physics ch20 ism

1951

Chapter 20 Thermal Properties and Processes Conceptual Problems 1 • Why does the mercury level of a thermometer first decrease slightly when 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 the sheet 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 the area of the hole always increases. )(b is correct. 3 • [SSM] Why is it a bad idea to place a tightly sealed glass bottle that is 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 no room for the expansion to take place. Because the glass contracts as it gets cold, making a slightly smaller volume available for the freezing water, ultimately the bottle will shatter. 4 • The windows of your physics laboratory are left open on a night when the temperature of the outside dropped well below freezing. A steel ruler and a wooden ruler were left on the window sill, and when you arrive in the morning they are both very cold. Which ruler should you use to make the most accurate length measurements? Explain your answer. Determine the Concept You should use the wooden ruler. Because the coefficient of expansion for wood is much less than that for metal, the metal ruler will have 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 have dissimilar coefficients of linear expansion and are bonded together to form one doubly thick strip. Suppose a bimetallic strip is constructed out of one steel strip and one copper strip, and suppose the bimetallic strip is curled in the shape of a circular arc with the steel strip on the outside. If the temperature of the strip is decreased, will the strip straighten out or curl more tightly?

Chapter 20

1952

Determine the Concept The strip will curl more tightly. Because the coefficient of linear expansion for copper is greater than the coefficient of linear expansion for steel, 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 volume expansion β compare? (a) βA = βB , (b) βA = 3βB , (c) βA = 6βB , (d) βA = 9βB , (e) You cannot tell from the data given. Determine the Concept We know that the coefficient of volume expansion is three times the coefficient of linear expansion and so can use this fact to express the ratio of Aβ to Bβ . Express the coefficient of volume expansion of metal A in terms of its coefficient of linear expansion:

AA 3αβ =

Express the coefficient of volume expansion of metal B in terms of its coefficient of linear expansion:

BB 3αβ =

Dividing the first of these equations by the second yields:

B

A

B

A

B

A

33

αα

αα

ββ

==

Because BA 3αα = :

33

B

B

B

A ==αα

ββ

⇒ BA 3ββ =

( )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 statement is true because at the summit of Mount Rainier (a) the air temperature is too low to boil water, (b) the air pressure is too low for alcohol fuel to burn, (c) the temperature of boiling water is not hot enough to hard boil the egg, (d) the oxygen content 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 pressure at 20ºC? Explain your answer.

Thermal Properties and Processes

1953

Determine the Concept Gases that cannot be liquefied by applying pressure at 20 °C are those for which Tc < 293 K. These are He, Ar, Ne, H2, O2, NO.

9 •• [SSM] The phase diagram in Figure 20-15 can be interpreted to yield information on how the boiling and melting points of water change with altitude. (a) Explain how this information can be obtained. (b) How might this information affect cooking procedures in the mountains? Determine the Concept (a) With increasing altitude, P decreases; from curve OC, the temperature T 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 from Section 20-3. Determine the Concept The following phase diagram for carbon dioxide was constructed using information in Section 20-3.

atm ,P

K ,T

5.1

216.6 304.2

Gas

Gas

SolidVapor

Liquid

point Critical

11 •• Explain why the carbon dioxide on Mars is found in the solid state in the polar regions even though the atmospheric pressure at the surface of Mars is only about 1 percent of the atmospheric pressure at the surface of Earth. Determine the Concept At pressures below 5.1 atm, carbon dioxide can exist only as a solid or gas (or vapor above the gas). Mars, on average, is warm enough so that the atmosphere is mostly gaseous carbon dioxide. Mars’ polar caps, however, must be cold enough to enable solid carbon dioxide (dry ice) to exist.

Chapter 20

1954

12 •• Explain why decreasing the temperature of your house at night in winter can save money on heating costs. Why doesn’t the cost of the fuel consumed to heat the house back to the daytime temperature in the morning equal the savings realized by cooling it down in the evening and keeping it cool throughout 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 the same lengths; their diameters are related by dA = 2dB. When the same temperature difference is maintained between the ends of the cylinders, they conduct heat at the same rate. Their thermal conductivities are therefore related by which of the following equations? (a) kA = kB/4, (b) kA = kB/2, (c) kA = kB, (d) kA = 2kB, (e) kA = 4kB Picture the Problem The rate at which heat is conducted through a cylinder is given by xTkAdtdQI ΔΔ== // where A is the cross-sectional area of the cylinder.

Express the rate at which heat is conducted through cylinder A:

xTdk

xTkAI

ΔΔ

ΔΔ 2

AA41

AA π==

Express the rate at which heat is conducted through cylinder B:

xTdk

xTkAI

ΔΔ

ΔΔ 2

BB41

BB π==

Equate these expressions and simplify to obtain: x

TdkxTdk

ΔΔ

ΔΔ 2

BB412

AA41 ππ =

or 2BB

2AA dkdk =

Because dA = 2dB: ( ) 2

BB2

BA 2 dkdk = and

4/BA kk = ⇒ )(a is correct. 14 •• Two solid cylinders made of materials A and B have the same diameter; their lengths are related by LA = 2LB. When the same temperature difference is maintained between the ends of the cylinders, they conduct heat at the same rate. Their thermal conductivities are therefore related by which of the following equations?


1955

(a) kA = kB/4, (b) kA = kB/2, (c) kA = kB, (d) kA = 2kB, (e) kA = 4kB. Determine the Concept We can use the expression for the heat current in a conductor, Equation 20-7, to relate the heat current in each cylinder to its thermal conductivity, cross-sectional area, temperature difference, and length. The heat current in cylinder A is given by Equation 20-7:

( )A

AAA

AAAA L

TAkxTAkI Δ

=⎟⎠⎞

⎜⎝⎛

ΔΔ

=

The heat current in cylinder B is given by:

( )B

BBB

BBBB L

TAkxTAkI Δ

=⎟⎠⎞

⎜⎝⎛

ΔΔ

=

Because the cylinders conduct heat at the same rate, we can equate these currents to obtain:

( ) ( )B

BBB

A

AAA L

TAkLTAk Δ

=Δ

Simplifying this expression yields: B

B

BB

B

ABA 22 k

LLk

LLkk ===

( )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 the window (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 almost the same. Because glass is an excellent conductor of heat, there need not be a huge temperature difference. Thus the temperature must drop quickly as you near the pane on the warm side and the same on the outside. This is sketched qualitatively in the following diagram. Convection and radiation are primarily responsible for heat transfer on the inside and outside, and it is mainly conduction through the glass. Conduction through the interior and exterior air is minimal.

