat current in cylinder A is the

same as the heat current in cylinder

B:

BA I I =

Substituting for the heat currents

yields: B

B

A

A

L

T Ak

L

T Ak

Δ=

Δ⇒

B

ABA

L

Lk k =

Because LA = 2 LB:B

B

BBA 2

2k

L

Lk k == ⇒ ( )d is correct.

15 •• If you feel the inside of a single pane window during a very cold day,it is cold, even though the room temperature can be quite comfortable. Assuming

the room temperature is 20.0°C and the outside temperature is 5.0°C, Construct a plot of temperature versus position starting from a point 5.0 m in behind thewindow (inside the room) and ending at a point 5.0 m in front of the window.Explain the heat transfer mechanisms that occur along this path.

Determine the Concept The temperatures on both sides of the glass are almostthe same. Because glass is an excellent conductor of heat, there need not be ahuge temperature difference. Thus the temperature must drop quickly as you nearthe pane on the warm side and the same on the outside. This is sketchedqualitatively in the following diagram. Convection and radiation are primarilyresponsible for heat transfer on the inside and outside, and it is mainly conductionthrough the glass. Conduction through the interior and exterior air is minimal.

C, °T

20

5

Inside Outside

m, x

50

16 •• During the thermal retrofitting of many older homes in California, itwas found that the 3.5-in-deep spaces between the wallboards and the outer

sheathing were filled with just air (no insulation). Filling the space with insulatingmaterial certainly reduces heating and cooling costs; although, the insulatingmaterial is a better conductor of heat than air is. Explain why adding theinsulation is a good idea.

Determine the Concept The tradeoff is the reduction of convection cells betweenthe walls by putting in the insulating material, versus a slight increase inconductivity. The net reduction in convection results in a higher R value.

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Chapter 201868

Estimation and Approximation

17 •• [SSM] You are using a cooking pot to boil water for a pasta dish.The recipe calls for at least 4.0 L of water to be used. You fill the pot with 4.0 Lof room temperature water and note that this amount of water filled the pot to the brim. Knowing some physics, you are counting on the volume expansion of thesteel pot to keep all of the water in the pot while the water is heated to a boil. Isyour assumption correct? Explain. If your assumption is not correct, how muchwater runs over the sides of the pot due to the thermal expansion of the water?

Determine the Concept The volume of water overflowing is the difference between the change in volume of the water and the change in volume of the pot.See Table 20-1 for the coefficient of volume expansion of water and thecoefficient of linear expansion of steel.

Express the volume of water that

overflows when the pot and the waterare heated:( ) T V

T V T V

V V V

Δ

ΔΔ

ΔΔ

0steelOH

0steel0OH

potOHovefrlow

2

2

2

β β

β β

−=−=

−=

Because the coefficient of volumeexpansion of steel is three times itscoefficient of linear expansion:

steelsteel 3α β =

Substituting for andOH2 β steel β yields: T V V Δ3 0steelOHoverflow 2

α α −=

Substitute numerical values and evaluate :overflowV

( )( )( )( ) mL56C20C100L4.0K 10113K 10207.0 1613

overflow =°−°×−×= −−−−V

Your assumption was not correct and 56 mL of water overflowed.

18 •• Liquid helium is stored in containers fitted with 7.00-cm-thick″superinsulation″ consisting of numerous layers of very thin aluminized Mylarsheets. The rate of evaporation of liquid helium in a 200-L container is about0.700 L per day when the container is stored at room temperature (20ºC). Thedensity of liquid helium is 0.125 kg/L and the latent heat of vaporization is

21.0 kJ/kg. Estimate the thermal conductivity of the superinsulation.

Picture the Problem We can express the heat current through the insulation

in terms of the rate of evaporation of the liquid helium and in terms of the

temperature gradient across the superinsulation. Equating these equations will

lead to an expression for the thermal conductivity k of the superinsulation. Note

that the boiling temperature of liquid helium is 4.2 K.

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Express the heat current in terms of

the rate of evaporation of the liquid

helium:

dt

dm L I v=

Express the heat current in terms of

the temperature gradient across thesuperinsulation and the conductivity

of the superinsulation:

x

T kA I

Δ

Δ=

Equate these expressions and solve

for k to obtain:T A

dt

dm x L

k Δ

Δ=

v

Using the definition of density,

express the rate of loss of liquid

helium:

dt

dV

dt

dm ρ =

Substitute fordt

dmto obtain:

T A

dt

dV x L

k Δ

Δ=

ρ v

Express the ratio of the area of the

spherical container to its volume: 3

34

24

r

r

V

A

π

π = ⇒ 3 236 V A π =

Substituting for A yields:

T V dt

dV x L

k Δ

Δ= 3 2

v

36π

ρ

Substitute numerical values and evaluate k :

( )

( ) ( ) K m

W1012.3

K 289m1020036

s86400

m100.700

m

kg125m107.00

kg

kJ21.0

6

3 233

33

3

2

⋅×=

×

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ×⎟ ⎠

⎞⎜⎝

⎛ ×⎟⎟ ⎠

⎞⎜⎜⎝

⎛

= −

−

−−

π

k

19 •• [SSM] Estimate the thermal conductivity of human skin.

Picture the Problem We can use the thermal current equation for the thermalconductivity of the skin. If we model a human body as a rectangular

parallelepiped that is 1.5 m high × 7 cm thick × 50 cm wide, then its surface areais about 1.8 m2. We’ll also assume that a typical human, while resting, producesenergy at the rate of 120 W, that normal internal and external temperatures are

33°C and 37°C, respectively, and that an average skin thickness is 1.0 mm.

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Chapter 201870

Use the thermal current equation toexpress the rate of conduction ofthermal energy:

I = kAΔT

Δ ⇒

x

T A

I k

ΔΔ

=

Substitute numerical values and

evaluate k : ( ) K m

mW

17

m100.1

C33C37m8.1

W120

3

2 ⋅=×

°−°=−

k

20 •• You are visiting Finland with a college friend and have met someFinnish friends. They talk you into taking part in a traditional Finnish after-saunaexercise which consists of leaving the sauna, wearing only a bathing suit, andrunning out into the mid-winter Finnish air. Estimate the rate at which youinitially lose energy to the cold air. Compare this rate of initial energy loss to theresting metabolic rate of a typical human under normal temperature conditions.Explain the difference.

Picture the Problem We can use the Stefan-Boltzmann law to estimate the rate atwhich you lose energy when you first step out of the sauna. If we model a human

body as a rectangular parallelepiped that is 1.5 m high × 7 cm thick × 50 cm wide,then its surface area is about 1.8 m2. Assume that your skin temperature is initially

37°C (310 K), that the mid-winter outside temperature is −10°C (263 K), and thatthe emissivity of your skin is 1.

Use the Stefan-Boltzmann law toexpress the net rate at which youradiate energy to the cold air:

( )4

air

4

skinnet T T A P −= εσ

Substitute numerical values and evaluate P net:

( ) ( ) ( ) ( )( ) W450K 632K 310m8.1K m

W 10670.51

442

42

8

net =−⎟ ⎠

⎞⎜⎝

⎛ ⋅

×= − P

This result is almost four times greater than the basal metabolic rate of 120 W.We can understand the difference in terms of the temperature of your skin whenyou first step out of the sauna and the fact that radiation loses are dependent onthe fourth power of the absolute temperature.

Remarks: The emissivity of your skin is, in fact, very close to 1.

21 •• Estimate the rate of heat conduction through a 2.0-in-thick woodendoor during a cold winter day in Minnesota. Include the brass doorknob. What isthe ratio of the heat that escapes through the doorknob to the heat that escapesthrough the whole door? What is the total overall R-factor for the door, including

the knob? The thermal conductivity of brass is ( )K mW85 ⋅ .

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Picture the Problem We can use the thermal current equation (Equation 20-7) toestimate the rate of heat loss through the door and its knob. We’ll assume that thedoor is made of oak with an area of 2.0 m2, that the knob is made of brass, that the

conducting path has a diameter of 3.0 cm, and that the inside temperature is 20°C.

Take the outside temperature to be −20°C. The reciprocal of the equivalent R-factor is the sum of the reciprocals of the R-values of the door and the knob.

