THE CHAPTER
13 TRANSFER OF HEAT
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION Convection is the process in which heat is carried from one place to another by the bulk movement of the medium. In liquids and gases, the molecules are free to move; hence, convection occurs as a result of bulk molecular motion. In solids, however, the molecules are generally bound to specific locations (lattice sites). While the molecules in a solid can vibrate about their equilibrium locations, they are not free to move from place to place within the solid. Therefore, convection does not generally occur in solids. REASONING AND SOLUTION A heavy drape, hung close to a cold window, reduces heat loss through the window by interfering with the process of convection. Without the drape, convection currents bring the warm air of the room into contact with the cold window. With the drape, convection currents are less prominent, and less room air is circulated directly past the cold surface of the window. REASONING AND SOLUTION Forced convection plays the principal role in the wind chill factor. The wind mixes the cold ambient air with the warm layer of air that immediately surrounds the exposed portions of your body. The forced convection removes heat from your exposed body surfaces, thereby making you feel colder than you would otherwise feel if there were no wind. REASONING AND SOLUTION A road surface is exposed to the air on its upper surface and to the earth on its lower surface. Even when the air temperature is at the freezing point, the road surface may be above this temperature as heat flows through the road from the earth. In order for a road to freeze, sufficient heat must be lost from the earth by conduction through the road surface. The temperature of the earth under the road must be reduced at least to the freezing point. A bridge is exposed to the air on both its upper and lower surfaces. It will, therefore, lose heat from both surfaces and reach thermal equilibrium with the air much more quickly than an ordinary roadbed. It is reasonable, then, that the bridge surface will usually freeze before the road surface. SSM REASONING AND SOLUTION A piece of Styrofoam and a piece of wood are sandwiched together to form a layered slab. The two pieces have the same thickness and cross-sectional area. The exposed surfaces have constant temperatures. The temperature of the exposed Styrofoam surface is greater than the temperature of the exposed wood surface. The rate of heat flow through either layer can be determined from Equation 13.1: Q / t k A T / L , where k is the thermal conductivity of the layer, A and L are the crosssectional area and thickness of the layer, respectively, and T is the temperature difference
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2.
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3.
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4.
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5.
Chapter 13 Conceptual Questions
671
between the ends of the layer. Since heat is not trapped within the sandwich, the rate at which heat flows through the sandwich, Q/t, must be uniform throughout both layers. Therefore, (k A T / L) Styrofoam (k A T / L) wood . Since both layers have the same crosssectional area and thickness, A and L are the same for both layers. Therefore, kStyrofoam T Styrofoam kwood T wood . From Table 13.1, we see that the thermal conductivity of Styrofoam is less than the thermal conductivity of wood; therefore, the temperature difference between the two ends of the wood layer must be smaller than the temperature difference between the two ends of the Styrofoam layer. From this, we can conclude that the temperature at the Styrofoam-wood interface must be closer to the lower temperature of the exposed wood surface.
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6.
REASONING AND SOLUTION When heat is transferred from place to place inside the human body by the flow of blood, the main method of heat transfer is forced convection, similar to that illustrated for the radiator fluid in Figure 13.7. The heart is analogous to the water pump in the figure. REASONING AND SOLUTION Some animals have hair, the strands of which are hollow, air-filled tubes. Other animals have hair that is composed of solid, tubular strands. For animals that live in very cold climates, hair that is composed of hollow air-filled tubes would be advantageous for survival. Since air has a small thermal conductivity, hair shafts composed of hollow air-filled tubes would reduce the loss of body heat by conduction. Since hair shafts are small, no appreciable convection would occur within them. Thus, the hollow air-filled structure of the hair shaft inhibits the loss of heat by conduction. REASONING AND SOLUTION A poker used in a fireplace is held at one end, while the other end is in the fire. Such pokers are made of iron rather than copper because the thermal conductivity of iron is roughly smaller by a factor of five than the thermal conductivity of copper. Therefore, the transfer of heat along the poker by conduction is considerably reduced by using iron. Hence, one end of the poker can be placed in the fire, and the other end will remain cool enough to be comfortably handled. REASONING AND SOLUTION Snow, with air trapped within it, is a thermal insulator, because air has a relatively low thermal conductivity and the small, dead-air spaces inhibit heat transfer by convection. Therefore, a lack of snow allows the ground to freeze at depths greater than normal.