C , °T

20

5

Inside Outside

m ,x50

Chapter 20

1956

16 •• During the thermal retrofitting of many older homes in California, it was found that the 3.5-in-deep spaces between their wallboards and outer sheathing was filled with just air (no insulation). Filling the space with insulating material certainly reduces heating and cooling costs; although, the insulating material is a better conductor of heat than air is. Explain why adding the insulation is a good idea. Determine the Concept The tradeoff is the reduction of convection cells between the walls by putting in the insulating material, versus a slight increase in conductivity. The net reduction in convection results in a higher R value. Estimation and Approximation 17 •• [SSM] You are using a 4.0-L 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 L of 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 the steel pot to keep all of the water in the pot while the water is heated to a boil. Is your assumption correct? Explain. If your assumption is not correct, how much water 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 the coefficient of linear expansion of steel. Express the volume of water that overflows when the pot and the water are heated:

( ) TV

TVTV

VVV

Δ

ΔΔ

ΔΔ

0steelOH

0steel0OH

potOHovefrlow

2

2

2

ββ

ββ

−=

−=

−=

Because the coefficient of volume expansion of steel is three times its coefficient of linear expansion:

steelsteel 3αβ =

Substituting for OH2β and steelβ yields: ( ) TVV Δ3 0steelOH overflow 2

αα −=

Substitute numerical values and evaluate overflowV :

( )( )( )( ) mL 56C20C100L 4.0K 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 Mylar


1957

sheets. The rate of evaporation of liquid helium in a 200-L container is about 0.700 L per day when the container is stored at room temperature (20ºC). The density 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.

Express the heat current in terms of the rate of evaporation of the liquid helium:

dtdmLI v=

Express the heat current in terms of the temperature gradient across the superinsulation and the conductivity of the superinsulation:

xTkAI

ΔΔ

=

Equate these expressions and solve for k to obtain: TA

dtdmxL

kΔ

Δ=

v

Using the definition of density, express the rate of loss of liquid helium:

dtdV

dtdm ρ=

Substitute for dtdm to obtain:

TAdtdVxL

kΔ

Δ=

ρv

Express the ratio of the area of the spherical container to its volume:

334

24rr

VA

ππ

= ⇒ 3 236 VA π=

Substituting for A yields:

TVdtdVxL

kΔ

Δ=

3 2

v

36π

ρ

Chapter 20

1958

Substitute numerical values and evaluate k:

( )

( ) ( ) KmW1012.3

K289m1020036

s86400m100.700

mkg125m107.00

kgkJ21.0

6

3 233

33

32

⋅×=

×

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×⎟⎠⎞

⎜⎝⎛×⎟⎟

⎠

⎞⎜⎜⎝

⎛

= −

−

−−

πk

19 •• [SSM] Estimate the thermal conductivity of human skin. Picture the Problem We can use the thermal current equation for the thermal conductivity 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 area is about 1.8 m2. We’ll also assume that a typical human, while resting, produces energy 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. Use the thermal current equation to express the rate of conduction of thermal energy:

I = kAΔTΔx

⇒

xTA

Ik

ΔΔ

=

Substitute numerical values and evaluate k: ( ) Km

mW17

m100.1C33C37m8.1

W120

32 ⋅

=

×°−°

=

−

k

20 •• You are visiting Finland with a college friend and have met some Finnish friends. They talk you into taking part in a traditional Finnish after-sauna exercise which consists of leaving the sauna, wearing only a bathing suit, and running out into the mid-winter Finnish air. Estimate the rate at which you initially lose energy. Compare this rate of initial energy loss to the resting 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 at which 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 that the emissivity of your skin is 1. Use the Stefan-Boltzmann law to express the net rate at which you radiate energy to the cold air:

( )4air

4skinnet TTAP −= εσ


1959

Substitute numerical values and evaluate Pnet:

( ) ( ) ( ) ( )( ) W450K 632K 310m 8.1Km

W 10670.51 44242

8net =−⎟

⎠⎞

⎜⎝⎛

⋅×= −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 when you first step out of the sauna and the fact that radiation loses are dependent on the 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 door during a cold winter day in Minnesota. Include the brass doorknob. What is the ratio of the heat that escapes through the doorknob to the heat that escapes through the whole door? What is the total overall R-factor for the door, including the knob? The thermal conductivity of brass is ( )KmW 85 ⋅ . Picture the Problem We can use the thermal current equation (Equation 20-7) to estimate the rate of heat loss through the door and its knob. We’ll assume that the door is made of oak with an area of 2.5 m2, that the knob is made of brass and has an area of 7.0 × 10−6 m2, and that the inside temperature is 20°C and the outside temperature is −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 the door is the sum of the thermal currents through the door and the knob: knob

knobknob

door

doordoor

knobdoortot

ΔΔ

ΔΔ

xTAk

xTAk

III

+=

+=

Assume that xxx ΔΔΔ knobdoor == to obtain:

( )xTAkAkIΔΔ

knobknobdoordoortot +=

Substitute numerical values and evaluate Itot:

( ) ( ) ( )( )

kW 30.0

in 39.37m 1in 0.2

C20C20m 100.7Km

W85m 5.2Km

W15.0 262tot

=

⎟⎠⎞

⎜⎝⎛

°−−°⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎠⎞

⎜⎝⎛

⋅+⎟

⎠⎞

⎜⎝⎛

⋅= −I

Chapter 20

1960

Express the ratio of the thermal currents through the knob and the door:

knobknobknob

doordoordoor

door

knob

ΔΔΔΔ

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

xTAk

xTAk

II

Because the temperature difference across the door and the knob are the same and we’ve assumed that the thickness of the door and length of the knob are the same:

knobknob

doordoor

door

knob

AkAk

II

=

Substitute numerical values to obtain: ( )

( )2

26

2

door

knob

103.6

m 100.7Km

W85

m 5.2Km

W15.0

×=

×⎟⎠⎞

⎜⎝⎛

⋅

⎟⎠⎞

⎜⎝⎛

⋅=−I

I

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

knobdooreq

111RRR

+=

Substitute for Rdoor and Rknob to obtain:

xAkAkAkx

AkxR

Δ

Δ1

Δ11

knobknobdoordoor

knobknobdoordoor

eq

+=

+=

Solving for Req yields:

knobknobdoordooreq

ΔAkAk

xR+

=

Substitute numerical values and evaluate Req:

( )

( ) ( ) WK14.0

m 107Km

W85m 5.2Km

W15.0

in 39.37m 1in 0.2

262eq =

×⎟⎠⎞

⎜⎝⎛

⋅+⎟

⎠⎞

⎜⎝⎛

⋅

⎟⎠⎞

⎜⎝⎛

=−

R

22 •• Estimate the effective emissivity of Earth, given the following information. The solar constant, which is the intensity of radiation incident on Earth from the Sun, is about 1.37 kW/m2. 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 πR2, where R is Earth’s radius, while the blackbody-emission area is 4πR2.)


1961

Picture the Problem The amount of heat radiated by Earth must equal the solar flux 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 ATeP σ= . Using the Stefan-Boltzmann law, express the rate at which Earth radiates energy as a function of its emissivity e and temperature T:

4r A'TeP σ= ⇒ 4

r

A'TPe

σ=

where A′ is the surface area of Earth.

Use its definition to express the intensity of the radiation received by Earth:

API absorbed=

where A is the cross-sectional area of Earth.