The total thermal current through thedoor is the sum of the thermalcurrents through the door and theknob: knob

knobknob

door

door door

knobdoor tot

Δ

Δ

Δ

Δ

x

T Ak

x

T Ak

I I I

+=

+=

Assume that to

obtain:

x x x ΔΔΔ knobdoor ==

( ) x

T Ak Ak I

Δ

Δ

knobknobdoor door tot +=

Substitute numerical values and evaluate I tot:

( ) ( ) ( )

( )

kW28.0

in39.37

m1in0.2

C20C20m100.3

4K m

W85m0.2

K m

W15.0

222

tot

=

⎟ ⎠

⎞⎜⎝

⎛ °−−°

⎥⎦

⎤⎢⎣

⎡ ×⎟ ⎠ ⎞

⎜⎝ ⎛

⋅+⎟

⎠ ⎞

⎜⎝ ⎛

⋅= −π

I

Express the ratio of the thermalcurrents through the knob and thedoor:

knob

knobknob

door

door door

door

knob

Δ

Δ

Δ

Δ

⎟ ⎠ ⎞

⎜⎝ ⎛

⎟ ⎠

⎞⎜⎝

⎛

=

x

T Ak

x

T Ak

I

I

Because the temperature differenceacross the door and the knob are thesame and we’ve assumed that thethickness of the door and length ofthe knob are the same:

knobknob

door door

door

knob

Ak

Ak

I

I =

Substitute numerical values toobtain: ( )

( )0.5

m100.34K m

W85

m0.2K m

W15.0

22

2

door

knob

=

×⎟ ⎠

⎞⎜⎝

⎛

⋅

⎟ ⎠

⎞⎜⎝

⎛ ⋅=

−π I

I

Relate the equivalent R-factor to the R-factors of the door and knob;

knobdoor eq

111

R R R+=

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Chapter 201872

Substitute for Rdoor and Rknob toobtain:

x

Ak Ak

Ak

x

Ak

x R

Δ

Δ

1

Δ

11

knobknobdoor door

knobknobdoor door

eq

+=

+=

Solving for Req yields:

knobknobdoor door

eq

Δ

Ak Ak

x R

+=

Substitute numerical values and evaluate Req:

( )

( ) ( ) W

K 14.0

m100.34K m

W85m0.2

K m

W15.0

in39.37

m1in0.2

222eq =

×⎟ ⎠

⎞⎜⎝

⎛ ⋅

+⎟ ⎠

⎞⎜⎝

⎛ ⋅

⎟ ⎠ ⎞

⎜⎝ ⎛

=−π

R

22 •• Estimate the effective emissivity of Earth, given the followinginformation. The solar constant, which is the intensity of radiation incident onEarth from the Sun, is about 1.37 kW/m

2. Seventy percent of this energy is

absorbed by Earth, and Earth’s average surface temperature is 288 K. (Assume

that the effective area that is absorbing the light is π R2, where R is Earth’s radius,

while the blackbody-emission area is 4π R2.)

Picture the Problem The amount of heat radiated by Earth must equal the solarflux from the Sun, or else the temperature on Earth would continually increase.The emissivity of Earth is related to the rate at which it radiates energy into space

by the Stefan-Boltzmann law . 4r AT e P σ =

Using the Stefan-Boltzmann law,express the rate at which Earthradiates energy as a function of itsemissivity e and temperature T :

4

r A'T e P σ = ⇒4

r

A'T

P e

σ =

where A′ is the surface area of Earth.

Use its definition to express theintensity of the radiation received byEarth:

A

P I absorbed=

where A is the cross-sectional area of

Earth.

For 70% absorption of the Sun’sradiation incident on Earth:

( ) A

P I r 70.0

=

Substitute for P r and A in theexpression for e and simplify toobtain:

( ) ( ) ( )442

2

4 4

70.0

4

70.070.0

T

I

T R

I R

A'T

AI e

σ σ π

π

σ ===

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Substitute numerical values andevaluate e:

( )( )( )( )

61.0

K 288K W/m10670.54

kW/m37.170.04428

2

=

⋅×=

−e

23 •• Black holes are highly condensed remnants of stars. Some blackholes, together with a normal star, form binary systems. In such systems the black hole and the normal star orbit about the center of mass of the system. Oneway black holes can be detected from Earth is by observing the frictional heatingof the atmospheric gases from the normal star that fall into the black hole. These

gases can reach temperatures greater than 1.0 × 106 K. Assuming that the falling

gas can be modeled as a blackbody radiator, estimate λ max for use in anastronomical detection of a black hole. (Remark: This is in the X-ray region ofthe electromagnetic spectrum.)

Picture the Problem The wavelength at which maximum power is radiated bythe gas falling into a black hole is related to its temperature by Wien’sdisplacement law.

Wien’s displacement law relates thewavelength at which maximum poweris radiated by the gas to its temperature:

T

K mm898.2max

⋅=λ

Substitute for T and evaluate λ max: nm9.2K 100.1

K mm898.26max =

×⋅

=λ

24 ••• Your cabin in northern Michigan has walls that consist of pine logs

that have average thicknesses of about 20 cm. You decide to finish the interior ofthe cabin to improve the look and to increase the insulation of the exterior walls.You choose to buy insulation with an R-factor of 31 to cover the walls. Inaddition, you cover the insulation with 1.0-in-thick gypsum wallboard. Assumingheat transfer is only due to conduction, estimate the ratio of thermal currentthrough the walls during a cold winter night before the renovation to the thermalcurrent through the walls following the renovation.

Picture the Problem We can use the thermal current equation to find the thermalcurrent per square meter through the walls of the cabin both before and after thewalls have been insulated. See Table 20-5 for the R-factor of gypsum wallboard

and Table 20-4 for the thermal conductivity of white pine.

The rate at which heat is conductedthrough the walls is given by thethermal current equation:

before

before before

Δ

Δ

Δ

R

T

x

T Ak I ==

With the insulation in place:

after

after after

Δ

Δ

Δ

R

T

x

T Ak I ==

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Chapter 201874

Divide the second of these equations by the first to obtain:

after

before

before

after

before

after

Δ

Δ

R

R

R

T

R

T

I

I ==

before R is the R-factor for pine andis the sum of the R-factors for

pine, the insulating material, and thegypsum board:

after R

pine before R R = and

gypsum31 pineeqafter R R R R R ++==

Substitute for and to

obtain: before R after R

gypsum31 pine

pine

before

after

R R R

R

I

I

++=

Because pine

pinek

x R

Δ= :

gypsum31

pine

pine

before

after

R Rk

x

k

x

I

I

++Δ

Δ

=

Substitute numerical values and evaluate the ratio : beforeafter / I I

%24

Btu

Ffth32.0

in0.375

in1

Btu

Ffth31

Ffth

inBtu 78.0

cm2.54

in1 cm20

Ffth

inBtu 78.0

cm2.54

in1 cm20

22

2

2

before

after =

°⋅⋅×+°⋅⋅+

°⋅⋅⋅

×

°⋅⋅⋅

×

=

I

I

25 ••• [SSM] You are in charge of transporting a liver from New York, New York to Los Angeles, California for a transplant surgery. The liver is keptcold in a Styrofoam ice chest initially filled with 1.0 kg of ice. It is crucial that theliver temperature is never warmer than 5.0°C. Assuming the trip from the hospitalin New York to the hospital in Los Angeles takes 7.0 h, estimate the R-value the

Styrofoam walls of the ice chest must have.

Picture the Problem The R factor is the thermal resistance per unit area of a slabof material. We can use the thermal current equation to express the thermalresistance of the styrofoam in terms of the maximum amount of heat that can

enter the chest in 7.0 h without raising the temperature above 5.0°C. We’llassume that the surface area of the ice chest is 1.0 m2 and that the ambient

temperature is 25°C

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The R-factor needed for thestyrofoam walls of the ice chest is the product of their thermal resistanceand area:

RA R =f (1)

Use the thermal current equation to

express R: tottot

ΔΔ

Δ

ΔΔ

Q

t T

t

Q

T

I

T

R ===

Substitute for R in equation (1) toobtain:

tot

f

ΔΔ

Q

t T A R =

The total heat entering the chest in7 h is given by:

OHOHicef ice

water icewarm

icemelttot

22ΔT cm Lm

QQQ

+=

+=

Substitute for Qtot and simplify toobtain: ( )OHOHf ice

f

22Δ

ΔΔ

T c Lm

t T A R+

=

Substitute numerical values and evaluate Rf :

( )

( ) ( )Btu

fthF8

C5K kg

kJ18.4

kg

kJ5.333kg1

Btu

J35.1054h7

C5

F9C20

m1029.9

ft1m1.0

222

22

f

⋅⋅°≈

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ °⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⋅

+

⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ °°

×°⎟⎟ ⎠

⎞⎜⎜⎝

⎛

××

=−

R

Thermal Expansion

26 •• You have inherited your grandfather’s grandfather clock that wascalibrated when the temperature of the room was 20ºC. Assume that the pendulumconsists of a thin brass rod of negligible mass with a compact heavy bob at itsend. (a) During a hot day, the temperature is 30ºC, does the clock run fast orslow? Explain. (b) How much time does it gain or lose during this day?

Picture the Problem We can determine whether the clock runs fast or slow from

the expression for the period of a simple pendulum and the dependence of its

length on the temperature. We can use the expression for the period of a simple

pendulum and the equation describing its length as a function of temperature to

find the time gained or lost in a 24-h period.

