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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. The bottle will be broken.
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. The coefficient of linear expansion of wood is about 5 × 10 −^6 K−^1. 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 about half 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 different 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?
1864 Chapter 20
Determine the Concept The strip will curl more tightly. Because the coefficient of linear expansion for copper (17 × 10 −^6 K−^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 volume
expansion β compare? ( a ) βA = βB , ( b ) βA = 3 βB , ( c ) βA = 6 βB , ( d ) βA = 9 βB ,
( 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 β Ato β B.
Express the coefficient of volume expansion of metal A in terms of its coefficient of linear expansion:
β A = 3 α A
Express the coefficient of volume expansion of metal B in terms of its coefficient of linear expansion:
β B = 3 α B
Dividing the first of these equations by the second yields: (^) B
A B
A B
A 3
Because αA = 3 αB:
B
B B
A = =
⇒ ( 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.
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, H 2 , O 2 , NO.
1866 Chapter 20
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 d A = 2 d B. 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 ) k A = k B/4, ( b ) k A = k B/2, ( c ) k A = k B, ( d ) k A = 2 k B, ( e ) k A = 4 k BB
Picture the Problem The rate at which heat is conducted through a cylinder is given by I = dQ / dt = kA Δ T /Δ x (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:
I A = I B
Substituting for the heat currents
yields: L
T
k A L
T
k A
A A B B ⇒
A
B A B A
A
k = k
Because d A = 2 d B: ⎟⎟ ⎠
B
B A B 4 A
A
k k ⇒ k (^) A = 41 k B
( a ) is correct.
14 •• Two solid cylinders made of materials A and B have the same diameter; their lengths are related by L A = 2 L B. 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 ) k A = k B/4, ( b ) k A = k B/2, ( c ) k A = k B, ( d ) k A = 2 k B, ( e ) k A = 4 k B.B
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.
Thermal Properties and Processes 1867
The heat current in cylinder A is the
same as the heat current in cylinder
B:
I A = I B
Substituting for the heat currents yields: A^ A B L B
T
k A L
T
k A
B
A A B L
L
k = k
Because L A = 2 L B: B B
B A B^2
k L
L
k = 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 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. T , ° C
20
5
Inside Outside
x ,m 0 5
16 •• During the thermal retrofitting of many older homes in California, it was 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 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.
Thermal Properties and Processes 1869
Express the heat current in terms of
the rate of evaporation of the liquid
helium:
dt
dm I = L v
Express the heat current in terms of the temperature gradient across the
superinsulation and the conductivity of the superinsulation:
x
T
I kA Δ
Equate these expressions and solve
for k to obtain: A T
dt
dm L x k Δ
v
Using the definition of density, express the rate of loss of liquid
helium:
dt
dV dt
dm
Substitute for dt
dm to obtain:
A T
dt
dV L x k Δ
v ρ
Express the ratio of the area of the
spherical container to its volume: 34 3
r
r V
A
= ⇒ A =^3 36 π V^2
Substituting for A yields:
V T
dt
dV L x k Δ
v
Substitute numerical values and evaluate k :
( )
( ) ( ) m K
W
36 200 10 m 289 K
86400 s
0.700 10 m m
kg 7.00 10 m 125 kg
kJ
6 3 3 32
3 3 3
2
= ×
×
⎛ ×
⎟⎟ ×
−
− −
k
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 m^2. 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.
1870 Chapter 20
Use the thermal current equation to express the rate of conduction of thermal energy:
I = kA
Δ T
Δ x
x
T
A
I
k
Δ
Substitute numerical values and evaluate k : ( ) m K
mW 17
- 0 10 m
37 C 33 C
- 8 m
120 W
3
×
−
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 to the cold air. 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 m^2. 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:
P net (^) =εσ A^ ( T^ skin^4 − T air^4 )
Substitute numerical values and evaluate P net:
( ) ( 1. 8 m ) (( 310 K) ( 263 K)) 450 W m K
W
net 1 5.^67010824 ⎟^24 −^4 = ⎠
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 wooden 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 85 W(m ⋅ K).
1872 Chapter 20
Substitute for R door and R knob to obtain:
x
k A k A
k A
x k A
R x
door door knob knob
door door knob knob
eq
Solving for R eq yields:
door door knob knob
eq
k A k A
x R
Substitute numerical values and evaluate R eq:
( )
( ) ( )
W
K
- 0 10 m m K 4
W
- 0 m 85 m K
W
39.37in
1 m
- 0 in eq (^222)
⎟ ×
π −
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/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 R^2 , where R is Earth’s radius, while the blackbody-emission area is 4 π R^2 .)
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 P r = e σ AT^4.
Using the Stefan-Boltzmann law, express the rate at which Earth radiates energy as a function of its emissivity e and temperature T :
4 P r = e σ A'T ⇒ (^4) r A'T
P
e
where A ′ is the surface area of Earth.
Use its definition to express the intensity of the radiation received by Earth:
A
P
I = absorbed
where A is the cross-sectional area of Earth.
For 70% absorption of the Sun’s radiation incident on Earth:
( ) A
P
I r
Substitute for P r and A in the expression for e and simplify to
( ) ( ) ( ) 2 4 4
2 (^4 )
T
I
R T
R I
A'T
AI
e
Thermal Properties and Processes 1873
obtain: Substitute numerical values and evaluate e :
( )( ) ( )(^ )
61
670 10 W/m K 288 K
70 1. 37 kW/m 8 2 4 4
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:
T
- 898 mm K max
Substitute for T and evaluate λmax:
9 nm
0 10 K
898 mm K max (^) × 6 =
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 of 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 20-5 for the R -factor of gypsum wallboard and Table 20-4 for the thermal conductivity of white pine.