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

( )A

PI r70.0=

Substitute for Pr and A in the expression for e and simplify to obtain:

( ) ( ) ( )442

2

4 470.0

470.070.0

TI

TRIR

A'TAIe

σσππ

σ===

Substitute numerical values and evaluate e:

( )( )( )( )

61.0

K288KW/m10670.54kW/m37.170.0

4428

2

=

⋅×=

−e

23 •• Black holes are highly condensed remnants of stars. Some black holes, 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. One way black holes can be detected from Earth is by observing the frictional heating of 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 λmax for use in an astronomical detection of a black hole. (Remark: This is in the X-ray region of the electromagnetic spectrum.) Picture the Problem The wavelength at which maximum power is radiated by the gas falling into a black hole is related to its temperature by Wien’s displacement law. Wien’s displacement law relates the wavelength at which maximum power is radiated by the gas to its temperature:

TKmm898.2

max⋅

=λ

Chapter 20

1962

Substitute for T and evaluate λmax: nm9.2K100.1

Kmm898.26max =

×⋅

=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 of the 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. In addition, you cover the insulation with 1.0-in-thick gypsum wallboard. Assuming heat transfer is only due to conduction, estimate the ratio thermal current through the walls during a cold winter night before the renovation to the thermal current through the walls following the renovation. Picture the Problem We can use the thermal current equation to find the thermal current per square meter through the walls of the cabin both before and after the walls have been insulated. See Table 20-4 for the R-factors of building materials. The rate at which heat is conducted through the walls is given by the thermal current equation:

beforebeforebefore

ΔΔΔ

RT

xTAkI ==

With the insulation in place:

afterafterafter

ΔΔΔ

RT

xTAkI ==

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

after

before

before

after

before

after

Δ

Δ

RR

RT

RT

II

==

beforeR is the R-factor for pine and

afterR is the sum of the R-factors for pine, the insulating material, and the gypsum board:

pinebefore RR = and

gypsum31pineeqafter RRRRR ++==

Substitute for beforeR and afterR to obtain: gypsum31pine

pine

before

after

RRRR

II

++=

Because Table 20-4 does not include a value for pine, use the value for Douglas fir. Substitute numerical values and evaluate the ratio beforeafter / II :


1963

%1Btu

Ffth32.0in 0.375

in 1Btu

Ffth31Btu

Ffth62.0in 0.5

cm 2.54in 1cm 20

BtuFfth62.0

222

2

before

after

≈

°⋅⋅×+

°⋅⋅+

°⋅⋅×

×

°⋅⋅

=II

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 kept cold in a Styrofoam ice chest initially filled with 1.0 kg of ice. It is crucial that the liver temperature is never warmer than 5.0°C. Assuming the trip from the hospital in 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 slab of material. We can use the thermal current equation to express the thermal resistance 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’ll assume that the surface area of the ice chest is 1.0 m2 and that the ambient temperature is 25°C The R-factor needed for the styrofoam walls of the ice chest is the product of their thermal resistance and area:

RAR =f (1)

Use the thermal current equation to express R:

tottot

ΔΔ

Δ

ΔΔQ

tT

tQ

TITR ===

Substitute for R in equation (1) to obtain:

totf

ΔΔQ

tTAR =

The total heat entering the chest in 7 h is given by:

OHOHicefice

watericewarm

icemelttot

22ΔTcmLm

QQQ

+=

+=

Substitute for Qtot and simplify to obtain: ( )OHOHfice

f22

ΔΔΔ

TcLmtTAR

+=

Chapter 20

1964

Substitute numerical values and evaluate Rf:

( )

( ) ( ) BtufthF8

C 5Kkg

kJ18.4kgkJ5.333kg 1

BtuJ 35.1054h 7

C 5F 9C 20

m 1029.9ft 1m 1.0

222

22

f⋅⋅°

≈

⎟⎟⎠

⎞⎜⎜⎝

⎛°⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

+

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

°°

×°⎟⎟⎠

⎞⎜⎜⎝

⎛×

×=

−

R

Thermal Expansion 26 •• You have inherited your grandfather’s grandfather clock that was calibrated when the temperature of the room was 20ºC. Assume that the pendulum consists of a thin brass rod of negligible mass with a compact heavy bob at its end. (a) During a hot day, the temperature is 30ºC, does the clock run fast or slow? 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 π2P =

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:

gLT 20

20 2π= (1)

When the temperature increases to 30°C, the period of the pendulum increases due to the increase in its length:

( )

C20

C20

1

122

tT

gtL

gLT

Δ+=

Δ+==

α

αππ

The daily fractional gain or loss is given by : 20

20

20 TTT

TT −

=Δ

=Δττ

Substituting for T and simplifying yields:

11

1

C

20

20C20

−Δ+=

−Δ+=

Δ

t

TTtT

α

αττ


1965

Solve for Δτ to obtain:

( )τατ 11 C −Δ+=Δ t

Substitute numerical values and Δτ:

( )( )( ) s2.8h

s3600h241C20C30K10191 16 =⎟⎠⎞

⎜⎝⎛ ×−°−°×+=Δ −−τ

27 •• [SSM] You need to fit a copper collar tightly around a steel shaft that has a diameter of 6.0000 cm at 20ºC. The inside diameter of the collar at that temperature is 5.9800 cm. What temperature must the copper collar have so that it will 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 linear expansion.

Express the temperature to which the copper collar must be raised in terms of its initial temperature and the increase in its temperature:

TTT Δ+= i

Apply the definition of the coefficient of linear expansion to express the change in temperature required for the collar to fit on the shaft:

DDD

D

TααΔ

=⎟⎠⎞

⎜⎝⎛ Δ

=Δ

Substitute for ΔT to obtain: DDTT

αΔ

+= i

Substitute numerical values and evaluate T: ( )( )

C220

cm5.9800K1017cm 9800.5cm0000.6C20 16

°=

×−

+°= −−T

28 •• You have a copper collar and a steel shaft. At 20°C, the collar has an inside diameter of 5.9800 cm and the steel shaft has diameter of 6.0000 cm. The copper collar was heated. When its inside diameter exceeded 6.0000 cm is was slipped on the shaft. The collar fitted tightly on the shaft after they cooled to room

Chapter 20

1966

temperature. Now, several years later, you need to remove the collar from the shaft. 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 the shaft? 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 in temperature 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:

TTT Δ+= i (1)

Express the diameter of the steel shaft when its temperature has been increased by ΔT:

( )TDD Δ+= ° steelCsteel,20steel 1 α

Express the diameter of the copper collar when its temperature has been increased by ΔT:

( )TDD Δ+= ° CuCCu,20Cu 1 α

If the collar is to fit over the shaft when the temperature of both has been increased by ΔT:

( )( )TD

TDΔ+=

Δ+

°

°

steelCsteel,20

CuCCu,20

11

α

α

Solving for ΔT yields: steelCsteel,20CuCCu,20

CCu,20Csteel,20

αα °°

°°

−−

=ΔDD

DDT

Substitute in equation (1) to obtain:

steelCsteel,20CuCCu,20

CCu,20Csteel,20i αα °°

°°

−−

+=DD

DDTT

Substitute numerical values and evaluate T:

( )( ) ( )( ) C580/K1011cm6.0000/K1017cm5.9800

cm5.9800cm6.0000C20 66 °=×−×

−+°= −−T

29 •• A container is filled to the brim with 1.4 L of mercury at 20ºC. As the temperature of container and mercury is increased to 60ºC, a total of 7.5 mL of


1967

mercury spill over the brim of the container. Determine the linear expansion coefficient 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 βc. Relate the linear expansion coefficient of the container to its coefficient of volume expansion:

c31

c βα = (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 VV (2)

Express HgVΔ using the definition of

the coefficient of volume expansion:

TVV Δ=Δ HgHgHg β

Express cVΔ using the definition of the coefficient of volume expansion:

TVV Δ=Δ ccc β

Substitute for HgVΔ and cVΔ in

equation (2) to obtain:

mL5.7ccHgHg =Δ−Δ TVTV ββ

Solving for βc yields: TV

TVΔ

mL5.7Δ

c

HgHgc

−=

ββ

or, because V = VHg = Vc,

TV

TVTV

ΔmL5.7

ΔmL5.7Δ

Hg

Hgc

−=

−=

β

ββ

Substitute for cβ in equation (1) to obtain:

TVTV Δ3mL5.7

Δ3mL5.7

HgHg31

c −=−= αβα

Substitute numerical values and evaluate αc:

( ) ( )( )16

1331

c

K1015

K40L4.13mL5.7K1018.0

−−

−−

×=

−×=α

Chapter 20

1968

30 •• A car has a 60.0-L steel gas tank filled to the brim with 60.0-L of gasoline when the temperature of the outside is 10ºC. How much gasoline spills out of the tank when the outside temperature increases to 25ºC? Take the expansion of 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:

tankgasspill VVV Δ−Δ=

Relate gasVΔ to the coefficient of

volume expansion for gasoline:

TVV Δ=Δ gasgas β

Relate tankVΔ to the coefficient of linear expansion for steel:

TVV Δ=Δ tanktank β or, because βsteel = 3αsteel,

TVV Δ=Δ steeltank 3α

Substitute for gasVΔ and tankVΔ and

simplify to obtain:

( )steelgas

steelgasspill

3

3

αβ

αβ

−Δ=

Δ−Δ=

TV

TVTVV

Substitute numerical values and evaluate spillV :

( )( )[ ( )] L8.0K10113K10950.0C01C25L60.0 1613

spill ≈×−×°−°= −−−−V

31 ••• What is the tensile stress in the copper collar of Problem 27 when its temperature 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 ΔL/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.

C20

StrainStress°

Δ=×=

LLYY


1969

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 to obtain:

Substitute numerical values and evaluate the stress:

( ) 212210 N/m107.3cm5.9800

cm 9800.5cm0000.6N/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 temperature of 100ºC and a pressure of 1.00 atm. (b) Calculate the volume of 1.00 mol of steam at a temperature of 100ºC and a pressure of 1.00 atm. Use a = 0.550 Pa⋅m6/mol2 and b = 30.0 cm3/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) Use the ideal-gas law to find the volume: ( )( )( )

L6.30

m10L1m1006.3

atmkPa101.325atm00.1

K373KJ/mol314.8mol00.1

3332

=

××=

×

⋅=

=

−−

PnRTV

TT dL π=

C20C20 °° = dL π

C20

C20

C20

C20Stress

°

°

°

°

−=

−=

dddY

dddY

T

T

πππ

Chapter 20

1970

(b) Solve van der Waals equation for T to obtain:

( )

nR

bnVVanP

T−⎟⎟

⎠

⎞⎜⎜⎝

⎛+

=2

2


( )( )( )

( ) ( )( ) ( )

K375

KJ/mol8.314mol1.00mol00.1/molm1030.0m103.06

m103.06mol1.00/molmPa0.550kPa325.101

3632

232

226

=

⋅×−×

×

⎟⎟⎠

⎞⎜⎜⎝

⎛

×

⋅+=

−−

−T

33 •• [SSM] Using Figure 20-16, find the following quantities. (a) The temperature at which water boils on a mountain where the atmospheric pressure is 70.0 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 C82° .

(c) The pressure at which water boils at 115°C is approximately kPa170 .

34 •• The van der Waals constants for helium are a = 0.03412 L2⋅atm/mol2 and b = 0.0237 L/mol. Use these data to find the volume in cubic centimeters occupied 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 radius from 3

34 rV π= and its volume from the van der Waals equation.

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:

atomcm 1094.3

molatoms 106.022

Lcm10

molL 0237.0

323

23

33

−×=

×

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

=b


1971

The volume of a spherical helium atom is given by:

334 rV π= ⇒ 33

43

43

ππbVr ==

Substitute numerical values and evaluate r:

( ) nm211.04

cm 1094.333

323

=×

=−

πr

Conduction 35 • [SSM] A 20-ft × 30-ft slab of insulation has an R factor of 11. At what rate is heat conducted through the slab if the temperature on one side is a constant 68ºF and the temperature of the other side is a constant 30ºF? 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 through the slab in terms of the temperature difference across it and its thermal resistance:

RTI Δ

=

Substitute to express R in terms of the insulation’s R factor:

ff / RTA

ARTI Δ

=Δ

=

Substitute numerical values and evaluate I:

( )( )( )

hkBtu2.1

BtuFfth11

F30F68ft30ft202

=

°⋅⋅°−°

=I

36 •• A copper cube and an aluminum, cube each with 3.00-cm-long edges, are arranged as shown in Figure 20-18. Find (a) the thermal resistance of each cube, (b) the thermal resistance of the two-cube combination, (c) the thermal current I, and (d) the temperature at the interface of the two cubes. Picture the Problem We can use kAxR Δ= to find the thermal resistance of each cube and the fact that they are in series to find the thermal resistance of the two-cube system. We can use RTI Δ= to find the thermal current through the cubes and the temperature at their interface. See Table 20-3 for the thermal conductivities of copper and aluminum.

Chapter 20

1972

(a) Using its definition, express the thermal resistance of each cube:

kAxR Δ

=

Substitute numerical values and evaluate the thermal resistance of the copper cube:

( )

K/W0831.0K/W08313.0

cm3.00Km

W401

cm3.002

Cu

==

⎟⎠⎞

⎜⎝⎛

⋅

=R

Substitute numerical values and evaluate the thermal resistance of the aluminum cube:

( )

K/W141.0K/W1406.0

cm3.00Km

W372

cm3.002

Al

==

⎟⎠⎞

⎜⎝⎛

⋅

=R

(b) Because the cubes are in series, their thermal resistances are additive: K/W0.224K/W0.2237

K/W0.1406K/W0.08313AlCu

==

+=+= RRR

(c) Using its definition, find the thermal current: kW0.36 W6.357

K/W0.2237C20C100Δ

==

°−°==

RTI

(d) Express the temperature at the interface between the two cubes:

Cuinterface ΔC100 TT −°=

Express the temperature differential across the copper cube:

CuCuCuCu IRRIT ==Δ

Substitute for ΔTCu to obtain: Cuinterface C100 IRT −°=

Substitute numerical values and evaluate Tinterface:

( )( )C70

K/W08313.0W357.6C100interface

°≈

−°=T

37 •• A copper tube and an aluminum tube, each with 3.00-cm-long edges, are arranged in parallel, as shown in Figure 20-18. Find (a) the thermal current in each cube, (b) the total thermal current, and (c) the thermal resistance of the two-cube combination.