(a) Express the period of the

pendulum in terms of its length: g

LT π 2P =

8/12/2019 Cap20 Prob


Chapter 201876

Because LT ∝P and L is temperature dependent, the clock runs slow.

(b) The period of the pendulum when

the temperature is 20°C is given by: g

LT 20

20 2π = (1)

When the temperature increases to30°C, the period of the pendulum

increases due to the increase in its

length:

( )

C20

C20

1

122

t T

g

t L

g

LT

Δ+=

Δ+==

α

α π π

The daily fractional gain or loss is

given by :20

20

20 T

T T

T

T −=

Δ=

Δτ

τ

Substituting for T and simplifying

yields:

11

1

C

20

20C20

−Δ+=

−Δ+=

Δ

t T

T t T

α

α

τ

τ

Solve for Δτ to obtain: τ α τ 11 C −Δ+=Δ t

Substitute numerical values and Δτ :

( )( )( ) s2.8h

s3600h241C20C30K 10191 16 =⎟

⎠

⎞⎜⎝

⎛ ×−°−°×+=Δ −−τ

27 •• [SSM] You need to fit a copper collar tightly around a steel shaft thathas a diameter of 6.0000 cm at 20ºC. The inside diameter of the collar at thattemperature is 5.9800 cm. What temperature must the copper collar have so that itwill just slip on the shaft, assuming the shaft itself remains at 20ºC?

Picture the Problem Because the temperature of the steel shaft does not change,

we need consider just the expansion of the copper collar. We can express the

required temperature in terms of the initial temperature and the change in

temperature that will produce the necessary increase in the diameter D of the

copper collar. This increase in the diameter is related to the diameter at 20°C and

the increase in temperature through the definition of the coefficient of linearexpansion.

Express the temperature to which the

copper collar must be raised in terms

of its initial temperature and the

increase in its temperature:

T T T Δ+= i

8/12/2019 Cap20 Prob



Apply the definition of the

coefficient of linear expansion to

express the change in temperature

required for the collar to fit on the

shaft:

D

D D

D

T α α

Δ=

⎟ ⎠ ⎞

⎜⎝ ⎛ Δ

=Δ

Substitute for ΔT to obtain:

D

DT T

α

Δ+= i


evaluate T : ( )( )

C220

cm5.9800K 1017

cm9800.5cm0000.6C20

16

°=

×−

+°=−−

T

28 •• You have a copper collar and a steel shaft. At 20°C, the collar has aninside diameter of 5.9800 cm and the steel shaft has diameter of 6.0000 cm. Thecopper collar was heated. When its inside diameter exceeded 6.0000 cm is wasslipped on the shaft. The collar fitted tightly on the shaft after they cooled to roomtemperature. Now, several years later, you need to remove the collar from theshaft. To do this you heat them both until you can just slip the collar off the shaft.What temperature must the collar have so that the collar will just slip off theshaft?

Picture the Problem Because the temperatures of both the steel shaft and the

copper collar change together, we can find the temperature change required for the

collar to fit the shaft by equating their diameters for a temperature increase ΔT.

These diameters are related to their diameters at 20°C and the increase intemperature through the definition of the coefficient of linear expansion.

Express the temperature to which the

collar and the shaft must be raised in

terms of their initial temperature and

the increase in their temperature:

T T T Δ+= i (1)

Express the diameter of the steel

shaft when its temperature has been

increased by ΔT :

( )T D D Δ+= ° steelCsteel,20steel 1 α

Express the diameter of the copper

collar when its temperature has been

increased by ΔT :

( )T D D Δ+= ° CuCCu,20Cu 1 α

8/12/2019 Cap20 Prob


Chapter 201878

If the collar is to fit over the shaft

when the temperature of both has

been increased by ΔT :

( )

( )T D

T D

Δ+=

Δ+

°

°

steelCsteel,20

CuCCu,20

1

1

α

α

Solving for ΔT yields:

steelCsteel,20CuCCu,20

CCu,20Csteel,20

α α °°

°°

−

−=Δ

D D

D DT

Substitute in equation (1) to obtain:

steelCsteel,20CuCCu,20

CCu,20Csteel,20

iα α °°

°°

−

−+=

D D

D DT T

Substitute numerical values and evaluate T :

( )( ) ( )( )C580

/K 1011cm6.0000/K 1017cm5.9800

cm5.9800cm6.0000C20

66 °=

×−×−

+°=−−

T

29 •• A container is filled to the brim with 1.4 L of mercury at 20ºC. As thetemperature of container and mercury is increased to 60ºC, a total of 7.5 mL ofmercury spill over the brim of the container. Determine the linear expansioncoefficient of the material that makes up the container.

Picture the Problem The linear expansion coefficient of the container is one-

third its coefficient of volume expansion. We can relate the changes in volume of

the mercury and the container to their initial volumes, temperature change, and

coefficients of volume expansion, and, because we know the amount of spillage,

obtain an equation that we can solve for β c.

Relate the linear expansion

coefficient of the container to its

coefficient of volume expansion:

c31

c β α = (1)

Express the difference in the change

in the volume of the mercury and the

container in terms of the spillage:

mL5.7cHg =Δ−Δ V V (2)

Express using the definition of

the coefficient of volume expansion:Hg

V Δ

T V V Δ=ΔHgHgHg

β

Express using the definition of

the coefficient of volume expansion:

cV Δ

T V V Δ=Δ ccc β

8/12/2019 Cap20 Prob



Substitute for and in

equation (2) to obtain:

HgV Δ cV Δ

mL5.7ccHgHg =Δ−Δ T V T V β β

Solving for β c yields:

T V

T V

Δ

mL5.7Δ

c

HgHg

c

−= β

β

or, because V = V Hg = V c,

T V

T V

T V

Δ

mL5.7

Δ

mL5.7Δ

Hg

Hg

c

−=

−=

β

β β

Substitute for c β in equation (1) to

obtain: T V T V Δ3

mL5.7

Δ3

mL5.7HgHg3

1c −=−= α β α


evaluate α c:( )

( )( )16

13

31

c

K 1015

K 40L4.13

mL5.7K 1018.0

−−

−−

×=

−×=α

30 •• A car has a 60.0-L steel gas tank filled to the brim with 60.0-L ofgasoline when the temperature of the outside is 10ºC. The coefficient of volume

expansion for gasoline at 20°C is 0.950 × 10−3 K −1. How much gasoline spills outof the tank when the outside temperature increases to 25ºC? Take the expansionof the steel tank into account.

Picture the Problem The amount of gas that spills is the difference between the

change in the volume of the gasoline and the change in volume of the tank. We

can find this difference by expressing the changes in volume of the gasoline and

the tank in terms of their common volume at 10°C, their coefficients of volume

expansion, and the change in the temperature.

Express the spill in terms of the

change in volume of the gasoline and

the change in volume of the tank:

tank gasolinespill V V V Δ−Δ=

Relate to the coefficient

of volume expansion for

gasoline:

gasolineV Δ

T V V Δ=Δ gasolinegas β

8/12/2019 Cap20 Prob


Chapter 201880

Relate to the coefficient of

linear expansion for steel:

tank V Δ

T V V Δ=Δ tank tank β

or, because β steel = 3α steel,

T V V Δ=Δ steeltank 3α

Substitute for andand simplify to obtain:

gasolineV Δ tank V Δ

( )steelgasoline

steelgasspill

33

α β α β −Δ=

Δ−Δ=T V

T V T V V

Substitute numerical values and evaluate :spillV

( )( )[ ( )] L8.0K 10113K 10950.0C01C25L60.0 1613

spill ≈×−×°−°= −−−−V

31 ••• What is the tensile stress in the copper collar of Problem 27 when itstemperature returns to 20ºC?

Picture the Problem We can use the definition of Young’s modulus to express

the tensile stress in the copper in terms of the strain it undergoes as its

temperature returns to 20°C. We can show that Δ L/ L for the circumference of the

collar is the same as Δd /d for its diameter.

Using Young’s modulus, relate the

stress in the collar to its strain:

where L20°C is the circumference of the

collar at 20°C.

Express the circumference of the

collar at the temperature at which

it fits over the shaft:

Express the circumference of the

collar at 20 °C:

Substitute for and and

simplify to obtain:

T L C20° L

C20

StrainStress°

Δ=×=

L

LY Y

T T d L π =

C20C20 °° = d L π

C20

C20

C20

C20Stress

°

°

°

°

−=

−=

d

d d Y

d

d d Y

T

T

π

π π

8/12/2019 Cap20 Prob



Substitute numerical values and evaluate the stress:

( ) 212210 N/m107.3cm5.9800

cm9800.5cm0000.6 N/m1011Stress −− ×=

−×=

The van der Waals Equation, Liquid-Vapor Isotherms, and Phase

Diagrams

32 • (a) Calculate the volume of 1.00 mol of an ideal gas at a temperatureof 100ºC and a pressure of 1.00 atm. (b) Calculate the temperature at which1.00 mol of steam at a pressure of 1.00 atm has the volume calculated in Part (a).