The rate at which heat is conducted through the walls is given by the thermal current equation: before
before before
R
T
x
T
I = k A =
Thermal Properties and Processes 1875
assume that the surface area of the ice chest is 1.0 m^2 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:
R (^) f = RA (1)
Use the thermal current equation to express R : tot tot
Q
T t
t
Q
T
I
T
R = = =
Substitute for R in equation (1) to obtain: tot
f
Q
A T t R =
The total heat entering the chest in 7 h is given by:
ice f ice HO H O
icewater warm ice tot melt
m L m c 2 Δ T 2
Q Q Q
Substitute for Q tot and simplify to obtain: (^) ( ) ice f HO HO
f 2 Δ 2
m L c T
A T t R
Substitute numerical values and evaluate R f:
( )
( ) ( )
Btu
F h ft 8 5 C kg K
kJ
- 18 kg
kJ 1 kg 333. 5
Btu
1054. 35 J
7 h 5 C
9 F
20 C
- 29 10 m
1 ft 1.0 m (^222)
2 2
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.
1876 Chapter 20
( a ) Express the period of the pendulum in terms of its length: (^) g
L
T P = 2 π
Because T (^) P ∝ L 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
L
T 20 (^) = 2 π^20 (1)
When the temperature increases to 30 °C, the period of the pendulum increases due to the increase in its length:
20 C
20 C
T t
g
L t g
L
T
The daily fractional gain or loss is given by : (^20)
20 20 T
T T
T
T −
Substituting for T and simplifying yields:
1 1
C
20
20 C 20
t
T
T t T
Solve for Δτ to obtain: Δτ =( 1 +αΔ t C − 1 ) τ
Substitute numerical values and Δτ:
( ( )( ) ) 8. 2 s
h
3600 s 1 19 106 K^130 C 20 C 1 24 h ⎟= ⎠
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 T = T i+Δ T
1878 Chapter 20
increased by Δ T :
If the collar is to fit over the shaft when the temperature of both has been increased by Δ T :
( ) D ( T )
D T
°
° steel,20C steel
Cu,20C Cu 1
α
α
Solving for Δ T yields: Cu,20C Cu steel,20C steel
steel,20C Cu,20C
°α^ °^ α
° ° −
D D
D D
T
Substitute in equation (1) to obtain: Cu,20C Cu steel,20C steel
steel,20C Cu,20C
i α α
° °
° ° −
D D
D D
T T
Substitute numerical values and evaluate T :
( )( ) ( )( )
580 C
5.9800cm 17 10 /K 6.0000cm 11 10 /K
6.0000cm 5.9800cm 20 C 6 6 = ° × − ×
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 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:
3 c 1
α c= β (1)
Express the difference in the change
in the volume of the mercury and the container in terms of the spillage:
Δ V (^) Hg (^) −Δ V c= 7. 5 mL (2)
Express using the definition of
the coefficient of volume expansion:
Δ V Hg Δ V Hg= βHg V HgΔ T
Express using the definition of
the coefficient of volume expansion:
Δ V c Δ V c= βc V cΔ T
- Thermal Properties and Processes
Thermal Properties and Processes 1881
Relate to the coefficient of
linear expansion for steel:
Δ V tank Δ V tank= βtank V Δ T
or, because βsteel = 3 αsteel,
Δ V tank = 3 αsteel V Δ T
Substitute for and
and simplify to obtain:
Δ V (^) gasoline Δ V tank ( (^) gasoline steel)
spill gas steel 3
V T
V V T V T
Substitute numerical values and evaluate V spill:
V spill = ( 60.0 L)( 25 °C− 10 °C)[ 0. 950 × 10 −^3 K−^1 − 3 ( 11 × 10 −^6 K−^1 )] ≈ 0. 8 L
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 Δ L / L for the circumference of the collar is the same as Δ d / d for its diameter.
Using Young’s modulus, relate the stress in the collar to its strain: where L 20 °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:
L T L 20 °C
20 C
Stress Strain °
= × =
L
L
Y Y
LT = π d T
L 20 °C = π d 20 ° C
20 C
20 C
20 C
Stress^20 C
°
°
°
°
d
d d Y
d
d d Y
T
T
1882 Chapter 20
Substitute numerical values and evaluate the stress:
( 10 2 ) 3. 7 1012 N/m^2 5.9800cm
- 0000 cm 5. 9800 cm Stress 11 10 −^ N/m = × −
= ×
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 temperature at which 1.00 mol of steam at a pressure of 1.00 atm has the volume calculated in Part ( a ). Use a = 0.550 Pa⋅m^6 /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.
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 =
Substitute numerical values and
evaluate V :
( )( )( )
30. 6 L
10 m
1 L
- 06 10 m
atm
101.325kPa
00 atm
00 mol 8. 314 J/mol K 373 K
3 3
2 3
= × ×
×
−
−
V
( b ) Solve van der Waals equation for T to obtain: (^ )
nR
V bn V
an P T
⎟⎟^ −
2
2
Substitute numerical values and evaluate T :
( )( ) ( ) ( ) ( ) ( ) ( ) 375 K
1.00mol 8.314J/mol K
3.06 10 m 30.0 10 m /mol 1. 00 mol
3.06 10 m
0.550Pa m /mol 1.00mol
- 325 kPa
2 3 6 3
2 3 2
6 2 2
× − ×
×
×
− −
T −