1973

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 ITR Δ= .

(a) The thermal current in each cube is given by Equation 20-9:

xTkA

RTI

ΔΔ

=Δ

=

Substitute numerical values and evaluate the thermal current in the copper cube:

( ) kW96.0 W4.962cm3.00

C20C100cm3.00Km

W401 2Cu ==⎟⎟

⎠

⎞⎜⎜⎝

⎛ °−°⎟⎠⎞

⎜⎝⎛

⋅=I

Substitute numerical values and evaluate the thermal current in the aluminum cube:

( ) kW57.0 W8.568cm3.00

C20C100cm3.00Km

W372 2Al ==⎟⎟

⎠

⎞⎜⎜⎝

⎛ °−°⎟⎠⎞

⎜⎝⎛

⋅=I

(b) Because the cubes are in parallel, their total thermal currents are additive:

kW1.5kW 1.531

W8.685W4.629AlCu

==

+=+= III

(c) Use the relationship between the total thermal current, temperature differential and thermal resistance to find Req:

K/W0.052

kW1.531C20C100Δ

totaleq

=

°−°==

ITR

38 •• The cost of air conditioning a house is approximately proportional to the rate at which heat is absorbed by the house from its surroundings divided by the coefficient of performance (COP) of the air conditioner. Let us denote the temperature difference between the inside temperature and the outside temperature 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 inside the house. Picture the Problem The cost of operating the air conditioner is proportional to the energy used in its operation. We can use the definition of the COP to relate the rate at which the air conditioner removes heat from the house to rate at which it must do work to maintain a constant temperature differential between the interior and 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 the

Chapter 20

1974

maximum efficiency possible. Doing so will allow us to derive an expression for the rate at which energy is used by the air conditioner that we can integrate to obtain the energy (and hence the cost of operation) required. Relate the cost C of air conditioning the energy W required to operate the air conditioner:

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

Express the rate dQ/dt at which heat flows into a house provided the house is maintained at a constant temperature:

TkdtdQP Δ==

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

Use the definition of the COP to relate the rate at which the air conditioner must remove heat dW/dt to maintain a constant temperature:

dtdWdtdQ

=COP ⇒dtdQ

dtdW

COP1

=

Express the maximum value of the COP: T

TΔ

= cmaxCOP

where house

theinsidec TT = is the temperature

of the cold reservoir.

Letting COP = COPmax, substitute to obtain an expression for the minimum rate at which the air conditioner must do work in order to maintain a constant temperature:

TTdtdQ

dtdW Δ

c

=

Substituting for dQ/dt gives: ( )2

cc

TTkT

TTk

dtdW

Δ=ΔΔ

=

Separate variables and integrate this equation to obtain:

( ) ( ) tTTkdt'T

TkW

t

ΔΔ=Δ= ∫Δ

2

c0

2

c

Substitute in equation (1) to obtain:

( ) tTTkuC ΔΔ 2

c

= ⇒ ( )

c

2ΔTTC ∝

39 •• A spherical shell of thermal conductivity k has inside radius r1 and outside radius r2 (Figure 20-19). The inside of the shell is held at a temperature T1, and the outside of the shell is held at temperature T2, with


1975

T1 < T2. In this problem, you are to show that the thermal current through the shell is given by

I = −4πkr1r2

r2 − r1

T2 − T1( )

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, where r1 < r < r2; (2) explain why the thermal 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 = 4πr2, the thickness dr, and the temperature difference dT across the element; and (4) separate variables (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 gradient drdT is given by:

drdTkAI =

(2) Conservation of energy requires that the thermal current through each shell be the same. (3) For a shell of area A = 4π r2:

drdTkrI 24π=

(4) Separating the variables yields:

24 rdr

kIdTπ

=

Integrate from r = r1 to r = r2 and simplify to obtain:

∫∫ =2

1

2

1

24

r

r

T

T rdr

kIdTπ

and

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=⎥⎦

⎤⎢⎣⎡−=−

2112

114

14

2

1rrk

Irk

ITTr

r ππ

Solving for I gives: ( )12

12

214 TTrrrkrI −

−−=

π

Chapter 20

1976

Radiation

40 • Calculate λmax (the wavelength at which the emitted power is maximum) for a human skin. Assume the human skin is a blackbody emitter with a temperature of 33ºC. Picture the Problem We can apply Wein’s displacement law to find the wavelength at which the power is a maximum. Wein’s law relates the maximum wavelength of the radiation to the temperature of its source:

TKmm898.2

max⋅

=λ

Substitute numerical values and evaluate λmax:

m47.9K33K273Kmm2.898

max μλ =+

⋅=

41 • [SSM] The universe is filled with radiation that is believed to be remaining 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 λmax (the wavelength at which 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 this radiation. Wein’s law relates the maximum wavelength of the background radiation to the temperature of the universe:

TKmm2.898

max⋅

=λ

Substitute the universe’s temperature to obtain:

mm .31K3.2

Kmm2.898max =

⋅=λ

42 • What is the range of temperatures for star surfaces for which λmax (the wavelength at which the power of the emitted radiation is maximum) is in the visible range? Picture the Problem The visible portion of the electromagnetic spectrum extends from approximately 400 nm to 700 nm. We can use Wein’s law to find the range of temperatures corresponding to these wavelengths. Wein’s law relates the maximum wavelength of the radiation to the temperature of its source:

TKmm2.898

max⋅

=λ


1977

Solving for T yields:

max

Kmm2.898λ

⋅=T

For λmax = 400 nm: K 7250

nm400Kmm2.898

=⋅

=T

For λmax = 700 nm: K 4140

nm 700Kmm2.898

=⋅

=T

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

43 • The heating wires of a 1.00-kW electric heater are red hot at a temperature of 900ºC. Assuming that 100 percent of the heat released is due to radiation 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:

( )40

4net TTAeP −= σ ⇒ ( )4

04net

TTePA

−=

σ

Substitute numerical values and evaluate A:

( ) ( ) ( )[ ]2

4442

8cm5.93

K293K1173Km

W105.67031

kW1.00=

−⎟⎠⎞

⎜⎝⎛

⋅×

=−

A

44 •• A blackened, solid copper sphere that has a radius equal to 4.0 cm hangs in an evacuated enclosure whose walls have a temperature of 20ºC. If the sphere 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 energy is given by dtmcdTdtdQ // = and, from the Stephan-Boltzmann law,

( )40

4net TTAeP −= σ where A is the surface area of the sphere, T0 is its

temperature, and T is the temperature of the walls. We can solve the first equation for dT/dt and substitute Pnet for dQ/dt in order to find the initial rate at which the temperature of the sphere changes.