Use a = 0.550 Pa⋅m6/mol

2 and b = 30.0 cm

3/mol.

Picture the Problem We can apply the ideal-gas law to find the volume of 1.00

mol of steam at 100°C and a pressure of 1.00 atm and then use the van der Waals

equation to find the temperature at which the steam will this volume.

(a) Solving the ideal-gas law for the

volume gives: P

nRT V =


evaluate V :

( )( )( )

L6.30m10

L1m1006.3

atm

kPa101.325atm00.1

K 373K J/mol314.8mol00.1

33

32

=

××=

×

⋅=

−−

V

(b) Solve van der Waals equation

for T to obtain:( )

nR

bnV V

an P

T

−⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +

=2

2


( )( )( )

( ) ( )( ) ( )

K 375

K J/mol8.314mol1.00

mol00.1/molm1030.0m103.06

m103.06

mol1.00/molmPa0.550kPa325.101

3632

232

226

=

⋅×−×

×

⎟⎟ ⎠ ⎞

⎜⎜⎝ ⎛

×⋅+=

−−

−T

8/12/2019 Cap20 Prob


Chapter 201882

33 •• [SSM] Using Figure 20-16, find the following quantities. (a) Thetemperature at which water boils on a mountain where the atmospheric pressure is70.0 kPa, (b) the temperature at which water boils in a container where the pressure inside the container is 0.500 atm, and (c) the pressure at which water boils at 115ºC.

Picture the Problem Consulting Figure 20-16, we see that:

(a) At 70.0 kPa, water boils at approximately C90° .

(b) At 0.500 atm (about 51 kPa), water boils at approximately C78° .

(c) The pressure at which water boils at 115°C is approximately kPa127 .

34 •• The van der Waals constants for helium are a = 0.03412 L

2

⋅atm/mol

2

and b = 0.0237 L/mol. Use these data to find the volume in cubic centimetersoccupied by one helium atom. Then, estimate the radius of the helium atom.

Picture the Problem Assume that a helium atom is spherical. Then we can find

its volume from the van der Waals equation and its radius from 3

34 r V π = .

In the van der Waals equation, b is the

volume of 1 mol of molecules. For He,

1 molecule = 1 atom. Use Avogadro’s

number to express b in cm3

/atom:

atom

cm 1094.3

mol

atoms 106.022

L

cm10

mol

L 0237.0

323

23

33

−×=

×

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛

=b

The volume of a spherical helium

atom is given by:3

34 r V π = ⇒ 33

4

3

4

3

π π

bV r ==


evaluate r :( )

nm211.04

cm1094.333

323

=×

=−

π r

Conduction

35 • [SSM] A 20-ft × 30-ft slab of insulation has an R factor of 11. Atwhat rate is heat conducted through the slab if the temperature on one side is aconstant 68ºF and the temperature of the other side is a constant 30ºF?

8/12/2019 Cap20 Prob



Picture the Problem We can use its definition to express the thermal current in

the slab in terms of the temperature differential across it and its thermal resistance

and use the definition of the R factor to express I as a function of ΔT , the cross-

sectional area of the slab, and Rf .

Express the thermal current throughthe slab in terms of the temperature

difference across it and its thermal

resistance:

RT I Δ=

Substitute to express R in terms of

the insulation’s R factor: f f / R

T A

A R

T I

Δ=

Δ=


evaluate I :

( )( )( )

h

kBtu2.1

Btu

Ffth11

F30F68ft30ft202

=

°⋅⋅°−°

= I

36 •• A copper cube and an aluminum, cube each with 3.00-cm-long edges,are arranged as shown in Figure 20-17. Find (a) the thermal resistance of eachcube, (b) the thermal resistance of the two-cube combination, (c) the thermalcurrent I , and (d ) the temperature at the interface of the two cubes.

Picture the Problem We can use kA x R Δ= to find the thermal resistance of

each cube and the fact that they are in series to find the thermal resistance of thetwo-cube system. We can use RT I Δ= to find the thermal current through the

cubes and the temperature at their interface. See Table 20-4 for the thermal

conductivities of copper and aluminum.

(a) Using its definition, express the

thermal resistance of each cube: kA

x R

Δ=


evaluate the thermal resistance of

the copper cube: ( )

K/W0831.0K/W08313.0

cm3.00K m

W

401

cm3.00

2Cu

==

⎟ ⎠

⎞

⎜⎝

⎛

⋅

= R

8/12/2019 Cap20 Prob


Chapter 201884


evaluate the thermal resistance of

the aluminum cube:( )

K/W141.0K/W1406.0

cm3.00K m

W372

cm3.00

2Al

==

⎟ ⎠

⎞⎜⎝

⎛ ⋅

= R

(b) Because the cubes are in series,

their thermal resistances are additive:

K/W0.224K/W0.2237

K/W0.1406K/W0.08313

AlCu

==

+=

+= R R R

(c) Using its definition, find the

thermal current:

kW0.36W6.357

K/W0.2237

C20C100Δ

==

°−°==

R

T I

(d ) Express the temperature at the

interface between the two cubes:Cuinterface ΔC100 T T −°=

Express the temperature differential

across the copper cube:CuCuCuCu IR R I T ==Δ

Substitute for ΔT Cu to obtain: Cuinterface C100 IRT −°=


evaluate T interface: ( )( )C70

K/W08313.0W357.6

C100interface

°≈−

°=T

37 •• Two metal cubes, one copper and one aluminum, with 3.00-cm-longedges, are arranged in parallel, as shown in Figure 20-18. Find (a) the thermalcurrent in each cube, (b) the total thermal current, and (c) the thermal resistance ofthe two-cube combination.

Picture the Problem We can use Equation 20-9 to find the thermal current in

each cube. Because the currents are additive, we can find the equivalent resistance

of the two-cube system from totaleq I T R Δ= .

(a) The thermal current in each cube

is given by Equation 20-9: x

T kA

R

T I

ΔΔ

=Δ

=

8/12/2019 Cap20 Prob



Substitute numerical values and evaluate the thermal current in the copper

cube:

( ) kW96.0W4.962

cm3.00

C20C100cm3.00

K m

W401

2

Cu ==⎟⎟

⎠

⎞⎜⎜

⎝

⎛ °−°⎟

⎠

⎞⎜

⎝

⎛

⋅

= I

Substitute numerical values and evaluate the thermal current in the

aluminum cube:

( ) kW57.0W8.568cm3.00

C20C100cm3.00

K m

W372

2

Al ==⎟⎟ ⎠

⎞⎜⎜⎝

⎛ °−°⎟ ⎠

⎞⎜⎝

⎛ ⋅

= I

(b) Because the cubes are in parallel,

their total thermal currents are

additive:kW1.5kW1.531

W8.685W4.629AlCu

==

+=+= I I I

(c) Use the relationship between the

total thermal current, temperature

differential and thermal resistance to

find Req:K/W0.052

kW1.531

C20C100Δ

total

eq

=

°−°==

I

T R

38 •• The cost of air conditioning a house is approximately proportional tothe rate at which heat is absorbed by the house from its surroundings divided bythe coefficient of performance (COP) of the air conditioner. Let us denote the

temperature difference between the inside temperature and the outsidetemperature as ΔT . Assuming that the rate at which heat is absorbed by a house is

proportional to ΔT and that the air conditioner is operating ideally, show that the

cost of air conditioning is proportional to (ΔT )2 divided by the temperature insidethe house.

Picture the Problem The cost of operating the air conditioner is proportional tothe energy used in its operation. We can use the definition of the COP to relate therate at which the air conditioner removes heat from the house to rate at which itmust do work to maintain a constant temperature differential between the interiorand the exterior of the house. To obtain an expression for the minimum rate at

which the air conditioner must do work, we’ll assume that it is operating with themaximum efficiency possible. Doing so will allow us to derive an expression forthe rate at which energy is used by the air conditioner that we can integrate toobtain the energy (and hence the cost of operation) required.

Relate the cost C of air conditioningthe energy W required to operate theair conditioner:

uW C = (1)where u is the unit cost of the energy.

8/12/2019 Cap20 Prob


Chapter 201886

Express the rate dQ/dt at which heatflows into a house provided thehouse is maintained at a constanttemperature:

T k dt

dQ P Δ==

where ΔT is the temperature difference between the interior and exterior of thehouse.

Use the definition of the COP torelate the rate at which the airconditioner must remove heat dW /dt

to maintain a constant temperature:

dt dW

dt dQ=COP ⇒

dt

dQ

dt

dW

COP

1=

Express the maximum value of theCOP: T

T

Δ= c

maxCOP

wherehouse

theinsidec T T = is the temperature

of the cold reservoir.