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7.
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8.
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9.
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10. REASONING AND SOLUTION Table 13.1 indicates that the thermal conductivity of steel is 14 J/(s m C ) , while that of concrete is 1.1 J/(s m C ) . According to Equation 13.1, Q k A Tt / L , this implies that heat will flow more readily through a volume of steel than it will through an identically shaped volume of concrete. Therefore, while steel reinforcement bars can enhance the structural stability of concrete walls, they degrade the insulating value of the concrete.
672 THE TRANSFER OF HEAT ____________________________________________________________________________________________
11. REASONING AND SOLUTION A potato will bake faster if a nail is driven into it before it is placed in the oven. Since the nail is metal, we can assume that the thermal conductivity of the nail is greater than the thermal conductivity of the potato. The nail conducts more heat from the oven to the interior of the potato than does the flesh of the potato, thereby causing the potato to bake faster.
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12. REASONING AND SOLUTION Several days after a snowstorm, the roof on a house is uniformly covered with snow. On a neighboring house, the snow on the roof has completely melted. Since one of the houses still has snow on the roof, it is reasonable to conclude that the ambient temperature is still below the freezing point of water. Since the snow has melted from the roof of the neighboring house, we can conclude that the heat required to melt the snow must have come through the attic and the roof by conduction. Hence, the house which has the uniform layer of snow on the roof is probably better insulated. The better the insulation, the smaller is the amount of heat conducted through the roof to melt the snow.
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13. SSM REASONING AND SOLUTION One car has a metal body, while another car has a plastic body. On a cold winter day, these cars are parked side by side. The metal car feels colder to the touch of your bare hand even though both cars are at the same temperature. This is because your fingers are sensitive to the rate at which heat is transferred to or from them, rather than to the temperature itself. The metal car feels colder than the plastic car at the same temperature, because heat flows from your bare hand into the metal car more readily than it flows into the plastic car. The flow occurs into the metal more readily, because the thermal conductivity of the metal is greater than that of the plastic.
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14. REASONING AND SOLUTION Many high-quality pots have copper bases and polished stainless steel sides. Since copper has a high thermal conductivity, heat can readily enter the bottom of the pot by means of conduction. Since the temperature of the pot is greater than the temperature of its environment, the pot will lose heat by means of radiation. Polished stainless steel has a low emissivity; that is, it is a poor emitter of radiant energy. Hence, by making the sides of the pot polished stainless steel, the amount of heat that would be lost by radiation is minimized. This design is optimal. If the pot were constructed entirely of copper, the bottom would efficiently conduct heat into the pan; however, heat would also be conducted efficiently into the sides of the pot, raising their temperature and increasing the loss from the sides via radiation. If, on the other hand, the pot were constructed entirely of stainless steel, the loss of heat through radiant energy would be minimized; however, since stainless steel has a low thermal conductivity, heat would not efficiently enter the bottom of the pot through conduction.
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Chapter 13 Conceptual Questions
673
15. REASONING AND SOLUTION The radiant energy Q emitted in a time t by an object that has a Kelvin temperature T, a surface area A, and an emissivity e, is given by Equation 13.2, Q e T 4 At , where is the Stefan-Boltzmann constant. We now consider two objects that have the same size and shape. Object A has an emissivity of 0.3, and object B has an emissivity of 0.6. Since each object radiates the same 4 4 power, eA TA AA e B TB AB . The Stefan-Boltzmann constant is a universal constant, and since the objects have the same size and shape, AA
4 AB ; therefore, eA TA
4
4 e B TB , or
TA / TB 4 e B / eA 4 2 . Hence, the Kelvin temperature of A is Kelvin temperature of B, not twice the temperature of B.