Chapter 20

1978

Relate the rate at which the sphere absorbs radiant energy to the rate at which its temperature changes:

dtdTmc

dtdQP ==net

Solving for dtdT and substituting

for mc gives:

crP

VcP

mcP

dtdT

ρπρ 334

netnetnet ===

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

042

40

4net

4 TTer

TTAeP

−=

−=

σπ

σ

Substitute for netP to obtain: ( )

( )cr

TTe

crTTer

dtdT

ρσ

ρπσπ

40

4

334

40

42

3

4

−=

−=

Substitute numerical values and evaluate dT/dt:

( ) ( ) ( )[ ]( )

K/s102.2

KkgkJ0.386

mkg108.93m104.0

K273K293Km

W105.6703133

332

4442

8

−

−

−

×=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⎟⎠⎞

⎜⎝⎛ ××

−⎟⎠⎞

⎜⎝⎛

⋅×

−=dtdT

45 •• The surface temperature of the filament of an incandescent lamp is 1300ºC. 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

40

4net

1TTATe

TTAeP

σ

σ


1979

Evaluate 4

0 ⎟⎠⎞

⎜⎝⎛

TT

: 444

0 109K1573K273 −×≈⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

TT

Because 4

0 ⎟⎠⎞

⎜⎝⎛

TT is so small, we can

neglect the temperature of the surroundings. Hence:

4net ATeP σ≈ ⇒

41net ⎟

⎠⎞

⎜⎝⎛=

AePTσ

Express the temperature T ′when the electric power input is doubled:

41net2

⎟⎠⎞

⎜⎝⎛=

AePT'σ

Divide the second of these equations by the first:

( ) 412=TT'

⇒ ( ) TT' 412=

Substitute numerical values and evaluate T ′

( ) ( )C1598

K1871K15732 41

°=

==T'

46 •• Liquid helium is stored at its boiling point (4.2 K) in a spherical can that is separated by an evaluated region of space from a surrounding shield that is maintained at the temperature of liquid nitrogen (77 K). If the can is 30 cm in diameter 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:

( )

( )

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−=

−=

−==

404

2

40

42

40

4net

1TTT

Lde

LTTdeL

TTAeL

Pdtdm

σπ

σπ

σ

If T0 << T, then:

42

TL

dedtdm σπ

≈

Substitute numerical values and evaluate dm/dt:

Chapter 20

1980

( ) ( )( )

g/h97

hs3600

skg102.68K77

kgkJ21

m30.0Km

W105.6703154

242

8

=

××=⎟⎠⎞

⎜⎝⎛

⋅×

≈ −

− π

dtdm

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

Express the change in the radius of the circle defined by the steel tape:

TRR Δ=Δ α where R is the radius of the earth, α is the coefficient of linear expansion of steel, and ΔT is the increase in temperature.

Substitute numerical values and evaluate ΔR:

( )( )( ) km1.2m1010.2C0C30K1011m106.37Δ 3166 =×=°−°××= −−R

48 •• Show that change in the density ρ of an isotropic material due to an increase in temperature ΔT is given by Δρ = –βρΔT. 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 changes with respect to temperature. Once we have an expression for dρ in terms of dT, we can apply a differential approximation to obtain Δρ in terms of ΔT. Using its definition, relate the density of the material to its mass and volume:

Vm

=ρ


1981

Using its definition, relate the volume of the material to its coefficient of volume expansion:

TVV Δ=Δ β

Differentiate ρ with respect to T and simplify to obtain:

ρββρ

βρρ

−=−=

−==

VV

V

VVm

dTdV

dVd

dTd

2

2

or dTd ρβρ −=

Using the differential approximations

ρρ Δ≈d and TdT Δ≈ yields: TΔΔ ρβρ −=

49 •• [SSM] The solar constant is the power received from the Sun per unit area perpendicular to the Sun’s rays at the mean distance of Earth from the Sun. Its value at the upper atmosphere of Earth is about 1.37 kW/m2. 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:

4r ATeP σ= ⇒ 4 r

AePTσ

=

Express the surface area of the sun: 2S4 RA π=

Relate the intensity of the sun’s radiation in the upper atmosphere to the total power radiated by the sun:

24 RPI r

π= ⇒ IRPr

24π=

where R is the earth-sun distance.

Substitute for Pr and A in the expression for T and simplify to obtain:

42S

2

42S

2

44

ReIR

ReIRT

σπσπ

==

Chapter 20

1982


( )

( ) ( )K5800

m106.96Km

W105.671

mkW1.35m101.5

428

28

2

211

=×⎟

⎠⎞

⎜⎝⎛

⋅×

⎟⎠⎞

⎜⎝⎛×

=−

T

50 •• As part of your summer job as an engineering intern at an insulation manufacturer, 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-W heater inside the box. After thermal equilibrium has been attained, the temperature 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:

xTkAI

ΔΔ

=

Rewrite this expression in terms of the R factor of the material: ff R

TA

ART

kAxTI Δ

=Δ

=ΔΔ

=

Solving for the R factor gives:

ITA

ITAR

Δ=

Δ= sideone

f

6

Substitute numerical values and evaluate R:

( )

BtuhftF2.2

s3600h1

BtuJ1054

mft10.76

K5F9

sJmK0.390

WmK0.39C20C90

W100in

m102.54in126

2

2

22

2

22

f

⋅⋅°=×××

°×

⋅=

⋅=°−°

⎟⎟⎠

⎞⎜⎜⎝

⎛ ××

=

−

R


1983

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 an ideal 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:

dTdV

V1

=β

For an ideal gas: P

nRTV = and P

nRdTdV

=

Substitute fordTdV to obtain:

TP

nRV

11==β

(b) Express the percent difference between the experimental value and the theoretical value:

%3.0

K2731

K2731K0.003673

1

11

th

thexp

<

−=

−

−

−−

βββ

52 •• A rod of length LA, made from material A, is placed next to a rod of length LB, made of material B. The rods remain in thermal equilibrium with each other. (a) Show that even though the lengths of each rod will vary with changes in the ambient temperature, the difference between the two lengths will remain constant if the lengths LA and LB are chosen such that LA/LB = αB/αA, where αA and αB are the coefficients of linear expansion, respectively. (b) If material B is steel, material A is brass, and LA = 250 cm at 0ºC, what is the value of LB? Picture the Problem Let ΔL be the difference between LB and LA and express LB and LA in terms of the coefficients of linear expansion of materials A and B. Requiring that ΔL be constant will lead us to the condition that LA/LB = αB/αA.