Letting COP = COPmax, substitute toobtain an expression for theminimum rate at which the airconditioner must do work in order tomaintain a constant temperature:

T T

dt

dQ

dt

dW Δ

c

=

Substituting for dQ/dt gives:( )2

cc

T T

k T

T

T k

dt

dW Δ=Δ

Δ=

Separate variables and integrate

this equation to obtain: ( ) ( ) t T T

k

dt' T T

k

W

t

ΔΔ=Δ= ∫

Δ2

c0

2

c

Substitute in equation (1) to obtain:( ) t T

T

k uC ΔΔ

2

c

= ⇒ ( )

c

2Δ

T

T C ∝

39 •• A spherical shell of thermal conductivity k has inside radius r 1 andoutside radius r 2 (Figure 20-19). The inside of the shell is held at a temperatureT 1, and the outside of the shell is held at temperature T 2, with T 1 < T 2. In this problem, you are to show that the thermal current through the shell is given by

I = − 4π kr 1r 2r

2 − r

1

T 2 − T 1( )

where I is positive if heat is transferred in the +r direction. Here is a suggested procedure for obtaining this result: (1) obtain an expression for the thermal current I through a thin spherical shell of radius r and thickness dr when there is atemperature difference dT across the thickness of the shell; (2) explain why thethermal current is the same through each such thin shell; (3) express the thermal

current I through such a shell element in terms of the area A = 4π r 2, the thickness

8/12/2019 Cap20 Prob



dr , and the temperature difference dT across the element; and (4) separatevariables (solve for dT in terms of r and dr ) and integrate.

Picture the Problem We can follow the step-by-step instructions given in the

problem statement to obtain the differential equation describing the variation of T

with r . Integrating this equation will yield an equation that we can solve for the

current I .

(1) The thermal current through a

thin spherical shell surface of area A

and thickness dr due to a

temperature gradientdr

dT is given by:

dr

dT kA I =

(2) Conservation of energy requires that the thermal current through each shell be

the same.

(3) For a shell of area A = 4π r 2:

dr

dT kr I 24π =

(4) Separating the variables yields:24 r

dr

k

I dT

π =

Integrate from r = r 1 to r = r 2 and

simplify to obtain: ∫∫ =2

1

2

1

24

r

r

T

T r

dr

k

I dT

π

and

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −−=⎥⎦

⎤⎢⎣

⎡−=−21

12

11

4

1

4

2

1r r k

I

r k

I T T

r

r π π

Solving for I gives:( )12

12

214T T

r r

r kr I −

−−= π

Radiation

40 • Calculate λ max (the wavelength at which the emitted power ismaximum) for a human skin. Assume the human skin is a blackbody emitter witha temperature of 33ºC.

Picture the Problem We can apply Wein’s displacement law to find the

wavelength at which the power is a maximum.

8/12/2019 Cap20 Prob


Chapter 201888

Wein’s law relates the maximum

wavelength of the radiation to the

temperature of its source:

T

K mm898.2max

⋅=λ


evaluate λ max:

m47.9

K 33K 273

K mm2.898max μ λ =

+

⋅=

41 • [SSM] The universe is filled with radiation that is believed to beremaining from the Big Bang. If the entire universe is considered to be a

blackbody with a temperature equal to 2.3 K, what is the λ max (the wavelength atwhich the power of the radiation is maximum) of this radiation?

Picture the Problem We can use Wein’s law to find the peak wavelength of thisradiation.


wavelength of the background

radiation to the temperature of the

universe:

T

K mm2.898max

⋅

=λ

Substituting for T gives:mm.31

K 3.2

K mm2.898max =

⋅=λ

42 • What is the range of temperatures for star surfaces for which λ max (thewavelength at which the power of the emitted radiation is maximum) is in thevisible range?

Picture the Problem The visible portion of the electromagnetic spectrum extendsfrom approximately 400 nm to 700 nm. We can use Wein’s law to find the rangeof temperatures corresponding to these wavelengths.


wavelength of the radiation to the

temperature of its source:

T

K mm2.898max

⋅=λ

Solving for T yields:

max

K mm2.898

λ

⋅=T

For λ max = 400 nm:K 7250

nm400

K mm2.898=

⋅=T

For λ max = 700 nm:K 4140

nm700

K mm2.898=

⋅=T

8/12/2019 Cap20 Prob



The range of temperatures is K 7250K 4140 ≤≤ T .

43 • The heating wires of a 1.00-kW electric heater are red hot at atemperature of 900ºC. Assuming that 100 percent of the heat released is due toradiation and that the wires act as blackbody emitters, what is the effective area of

the radiating surface? (Assume a room temperature of 20ºC.)

Picture the Problem We can apply the Stefan-Boltzmann law to find the net

power radiated by the wires of its heater to the room.

Relate the net power radiated to the

surface area of the heating wires,

their temperature, and the room

temperature:

4

0

4

net T T Ae P −= σ ⇒( )4

0

4

net

T T e

P A

−=

σ

Substitute numerical values and evaluate A:

( ) ( ) ( )[ ]2

44

42

8

cm5.93

K 293K 1173K m

W105.67031

kW1.00=

−⎟ ⎠

⎞⎜⎝

⎛ ⋅

×=

−

A

44 •• A blackened, solid copper sphere that has a radius equal to 4.0 cmhangs in an evacuated enclosure whose walls have a temperature of 20ºC. If thesphere is initially at 0ºC, find the initial rate at which its temperature changes,assuming that heat is transferred by radiation only. (Assume the sphere is a

blackbody emitter.)

Picture the Problem The rate at which the copper sphere absorbs radiant energyis given by and, from the Stephan-Boltzmann law,dt mcdT dt dQ // =

( )4

0

4

net T T Ae P −= σ where A is the surface area of the sphere, T 0 is its

temperature, and T is the temperature of the walls. We can solve the first equation

for dT /dt and substitute P net for dQ/dt in order to find the initial rate at which the

temperature of the sphere changes.

Relate the rate at which the sphere

absorbs radiant energy to the rate atwhich its temperature changes:

dt

dT mc

dt

dQ P ==net

Solving fordt

dT and substituting

for mc gives:

cr

P

Vc

P

mc

P

dt

dT

ρ π ρ 3

34

netnetnet ===

8/12/2019 Cap20 Prob


Chapter 201890

Apply the Stefan-Boltzmann law to

relate the net power radiated to the

sphere to the difference in

temperature of the walls and the

blackened copper sphere:

( )( )4

0

42

4

0

4

net

4 T T er

T T Ae P

−=

−=

σ π

σ

Substitute for to obtain:net P ( )

( )cr

T T e

cr

T T er

dt

dT

ρ

σ

ρ π

σ π

4

0

4

3

34

4

0

42

3

4

−=

−=

Substitute numerical values and evaluate dT /dt :

( ) ( ) ( )[ ]

( )K/s102.2

K kg

kJ0.386

m

kg108.93m104.0

K 273K 293

K m

W105.670313

3

3

32

44

42

8

−

−

−

×=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⋅

⎟ ⎠ ⎞

⎜⎝ ⎛ ××

−⎟

⎠

⎞⎜

⎝

⎛

⋅

×

−=dt dT

45 •• The surface temperature of the filament of an incandescent lamp is1300ºC. If the electric power input is doubled, what will the new temperature be? Hint: Show that you can neglect the temperature of the surroundings.

Picture the Problem We can apply the Stephan-Boltzmann law to express the net

power radiated by the incandescent lamp to its surroundings.

Express the rate at which energy is

radiated to the surroundings:

( )

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟ ⎠ ⎞

⎜⎝ ⎛ −=

−=4

04

4

0

4

net

1T

T AT e

T T Ae P

σ

σ

Evaluate

4

0 ⎟ ⎠

⎞⎜⎝

⎛ T

T : 3

44

0 101K 1573

K 293 −×≈⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =⎟

⎠

⎞⎜⎝

⎛ T

T

Because

4

0 ⎟ ⎠ ⎞⎜⎝ ⎛ T

T is so small, we can

neglect the temperature of the

surroundings. Hence:

4net AT e P σ ≈ ⇒

41

net ⎟ ⎠ ⎞⎜⎝ ⎛ = Ae P T σ

Express the temperature T ′when the

electric power input is doubled:

41

net2⎟ ⎠ ⎞

⎜⎝

⎛ = Ae

P T'

σ

8/12/2019 Cap20 Prob



Dividing the second of these

equations by the first and solving for

T gives:

( ) 412=

T

T' ⇒ ( ) T T'

412=


evaluate T ′

( ) ( )

C1598

K 1871K 1573241

°=

==T'

46 •• Liquid helium is stored at its boiling point (4.2 K) in a spherical canthat is separated by an evacuated region of space from a surrounding shield that ismaintained at the temperature of liquid nitrogen (77 K). If the can is 30 cm indiameter and is blackened on the outside so that it acts as a blackbody emitter,how much helium boils away per hour?