2 or 1.19 times the
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16. SSM REASONING AND SOLUTION a. The highly reflective paint reduces the ability of the so-called "radiator" to deliver heat into the room via the mechanism of radiation. This is because the paint allows the surface of the device to reflect more radiation than it otherwise would. Being a better reflector means that the device has become a poorer absorber of radiation, and poor absorbers are also poor emitters. Since the painted "radiator" loses less heat by the mechanism of radiation, it becomes hotter than it would if it were unpainted. b. Since painting the device reduces its ability to radiate electromagnetic waves, we dismiss radiation as the primary mechanism by which "radiators" deliver heat. We also dismiss conduction, since air is not a good conductor of heat. That leaves convection. "Radiators" indeed function primarily via convection. The fact that the paint enables the device to become hotter for a given supply of hot water or steam is beneficial for convection. The hotter the device becomes, the more effectively it can generate the convection currents that distribute the heat around the room. 17. REASONING AND SOLUTION Two strips of material, A and B, are identical except that they have emissivities of 0.4 and 0.7, respectively. The strips are heated to the same temperature and have a bright glow. The emissivity is the ratio of the energy that an object actually radiates to the energy that the object would radiate if it were a perfect emitter. The strip with the higher emissivity will radiate more energy per second than the strip with the lower emissivity, other things being equal. Therefore, strip B will have the brighter glow.
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18. REASONING AND SOLUTION The thermal conductivity of the bottom of the pot is greater than the thermal conductivity of air; therefore, the portion of the heating element beneath the pot loses heat by conduction through the bottom of the pot. The exposed portion of the heating element loses some heat through convection, but the convective process is not as efficient as the conductive process through the bottom of the pot. The exposed portion of the heating element will, therefore, lose less heat and be at a higher temperature than the portion of the heating element beneath the pot. Thus, the exposed portion glows cherry red.
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674 THE TRANSFER OF HEAT
19. REASONING AND SOLUTION If we consider a glove and a mitten, each of the same "size" and made of the same material, we can deduce that the mitten has less surface area A exposed to the cold winter air. Thus, according to Equation 13.1, Q k A Tt / L , we can conclude that the mitten will conduct less heat per unit time from the hand to the winter air. Therefore, to keep your hands as warm as possible during skiing, you should wear mittens as opposed to gloves.
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20. REASONING AND SOLUTION Two identical hot cups of cocoa are sitting on a kitchen table. One has a metal spoon in it and one does not. After five minutes, the cocoa with the metal spoon in it will be cooler. The metal spoon conducts heat from the cocoa to the handle of the spoon. Convection currents in the air and radiation then remove the heat from the spoon handle. The conduction-convection-radiation process removes heat from the cocoa, thereby cooling it faster than the cocoa that does not have a spoon in it.
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21. REASONING AND SOLUTION The radiant energy Q emitted in a time t by an object that has a Kelvin temperature T, a surface area A, and an emissivity e, is given by Equation 13.2: Q e T 4 At , where is the Stefan-Boltzmann constant. a. A hot solid cube will cool more rapidly if it is cut in half, rather than if it is left intact. Since the cube is warmer than its environment, it will lose heat primarily through radiation. Convection currents will also remove some heat from the surface of the cube. When the cube has been cut in half, the surface area of the solid has been increased. If the length of one edge of the original cube is L, then cutting the cube in half increases the surface area from 6L2 to 8L2. From Equation 13.2, the amount of heat Q radiated in a time t is proportional to the surface area of the cube; therefore, the cube will radiate more rapidly and cool more rapidly if it is cut in half. b. One pound of spaghetti noodles has a larger effective surface area than one pound of lasagna noodles. Imagine cutting many spaghetti noodles from one large lasagna noodle, in a way similar to what was done to the cube in part (a). Since the effective surface area of the spaghetti noodles is greater than that of the lasagna, heat will be radiated from the surface of the spaghetti noodles more effectively than heat will be radiated from the surface of the lasagna noodles. Therefore, the spaghetti noodles will cool more rapidly from the same initial temperature than the lasagna noodles.