Chapter 20

1984

(a) Express the condition that ΔL does not change when the temperature of the materials changes:

constantAB =−=Δ LLL (1)

Using the definition of the coefficient of linear expansion, substitute for LB and LA:

( ) ( )( ) ( )

( ) TLLLTLLLL

TLLTLLL

Δ−+Δ=Δ−+−=

Δ+−Δ+=Δ

AABB

AABBAB

AAABBB

αααα

αα

or ( ) 0AABB =Δ− TLL αα

The condition that ΔL remain constant when the temperature changes by ΔT is:

0AABB =− LL αα

Solving for the ratio of LA to LB yields:

A

B

B

A

αα

=LL

(b) From (a) we have: steel

brassbrass

B

AAsteelB α

ααα LLLL ===

Substitute numerical values and evaluate LB: ( ) cm430

K1011K1019cm250 16

16

B =××

= −−

−−

L

53 •• [SSM] On the average, the temperature of Earth’s crust increases 1.0ºC 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 second due to conduction from the core? How does this heat loss compare with the average power received from the Sun (which is about 1.37 kW/m2)? 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:

xTkA

dtdQI

ΔΔ

== (1)

or

xTkR

dtdQ

ΔΔ

= 2E4π


1985

Substitute numerical values and evaluate dQ/dt:

( ) kW103.1m30C1.0

KsmJ0.74m1037.64 1026 ×=⎟⎟

⎠

⎞⎜⎜⎝

⎛ °⎟⎠⎞

⎜⎝⎛

⋅⋅×= π

dtdQ

Rewrite equation (1) to express the thermal current per unit area:

xTk

AI

ΔΔ

=

Substitute numerical values and evaluate I/A: ( )

2W/m0.0247

m30C1.0KsJ/m0.74

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ °⋅⋅=

AI

Express the ratio of I/A to the solar constant:

0 007

2

2

0.0247W/msolar constant 1.37 kW/m

. %

I A=

<

54 •• A copper-bottomed saucepan containing 0.800 L of boiling water boils dry 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, calculate the temperature of the outside of the copper bottom while some water is still in the 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:

TTTT ΔC100Δinout +°=+= (1)

Relate the thermal current through the bottom of the saucepan to its thermal conductivity, area, and the temperature gradient between its surfaces:

xTkA

tQ

ΔΔ

=ΔΔ

⇒ xtQ

kAT Δ

ΔΔ

=Δ1

Because the water is boiling: vmLQ =Δ

Chapter 20

1986

Substitute for ΔQ to obtain: tkA

xmLT

ΔΔ

Δ v=

Substituting for ΔT in equation (1) yields:

tkAxmLT

ΔΔC100 v

out +°=

Substitute numerical values and evaluate Tout:

( ) ( )

( ) ( )C101

s600m0.1504Km

W401

m103.00kgMJ2.26kg0.800

C1002

3

out °=

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛

⋅

×⎟⎟⎠

⎞⎜⎜⎝

⎛

+°=

−

πT

55 •• A steel hot-water tank of cylindrical shape has an inside diameter 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 0.0350 W/(m⋅K). The insulation is covered by a thin sheet-metal skin. The steel tank and the sheet-metal skin have thermal conductivities that are much greater than that of the glass wool. How much electrical power must be supplied to this tank in order to maintain the water temperature at 75.0ºC when the external temperature is 1.0ºC? 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:

sidebottomandtop III += (1)

Express I through the top and bottom surfaces:

xTkd

xTkAI

ΔΔ

ΔΔ2 2

21

bottomandtop π=⎟⎠⎞

⎜⎝⎛=

Substitute numerical values and evaluate Itop and bottom:

( )( )

W6.24m0.0500

C0.1C75.0Km

W0.0350m550.0 2

21

bottomandtop =°−°⎟

⎠⎞

⎜⎝⎛

⋅= πI


1987

Letting r represent the inside radius of the tank, express the heat current through the cylindrical side:

drdTkLr

drdTkAI π2side −=−=

where the minus sign is a consequence of the heat current being opposite the temperature gradient.

Separating the variables yields: rdr

kLIdTπ2

side−=

Integrate from r = r1 to r = r2 and T = T1 to T = T2 and simplify to obtain:

∫∫ −=2

1

2

12

sider

r

T

T rdr

kLIdTπ

and

]

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−=−

2

1side

1

2side

side12

ln2

ln2

ln2

2

1

rr

kLI

rr

kLI

rkL

ITT rr

π

π

π

Solving for Iside gives:

( )12

2

1side

ln

2 TT

rrkLI −

⎟⎟⎠

⎞⎜⎜⎝

⎛=

π

Substitute numerical values and evaluate Iside:

( )( ) W117C0.1C0.75

m0.275m0.325ln

m1.20Km

W0.03502side =°−°

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

⋅=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 = d0(1 + ax), where a is a constant and x is the distance from one end. If the thermal conductivity of the material is k what is the thermal resistance of the rod?

Chapter 20

1988

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:

ITR 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:

dxdTkAI −=

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 the area of the rod:

( )220

2

144

axddA +==ππ

Substitute for A to obtain:

( )dxdTaxdkI ⎥⎦

⎤⎢⎣⎡ +−= 22

0 14π

Separating variables yields: ( )

( )220

220

14

14

axdx

kdI

axdk

IdxdT

+−=

⎥⎦⎤

⎢⎣⎡ +

−=

π

π

Integrating T from T1 to T2 and x from 0 to L gives: ( )∫∫ +

−=LT

T axdx

kdIdT

022

0 142

1π

and

( )aLkdILTTT

+=Δ=−

1420

12 π

Substitute for ΔT and I in equation (1) and simplify to obtain: ( )

( )aLkdL

IaLkd

IL

R+

=+

=1

414

20

20

ππ

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 ω1 at temperature T1. The temperature of the disk decreases to T2. Express the angular


1989

velocity ω2, rotational kinetic energy K2, and angular momentum L2 in terms of their values at the temperature T1 and the linear expansion coefficient α of the disk. Picture the Problem Let ΔT = T2 – T1. We can apply Newton’s 2nd law to establish the relationship between L2 and L1 and angular momentum conservation to relate ω2 and ω1. We can express both E2 and E1 in terms of their angular momenta and rotational inertias and take their ratio to establish their relationship.

Apply tL

ΔΔ

=∑τ to the spinning

disk:

Because 0=∑τ , ΔL = 0

and

12 LL =

Apply conservation of angular momentum to relate the angular velocity of the disk at T2 to the angular velocity at T1:

1122 ωω II = ⇒ 12

12 ωω

II

= (1)

Express I2: ( )( )( )2

1

221

222

ΔΔ21

Δ1

TTI

TmrmrI

αα

α

++=

+==

Because (αΔT)2 is small compared to αΔT:

( )TII Δ2112 α+≈

Substitute for I2 in equation (1) to obtain:

( ) 11

12 Δ21

ωα

ωTI

I+

= (2)

Expanding ( ) 1Δ21 −+ Tα binomially yields:

( )sorder termhigher

Δ21Δ21 1

+−=+ − TT αα

Because the higher order terms are very small:

( ) TT Δ21Δ21 1 αα −≈+ −

Substituting in equation (2) yields: ( ) 12 21 ωαω TΔ−≈

Express E2 in terms of L2 and I2:

2

21

2

22

2 22 IL

ILE == because L2 = L1.