Picture the Problem We can differentiate Q = mL, where L is the latent heat of

boiling for helium, with respect to time to obtain an expression for the rate at

which the helium boils away.

Express the rate at which the helium

boils away in terms of the rate at

which its container absorbs radiant

energy:

( )

( )

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ −=

−=

−==

4

042

4

0

42

4

0

4

net

1T

T T

L

d e

L

T T d e

L

T T Ae

L

P

dt

dm

σπ

σπ

σ

If T 0 << T, then: 42

T L

d e

dt

dm σπ ≈

Substitute numerical values and evaluate dm/dt :

( ) ( )( )

g/h97h

s3600s

kg102.68

K 77

kg

kJ21

m30.0K m

W105.67031

5

4

2

42

8

=××=

⎟ ⎠

⎞⎜⎝

⎛ ⋅

×≈

−

− π

dt

dm

8/12/2019 Cap20 Prob


Chapter 201892

General Problems

47 • A steel tape is placed around Earth at the equator when the temperatureis 0ºC. What will the clearance between the tape and the ground (assumed to beuniform) be if the temperature of the tape increases to 30ºC? Neglect theexpansion of Earth.

Picture the Problem The distance by which the tape clears the ground equals

the change in the radius of the circle formed by the tape placed around Earth at the

equator.

Express the change in the radius of

the circle defined by the steel tape:

T R R Δ=Δ α

where R is the radius of Earth, α is the

coefficient of linear expansion of steel,

and ΔT is the increase in temperature.

Substitute numerical values and evaluate Δ R:

( )( )( ) km1.2m1010.2C0C30K 1011m106.37Δ3166 =×=°−°××= −− R

48 •• Show that change in the density ρ of an isotropic material due to an

increase in temperature ΔT is given by Δ ρ = – βρ ΔT .

Picture the Problem We can differentiate the definition of the density of an

isotropic material with respect to T and use the definition of the coefficient of

volume expansion to express the rate at which the density of the material changeswith respect to temperature. Once we have an expression for d ρ in terms of dT ,

we can apply a differential approximation to obtain Δ ρ in terms of ΔT.

Using its definition, relate the

density of the material to its

mass and volume:

V

m= ρ

Using its definition, relate the

volume of the material to its

coefficient of volume expansion:

T V V Δ=Δ β

8/12/2019 Cap20 Prob



Differentiate ρ with respect to T and

simplify to obtain:

ρβ β ρ

β ρ ρ

−=−=

−==

V V

V

V V

m

dT

dV

dV

d

dT

d

2

2

or dT d ρβ ρ −=

Using the differential approximations ρ ρ Δ≈d and yields:T dT Δ≈

T ΔΔ ρβ ρ −=

49 •• [SSM] The solar constant is the power received from the Sun perunit area perpendicular to the Sun’s rays at the mean distance of Earth from theSun. Its value at the upper atmosphere of Earth is about 1.37 kW/m

2. Calculate

the effective temperature of the Sun if it radiates like a blackbody. (The radius of

the Sun is 6.96 × 108 m.).

Picture the Problem We can apply the Stefan-Boltzmann law to express the

effective temperature of the Sun in terms of the total power it radiates. We can, in

turn, use the intensity of the Sun’s radiation in the upper atmosphere

of Earth to approximate the total power it radiates.

Apply the Stefan-Boltzmann law to

relate the energy radiated by the Sun

to its temperature:

4

r AT e P σ = ⇒ 4 r

Ae

P T

σ =

Express the surface area of the sun: 2

S4 R A π =

Relate the intensity of the Sun’s

radiation in the upper atmosphere

to the total power radiated by the

sun:

24 R

P I r

π = ⇒ I R P r

24π =

where R is the Earth-Sun distance.

Substitute for P r and A in the

expression for T and simplify to

obtain:

42

S

2

42

S

2

4

4

Re

I R

Re

I RT

σ π σ

π ==

8/12/2019 Cap20 Prob


Chapter 201894


( )

( ) ( )K 5800

m106.96K m

W105.671

m

kW1.35m101.5

4

282

8

2

211

=

×⎟ ⎠ ⎞⎜

⎝ ⎛

⋅×

⎟ ⎠

⎞⎜⎝

⎛ ×=

−

T

50 •• As part of your summer job as an engineering intern at an insulationmanufacturer, you are asked to determine the R factor of insulating material. This

particular material comes in 12

-in sheets. Using this material, you construct a

hollow cube that has 12-in long edges. You place a thermometer and a 100-Wheater inside the box. After thermal equilibrium has been attained, thetemperature inside the box is 90ºC when the temperature outside the box is 20ºC.Determine the R factor of the material.

Picture the Problem We can solve the thermal-current equation for the R factor

of the material.

Use the equation for the thermal

current to express I in terms of the

temperature gradient across the

insulation:

x

T kA I

ΔΔ

=

Rewrite this expression in terms of

the R factor of the material: f f R

T A

A R

T

kA x

T I

Δ=

Δ=

Δ

Δ=

Solving for the R factor gives:

I

T A

I

T A R

Δ=

Δ= sideone

f

6

Substitute numerical values and evaluate R:

( )

Btu

hftF2.2

s3600

h1

Btu

J1054

m

ft10.76

K 5

F9

s

J

mK 0.390

W

mK 0.39C20C90W100

in

m102.54in126

2

2

22

2

22

f

⋅⋅°=×××

°×

⋅=

⋅=°−°

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ××

=

−

R

8/12/2019 Cap20 Prob



51 • (a) From the definition of β , the coefficient of volume expansion (at

constant pressure), show that β = 1/T for an ideal gas. (b) The experimentally

determined value of β for N2 gas at 0ºC is 0.003673 K –1

. By what percent does

this measured value of β differ from the value obtained by modeling N2 as anideal gas?

Picture the Problem We can use the definition of the coefficient of volume

expansion with the ideal-gas law to show that β = 1/T.

(a) Use the definition of the

coefficient of volume expansion to

express β in terms of the rate of

change of the volume with

temperature:

dT

dV

V

1= β

For an ideal gas:

P

nRT V = and P

nR

dT

dV

=

Substitute for dT

dV to obtain:

T P

nR

V

11== β

(b) Express the percent difference

between the experimental value and

the theoretical value:

%3.0

K 273

1

K 273

1K 0.003673

1

11

th

thexp

<

−=

−

−

−−

β

β β

52 •• A rod of length LA, made from material A, is placed next to a rod oflength LB, made of material B. The rods remain in thermal equilibrium with eachother. (a) Show that even though the lengths of each rod will change with changesin the ambient temperature, the difference between the two lengths will remain

constant if the lengths LA and LB are chosen such that LA/ LB = α B/α A, where α A

and α B are the coefficients of linear expansion, respectively. (b) If material B issteel, material A is brass, and LA = 250 cm at 0ºC, what is the value of LB?

Picture the Problem Let Δ L be the difference between LB and LB A

and express LB and LA in terms of the coefficients of linear expansion of materialsA and B. Requiring that Δ L be constant will lead us to the condition that

LA/LB = α B/α A.

8/12/2019 Cap20 Prob


Chapter 201896

(a) Express the condition that Δ L

does not change when the

temperature of the materials

changes:

constantAB =−=Δ L L L (1)

Using the definition of thecoefficient of linear expansion,

substitute for LB and LA:

( ) ( )( ) ( )

( ) T L L L

T L L L LT L LT L L L

Δ−+Δ=

Δ−+−=Δ+−Δ+=Δ

AABB

AABBAB

AAABBB

α α

α α α α

or

( ) 0AABB =Δ− T L L α α

The condition that Δ L remain

constant when the temperature

changes by ΔT is:

0AABB =− L L α α

Solving for the ratio of LA to LB

yields:A

B

B

A

α

α =

L

L

(b) From (a) we have:

steel

brass brass

B

AAsteelB

α

α

α

α L L L L ===


evaluate LB:( ) cm430

K 1011

K 1019cm250

16

16

B =××

=−−

−−

L

53 •• [SSM] On the average, the temperature of Earth’s crust increases1.0ºC for every increase in depth of 30 m. The average thermal conductivity of

Earth’s crust material is 0.74 J/m⋅s⋅K. What is the heat loss of Earth per seconddue to conduction from the core? How does this heat loss compare with theaverage power received from the Sun (which is about 1.37 kW/m

2)?

Picture the Problem We can apply the thermal-current equation to calculate

the heat loss of Earth per second due to conduction from its core. We can

also use the thermal-current equation to find the power per unit area radiated

from Earth and compare this quantity to the solar constant.