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22. REASONING AND SOLUTION The black asphalt is a better absorber than the cement; the black asphalt will absorb more of the sun's radiant energy than the cement. Since the sun has been shining all day, the asphalt will be at a higher temperature than the cement. The temperature of the asphalt is apparently above the freezing point of water, while the temperature of the cement playground is below the freezing point of water. Therefore, when snow hits the asphalt, it melts immediately, while the snow collects on the cement.
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Chapter 13 Conceptual Questions
675
23. REASONING AND SOLUTION The radiant energy Q emitted in a time t by an object that has a Kelvin temperature T, a surface area A, and an emissivity e, is given by Equation 13.2: Q e T 4 At , where is the Stefan-Boltzmann constant. If you are stranded in the mountains in bitter cold weather, you could minimize energy losses from your body by curling up into the tightest possible ball. In doing so, you minimize your effective surface area. Therefore, A in Equation 13.2 is made smaller, and you would radiate less heat.
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676 THE TRANSFER OF HEAT
CHAPTER 13 THE TRANSFER OF HEAT
PROBLEMS
______________________________________________________________________________ 1. SSM REASONING The heat conducted through the iron poker is given by Equation 13.1, Q ( kA T ) t / L . If we assume that the poker has a circular cross-section, then its r 2 . Table 13.1 gives the thermal conductivity of iron as cross-sectional area is A 79 J / (s m C ) . SOLUTION The amount of heat conducted from one end of the poker to the other in 5.0 s is, therefore,
79 J / s m C 5.0 10 m 502 C 26 C 5.0 s (k A T)t Q 12 J L 1.2 m ______________________________________________________________________________
b
gc
3
hb
2
b g g
2.
REASONING AND SOLUTION The rate at which energy is gained through the refrigerator walls is
Q t kA T L 0.030 J/(s m C ) 5.3 m 2 25 C 5 C 0.075 m 42 J/s
Therefore, the amount of heat per second that must be removed from the unit to keep it cool is 42 J / s . ______________________________________________________________________________ 3. REASONING Since heat Q is conducted from the blood capillaries to the skin, we can use kA T t the relation Q (Equation 13.1) to describe how the conduction process depends L on the various factors. We can determine the temperature difference between the capillaries and the skin by solving this equation for T and noting that the heat conducted per second is Q/t. SOLUTION Solving Equation 13.1 for the temperature difference, and using the fact that Q/t = 240 J/s, yields
T
Q/t L kA
240 J/s 2.0 10 3 m 0.20 J/ s m C 1.6 m 2 1.5 C
We have taken the thermal conductivity of body fat from Table 13.1. ______________________________________________________________________________
Chapter 13 Problems
677
4.
REASONING AND SOLUTION The heat lost in each case is given by Q = (kA T)t/L. For the goose down jacket 0.025 J/(s m C ) A T t Qg 1.5 10 2 m For the wool jacket 0.040 J/(s m C ) A T t Qw 5.0 10 3 m Now
Qw / Qg
5.
4.8
______________________________________________________________________________ SSM REASONING The heat transferred in a time t is given by Equation 13.1, Q ( k A T ) t / L . If the same amount of heat per second is conducted through the two plates, then Q / t al Q / t st . Using Equation 13.1, this becomes
kal A T Lal
This expression can be solved for Lst . SOLUTION Solving for Lst gives
kst A T Lst
14 J/(s m C ) (0.035 m) = 2.0 103 m kal 240 J/(s m C ) ______________________________________________________________________________ Lst Lal
6. REASONING The heat Q conducted along the bar is given by the relation Q
kst
kA T t
L (Equation 13.1). We can determine the temperature difference between the hot end of the bar and a point 0.15 m from that end by solving this equation for T and noting that the heat conducted per second is Q/t and that L = 0.15 m.