Chapter 20

1990

Express E1 in terms of L1 and I1:

1

21

1 2ILE =

Express the ratio of E2 to E1 and simplify to obtain:

2

1

1

21

2

21

1

2

2

2II

ILI

L

EE

== ⇒2

112 I

IEE =

Substituting for the ratio of I1 to I2 yields:

( )TEEE Δ2111

212 α

ωω

−==

58 ••• Write a spreadsheet program to graph the average temperature of the surface of Earth as a function of emissivity, using the results of Problem 22. How much does the emissivity have to change in order for the average temperature to increase by 1 K? This result can be thought of as a model for the effect of increasing concentrations of greenhouse gases like methane and CO2 in Earth’s atmosphere. Picture the Problem The amount of heat radiated by Earth must equal the solar flux 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 ATeP σ= Using the Stefan-Boltzmann law, express the rate at which Earth radiates energy as a function of its emissivity e and temperature T:

4r A'TeP σ=

where A′ is the surface area of the earth.

Use its definition to express the intensity of the radiation Pa absorbed by Earth:

API a= or AIP =a

where A is the cross-sectional area of the earth.

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

( )AIP 70.0a =

Equate Pr and Pa and simplify:

( ) 470.0 A'TeAI σ= or ( ) ( )422 470.0 TReIR σππ =

Solve for T to obtain: ( ) 414470.0 −== Ce

eIT

σ (1)


1991

Substitute numerical values for I and σ and simplify to obtain:

( )( )( )

( ) 41

4428

2

K255

KW/m10670.54W/m137070.0

−

−

=

⋅×=

e

eT

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

Cell Formula/Content Algebraic FormB1 255 B4 0.4 e B5 B4+0.01 e + 0.1 C4 $B$1/(B4^0.25) ( ) 41K255 −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

)

Chapter 20

1992

Treating e as a variable, differentiate equation (1) to obtain: deCe

dedT 45

41 −−= (2)

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

ede

Ce

deCe

TdT

414

1

41

45

−=−

= −

−

Use a differential approximation to obtain:

ee

TT Δ

−=Δ

41

⇒TTee Δ

−=Δ 4

Substitute numerical values (e ≈ 0.615 for Tearth = 288 K) and evaluate Δe:

( ) 00854.0K288

K1615.04Δ −=−=e

or about a 1.4% change in e. 59 ••• [SSM] A small pond has a layer of ice 1.00 cm thick floating on its surface. (a) If the air temperature is –10ºC on a day when there is a breeze, find the rate in centimeters per hour at which ice is added to the bottom of the layer. The density of ice is 0.917 g/cm3. (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:

fmLQ = ⇒dtdmL

dtdQ

f=

Using the definition of density, relate the mass of the ice added to the bottom of the layer to its density and volume:

AxVm ρρ ==

Differentiate with respect to time to express the rate of build-up of the ice:

dtdxA

dtdm ρ=

Substitute for dtdm to obtain:

dtdxAL

dtdQ ρf=


1993

The thermal-current equation is: xTkA

dtdQ Δ

=

Equate these expressions and

solve for dtdx :

xTkA

dtdxAL Δ

=ρf ⇒xT

Lk

dtdx Δ

=ρf

(1)

Substitute numerical values and

evaluate dtdx :

( )( )

( )( )

cm/h0.70

cm/h0.697h

s 3600sm94.1

m0.0100kg/m917kgkJ333.5

C10C0Km

W0.592

3

=

=×=

⎟⎟⎠

⎞⎜⎜⎝

⎛

°−−°⎟⎠⎞

⎜⎝⎛

⋅=

μ

dtdx

(b) Separating the variables in equation (1) gives:

dtL

Tkxdxρf

Δ=

Integrate x from xi to xf and t′ from 0 to t: 'dt

LTkxdx

tx

x∫∫

Δ=

0f

f

iρ

and

( ) tLTkxxf

2i

2f2

1

ρΔ

=− ⇒( )

TkxxLt

Δ−

=2

2i

2ffρ

Substitute numerical values and evaluate t:

( )( )( ) ( )[ ]

d12h24

d1s3600

h1s1003.1

m0.010m0.200C10C0

KmW0.5922

kgkJ333.5

mkg917

6

223

=×××=

−°−−°⎟

⎠⎞

⎜⎝⎛

⋅

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

=t

60 ••• A blackened copper cube that has 1.00-cm-long edges is heated to a temperature of 300ºC, and then is placed in a vacuum chamber whose walls are at a temperature of 0ºC. In the vacuum chamber, the cube cools radiatively. (a) Show that the (absolute) temperature T of the cube follows the

differential equation:

dTdt

= −eσ A

CT 4 − T0

4( ) where C is the heat capacity of the

cube, A is its surface area, e the emissivity, and To the temperature of the vacuum

Chapter 20

1994

chamber. (b) Using Euler’s method (Section 5.4 of Chapter 5), numerically solve the differential equation to find T(t), and graph it. Assume e = 1.00. How long does it take the cube to cool to a temperature of 15ºC? Picture the Problem We can use the Stefan-Boltzmann equation and the definition of heat capacity to obtain the differential equation expressing the rate at which the temperature of the copper block decreases. We can then approximate the 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:

( )40

4 TTAedtdQ

−= σ

Using the definition of heat capacity, relate the thermal current to the rate at which the temperature of the cube is changing:

dtdTC

dtdQ

−=

Equate these expressions and solve

for dtdT to obtain:

( )40

4 TTC

AedtdT

−−=σ

Approximating the differential equation by the difference equation gives:

( )40

4 TTC

AetT

−−=ΔΔ σ

Solving for ΔT yields: ( ) tTTC

AeT Δ−−=Δ 40

4σ

or

( ) tTTC

AeTT nnn Δ−−=+4

04

1σ (1)

Use the definition of heat capacity to obtain:

VcmcC ρ==

Substitute numerical values (see Figure 13-1 for ρCu and Table 19-1 for cCu) and evaluate C:

( )KJ45.3

KkgkJ386.0m1000.1

mkg1093.8 36

33 =⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

×⎟⎠⎞

⎜⎝⎛ ×= −C

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


1995

Cell Formula/Content Algebraic Form B1 5.67×10−8 σ B2 6.00×10−4 A B3 3.45 C B4 273 T0 B5 10 Δt A9 A8+$B$5 t+Δt B9 B8-($B$1*$B$2/$B$3)

*(B8^4−$B$4^4)*$B$5 ( ) tTTC

AeT nn Δ−− 40

4σ

A B C 1 σ= 5.67E−08 W/m^2⋅K^4 2 A= 6.00E−04 m^2 3 C= 3.45 J/K 4 T0= 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

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

)

Chapter 20

1996

physics ch20 ism

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