Express the heat loss of Earth per

unit time as a function of the thermal

conductivity of Earth and its

temperature gradient:

x

T kA

dt

dQ I

ΔΔ

== (1)

or

x

T k R

dt

dQ

ΔΔ

= 2

E4π

8/12/2019 Cap20 Prob



Substitute numerical values and evaluate dQ/dt :

( ) kW103.1m30

C1.0

K sm

J0.74m1037.64 1026 ×=⎟⎟

⎠

⎞⎜⎜⎝

⎛ °⎟ ⎠

⎞⎜⎝

⎛ ⋅⋅

×= π dt

dQ

Rewrite equation (1) to express the

thermal current per unit area: x

T k

A

I

ΔΔ=


evaluate I / A:( )

2W/m0.0247

m30

C1.0K sJ/m0.74

=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ °⋅⋅=

A

I

Express the ratio of I / A to the solar

constant:

%002.0

kW/m1.37

W/m0.0247

constantsolar 2

2

<

= A I

54 •• A copper-bottomed saucepan containing 0.800 L of boiling water boilsdry in 10.0 min. Assuming that all the heat is transferred through the flat copper bottom, which has a diameter of 15.0 cm and a thickness of 3.00 mm, calculatethe temperature of the outside of the copper bottom while some water is still inthe pan.

Picture the Problem We can find the temperature of the outside of the copper

bottom by finding the temperature difference between the outside of the saucepan

and the boiling water. This temperature difference is related to the rate at which

the water is evaporating through the thermal-current equation.

Express the temperature outside the

pan in terms of the temperature

inside the pan:

T T T T ΔC100Δinout +°=+= (1)

Relate the thermal current through

the bottom of the saucepan to its

thermal conductivity, area, and the

temperature gradient between its

surfaces:

x

T kA

t

Q

ΔΔ

=ΔΔ

⇒ xt

Q

kAT Δ

ΔΔ

=Δ1

Because the water is boiling: vmLQ =Δ

8/12/2019 Cap20 Prob


Chapter 201898

Substitute for ΔQ to obtain:

t kA

xmLT

Δ

ΔΔ

v=

Substituting for ΔT in equation (1)

yields: t kA

xmLT

Δ

ΔC100 v

out +°=

Substitute numerical values and evaluate T out:

( ) ( )

( ) ( )C101

s600m0.1504K m

W401

m103.00kg

MJ2.26kg0.800

C1002

3

out °=

⎥⎦⎤

⎢⎣⎡

⎟ ⎠ ⎞

⎜⎝ ⎛

⋅

×⎟⎟ ⎠

⎞⎜⎜⎝

⎛

+°=

−

π

T

55 •• A cylindrical steel hot-water tank of cylindrical shape has an insidediameter of 0.550 m and inside height of 1.20 m. The tank is enclosed with a

5.00-cm-thick insulating layer of glass wool whose thermal conductivity is 0.0350W/(m⋅K). The insulation is covered by a thin sheet-metal skin. The steel tank andthe sheet-metal skin have thermal conductivities that are much greater than that ofthe glass wool. How much electrical power must be supplied to this tank in orderto maintain the water temperature at 75.0ºC when the external temperature is1.0ºC?

Picture the Problem Thermal energy is lost from this tank through its cylindrical

side and its top and bottom. The power required to maintain the temperature of

the water in the tank is equal to the rate at which thermal energy is conducted

through the insulation.

Express the total thermal current as

the sum of the thermal currents

through the top and bottom and the

side of the hot-water tank:

side bottomandtop I I I += (1)

Express I through the top and bottom

surfaces: x

T k d

x

T kA I

Δ

Δ

Δ

Δ2 2

21

bottomandtop π =⎟

⎠ ⎞

⎜⎝ ⎛ =

Substitute numerical values and evaluate : bottom

andtop I

( )( )

W6.24m0.0500

C0.1C75.0K m

W0.0350

m550.02

21

bottomandtop =

°−°⎟ ⎠

⎞⎜⎝

⎛ ⋅= π I

8/12/2019 Cap20 Prob



Letting r represent the inside radius

of the tank, express the heat current

through the cylindrical side:

dr

dT kLr

dr

dT kA I π 2side −=−=

where the minus sign is a consequence

of the heat current being opposite the

temperature gradient.

Separating the variables yields:

r

dr

kL

I dT

π 2side−=

Integrate from r = r 1 to r = r 2 and

T = T 1 to T = T 2 and simplify to

obtain:

∫∫ −=2

1

2

12

side

r

r

T

T r

dr

kL

I dT

π

and

]

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ −=

−=−

2

1side

1

2side

side12

ln2

ln2

ln2

2

1

r

r

kL

I

r r

kL I

r kL

I T T

r

r

π

π

π

Solving for I side gives: ( )12

2

1

side

ln

2T T

r

r

kL I −

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ = π

Substitute numerical values and evaluate I side:

( )( ) W117C0.1C0.75

m0.275

m0.325ln

m1.20K m

W0.03502

side =°−°

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

⎟ ⎠

⎞⎜⎝

⎛ ⋅

=π

I

Substitute numerical values in

equation (1) and evaluate I :W142W171W4.62 =+= I

56 ••• The diameter d of a tapered rod of length L is given by d = d 0(1 + ax),where a is a constant and x is the distance from one end. If the thermalconductivity of the material is k what is the thermal resistance of the rod?

8/12/2019 Cap20 Prob


Chapter 201900

Picture the Problem We can use R = ΔT / I and I = −kAdT /dt to express dT in

terms of the linearly increasing diameter of the rod. Integrating this expression

will allow us to find ΔT and, hence, R.

The thermal resistance of the rod is

given by: I

T R

Δ= (1)

where I the thermal current in the rod.

Relate the thermal current in the rod

to its thermal conductivity k , cross-

sectional area A, and temperature

gradient:

dx

dT kA I −=

where the minus sign is a consequence

of the heat current being opposite the

temperature gradient.

Using the dependence of the

diameter of the rod on x, express thearea of the rod:

( )22

0

2

144

axd d

A +== π π

Substitute for A to obtain: ( )dx

dT axd k I ⎥⎦

⎤⎢⎣⎡ +−= 22

0 14

π

Separating variables yields:

( )

( )22

0

22

0

1

4

14

ax

dx

kd

I

axd k

IdxdT

+−=

⎥⎦

⎤⎢⎣

⎡ +−=

π

π

Integrating T from T 1 to T 2 and x

from 0 to L gives: ( )∫∫ +−=

LT

T ax

dx

kd

I dT

0

22

0 1

42

1π

and

( )aLkd

ILT T T

+=Δ=−

1

42

0

12π

Substitute forΔT

and I

in equation

(1) and simplify to obtain: ( )( )aLkd

L

I

aLkd

IL

R+

=+

=1

41

4

2

0

2

0

π

π

57 ••• A solid disk of radius r and mass m is spinning about a frictionless

axis through its center and perpendicular to the disk, with angular velocity ω 1 attemperature T 1. The temperature of the disk decreases to T 2. Express the angular

velocity ω 2, rotational kinetic energy K 2, and angular momentum L2 in terms of

8/12/2019 Cap20 Prob



their values at the temperature T 1 and the linear expansion coefficient α of thedisk.

Picture the Problem Let ΔT = T 2 – T 1. We can apply Newton’s second law to

establish the relationship between L2 and L1 and angular momentum conservation

to relate ω 2 and ω

1. We can express both E

2 and E

1 in terms of their angular

momenta and rotational inertias and take their ratio to establish their relationship.

Applyt

L

ΔΔ

=∑τ to the spinning

disk:

Because 0=∑τ , Δ L = 0

and

12 L L =

Apply conservation of angular

momentum to relate the angular

velocity of the disk at T 2 to the

angular velocity at T 1:

1122 I I = ⇒ 1

2

12 ω ω

I

I = (1)

Express I 2: ( )

( )( )2

1

22

1

2

22

ΔΔ21

Δ1

T T I

T mr mr I

α α

α

++=

+==

Because (α ΔT )2 is small compared to

α ΔT :

( )T I I Δ2112 α +≈

Substitute for I 2 in equation (1)

to obtain: ( ) 11

1

2 Δ21 ω α ω T I

I

+= (2)

Expanding ( binomially

yields:

) 1Δ21

−+ T α

( )

sorder termhigher

Δ21Δ211

+

−=+ −T T α α

Because the higher order terms are

very small:

( ) T T Δ21Δ211

α α −≈+ −

Substituting in equation (2) yields: ( ) 12 21 ω α ω T Δ−≈

Express E 2 in terms of L2 and I 2:

2

2

1

2

2

22

22 I

L

I

L E == because L2 = L1.

Express E 1 in terms of L1 and I 1:

1

2

11

2 I

L E =

8/12/2019 Cap20 Prob


Chapter 201902

Express the ratio of E 2 to E 1 and

simplify to obtain:

2

1

1

2

1

2

2

1

1

2

2

2

I

I

I

L

I

L

E

E == ⇒

2

112 I

I E E =

Substituting for the ratio of I 1 to I 2

yields:( )T E E E Δ211

1

212 α ω

−==

58 ••• Write a spreadsheet program to graph the average temperature of thesurface of Earth as a function of emissivity, using the results of Problem 22. Howmuch does the emissivity have to change in order for the average temperature toincrease by 1 K? This result can be thought of as a model for the effect ofincreasing concentrations of greenhouse gases like methane and CO2 in Earth’satmosphere.

Picture the Problem The amount of heat radiated by Earth must equal the solarflux from the Sun, or else the temperature on Earth would continually increase.The emissivity of Earth is related to the rate at which it radiates energy into space

by the Stefan-Boltzmann law .4

r AT e P σ =

Using the Stefan-Boltzmann law,express the rate at which Earthradiates energy as a function of itsemissivity e and temperature T :

4

r A'T e P σ =

where A′ is the surface area of Earth.

Use its definition to express the

intensity of the radiation P a absorbed by Earth:

A

P I a= or AI P =

a

where A is the cross-sectional area ofEarth.

For 70% absorption of the Sun’sradiation incident on Earth:

( ) AI P 70.0a =

Equate P r and P a and simplify: ( ) 470.0 A'T e AI σ =

or

( ) ( )422 470.0 T Re I R σ π π =

Solve for T to obtain: ( ) 414

4

70.0 −== Cee

I T

σ (1)

Substitute numerical values for I

and σ and simplify to obtain:( )( )

( )( ) 41

4428

2

K 255

K W/m10670.54

W/m137070.0

−

−

=

⋅×=

e

eT

8/12/2019 Cap20 Prob



A spreadsheet program to evaluate T as a function of e is shown below. Theformulas used to calculate the quantities in the columns are as follows:

Cell Formula/Content Algebraic Form

B1 255

B4 0.4 e

B5 B4+0.01 e + 0.1C4 $B$1/(B4^0.25) ( ) 41K 255 −e

A B C D

1 T = 255 K

2

3 e T

4 0.40 321

5 0.41 319

6 0.42 317

7 0.43 315

23 0.59 291

24 0.60 290

25 0.61 289

26 0.62 287

A graph of T as a function of e follows.

285

290

295

300

305

310

315

320

325

0.40 0.45 0.50 0.55 0.60

e

T ( K

)

Treating e as a variable, differentiateequation (1) to obtain:

deCede

dT 45

4

1 −−= (2)

8/12/2019 Cap20 Prob


Chapter 201904

Divide equation (2) by equation (1)to obtain:

e

de

Ce

deCe

T

dT

4

14

1

41

45

−=−

= −

−

Use a differential approximation toobtain:

e

e

T

T Δ−=

Δ

4

1⇒

T

T

e

e Δ−=

Δ4

Substitute numerical values

(e ≈ 0.615 for T Earth = 288 K) and

evaluate eeΔ :

014.0K 288

K 14 −=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

Δe

e

or about a 1.4% change.

59 ••• [SSM] A small pond has a layer of ice 1.00 cm thick floating on itssurface. (a) If the air temperature is –10ºC on a day when there is a breeze, findthe rate in centimeters per hour at which ice is added to the bottom of the layer.The density of ice is 0.917 g/cm

3. (b) How long do you and your friends have to

wait for a 20.0-cm layer to be built up so you can play hockey?

Picture the Problem (a) We can differentiate the expression for the heat that

must be removed from water in order to form ice to relate dQ/dt to the rate of ice

build-up dm/dt. We can apply the thermal-current equation to express the rate at

which heat is removed from the water to the temperature gradient and solve this

equation for dm/dt . In Part (b) we can separate the variables in the differential

equation relating dm/dt and ΔT and integrate to find out how long it takes for a

20.0-cm layer of ice to be built up.

(a) Relate the heat that must be

removed from the water to freeze it

to its mass and heat of fusion:

f mLQ = ⇒ dt

dm

Ldt

dQf =

Using the definition of density,

relate the mass of the ice added to

the bottom of the layer to its density

and volume:

AxV m ρ ρ ==

Differentiate with respect to time

to express the rate of build-up of

the ice: dt

dx A

dt

dm ρ =

Substitute fordt

dmto obtain:

dt

dx A L

dt

dQ ρ f =

8/12/2019 Cap20 Prob



The thermal-current equation is:

x

T kA

dt

dQ Δ=

Equate these expressions and

solve fordt

dx:

x

T kA

dt

dx A L

Δ= ρ f ⇒

x

T

L

k

dt

dx Δ=

ρ f

(1)


evaluatedt

dx:

( )( )

( )( )

cm/h0.70

cm/h0.697h

s3600

s

m94.1

m0.0100kg/m917kg

kJ333.5

C10C0K m

W0.592

3

=

=×=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

°−−°⎟ ⎠ ⎞

⎜⎝ ⎛

⋅=

μ

dt

dx

(b) Separating the variables inequation (1) gives:

dt L

T k xdx ρ f

Δ=

Integrate x from xi to xf and t ′ from

0 to t :' dt

L

T k xdx

t x

x

∫∫ Δ=

0f

f

i ρ

and

( ) t L

T k x x

f

2

i

2

f 21

ρ

Δ=− ⇒

( )T k

x x Lt

Δ−

=2

2

i

2

f f ρ

Substitute numerical values and evaluate t :

( )( )( ) ( )[ ]

d12h24

d1

s3600

h1s1003.1

m0.010m0.200

C10C0K m

W0.5922

kg

kJ333.5

m

kg917

6

223

=×××=

−°−−°⎟

⎠

⎞⎜⎝

⎛ ⋅

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛

=t

60 ••• [SSM] A blackened copper cube that has 1.00-cm-long edges isheated to a temperature of 300ºC, and then is placed in a vacuum chamber whosewalls are at a temperature of 0ºC. In the vacuum chamber, the cube coolsradiatively. (a) Show that the (absolute) temperature T of the cube follows the

differential equation:dT

dt = −

eσ A

C T 4 − T

0

4( ) where C is the heat capacity of the

cube, A is its surface area, e the emissivity, and T o the temperature of the vacuumchamber. (b) Using Euler’s method (Section 5.4 of Chapter 5), numerically solve

8/12/2019 Cap20 Prob


Chapter 201906

the differential equation to find T (t ), and graph it. Assume e = 1.00. How longdoes it take the cube to cool to a temperature of 15ºC?Picture the Problem We can use the Stefan-Boltzmann equation and thedefinition of heat capacity to obtain the differential equation expressing the rate atwhich the temperature of the copper block decreases. We can then approximatethe differential equation with a difference equation for the purpose of solving for

the temperature of the block as a function of time using Euler’s method.

(a) The rate at which heat is radiated

away from the cube is given by:( )4

0

4 T T Aedt

dQ−= σ

Using the definition of heat capacity,relate the thermal current to the rateat which the temperature of the cubeis changing:

dt

dT C

dt

dQ−=

Equate these expressions and solve

fordt

dT to obtain: ( )

4

0

4

T T C

Ae

dt

dT

−−= σ

Approximating the differentialequation by the difference equationgives:

( )4

0

4 T T C

Ae

t

T −−=

ΔΔ σ

Solving for ΔT yields: ( ) t T T C

AeT Δ−−=Δ 4

0

4σ

or

( ) t T T

C

AeT T nnn Δ−−=+

4

0

4

1

σ (1)

Use the definition of heat capacity toobtain:

VcmcC ρ ==

Substitute numerical values (see Figure 13-1 for ρ Cu and Table 19-1 for cCu) andevaluate C :

( )K

J45.3

K kg

kJ386.0m1000.1

m

kg1093.8 36

3

3 =⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⋅

×⎟ ⎠

⎞⎜⎝

⎛ ×= −C

8/12/2019 Cap20 Prob



(b) A spreadsheet program to calculate T as a function of t using equation (1) isshown below. The formulas used to calculate the quantities in the columns are asfollows:

Cell Formula/Content Algebraic Form

B1 5.67E−08 σ B2 6.00E−04 A

B3 3.45 C

B4 273 T 0B5 10 Δt

A9 A8+$B$5 t +Δt

B9 B8-($B$1*$B$2/$B$3)

*(B8^4−$B$4^4)*$B$5 ( ) t T T

C

AeT nn Δ−− 4

0

4σ

A B C

1 σ = 5.67 × 10−8

W/m2

⋅K 4

2 A= 6.00 × 10−4 m2

3 C = 3.45 J/K

4 T 0= 273 K

5 Δt = 10 s

6

7 t (s) T (K)

8 0 573.00

9 10 562.92

10 20 553.56

11 30 544.85

248 2400 288.22

249 2410 288.08

250 2420 287.95

251 2430 287.82

8/12/2019 Cap20 Prob


Chapter 201908

From the spreadsheet solution, the time to cool to 15°C (288 K) is about 2420 s or

.min5.40 A graph of T as a function of t follows.

270

320

370

420

470

520

570

0 500 1000 1500 2000 2500

t (s)

T ( K

)

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Cap20 Prob