SOLUTION Solving Equation 13.1 for the temperature difference, using the fact that Q/t = 3.6 J/s, and taking the thermal conductivity of brass from Table 13.1, yield
T Q/t L kA 3.6 J/s 0.15 m 110 J/ s m C 2.6 10
4
m2
19 C
The temperature at a distance of 0.15 m from the hot end of the bar is
T 306 C 19 C = 287 C ______________________________________________________________________________
678 THE TRANSFER OF HEAT
7.
SSM WWW REASONING AND SOLUTION Values for the thermal conductivities of Styrofoam and air are given in Table 11.1. The conductance of an 0.080 mm thick sample of Styrofoam of cross-sectional area A is
ks A Ls 0.010 J/(s m C ) A 0.080 10
3
m
[125 J/(s m 2 C )] A
The conductance of a 3.5 mm thick sample of air of cross-sectional area A is
ka A La 0.0256 J/(s m C ) A 3.5 10
3
m
[7.3 J/(s m 2 C )] A
Dividing the conductance of Styrofoam by the conductance of air for samples of the same cross-sectional area A, gives
[125 J/(s m2 C )] A [7.3 J/(s m2 C )] A
17
Therefore, the body can adjust the conductance of the tissues beneath the skin by
a factor of 17 . ______________________________________________________________________________
8. REASONING To find the total heat conducted, we will apply Equation 13.1 to the steel portion and the iron portion of the rod. In so doing, we use the area of a square for the cross section of the steel. The area of the iron is the area of the circle minus the area of the square. The radius of the circle is one half the length of the diagonal of the square. SOLUTION In preparation for applying Equation 13.1, we need the area of the steel and the area of the iron. For the steel, the area is simply ASteel = L2, where L is the length of a side of the square. For the iron, the area is AIron = R2 L2. To find the radius R, we use the Pythagorean theorem, which indicates that the length D of the diagonal is related to the length of the sides according to D2 = L2 + L2. Therefore, the radius of the circle is R D/2 2 L / 2 . For the iron, then, the area is
2 2
AIron
R
L
2L 2
2
L2
2
1 L2
Taking values for the thermal conductivities of steel and iron from Table 13.1 and applying Equation 13.1, we find
Chapter 13 Problems
679
QTotal
QSteel QIron kA T t L T t L
kA T t L
kSteel L2 kIron
Iron
Steel
2
1 L2
14
J s m C
0.010 m
2
79
J s m C
2
1 0.010 m
2
78 C 18 C 120 s 85 J 0.50 m ______________________________________________________________________________
9. REASONING AND SOLUTION Using Equation 13.1, Q ( kA T ) t / L , we obtain
Q FI kT GJ L HK At
(1)
Before Equation (1) can be applied to the ice-aluminum combination, the temperature T at the interface must be determined. We find the temperature at the interface by noting that the heat conducted through the ice must be equal to the heat conducted through the aluminum: Qice = Qaluminum. Applying Equation 13.1 to this condition, we have
kA Tt L kA Tt L
(2)
aluminum
ice
or
2.2 J/(s m C ) A ( 10.0 C) T t 0.0050 m
240 J/(s m C ) A T ( 25.0 C) t 0.0015 m
The factors A and t can be eliminated algebraically. Solving for T gives T = 24.959 C for the temperature at the interface. a. Applying Equation (1) to the ice leads to
Q At
[2.2 J/(s m C )] ( 10.0 C) ( 24.959 C)
ice
0.0050 m
6.58 103 J/(s m2 )
Since heat is not building up in the materials, the rate of heat transfer per unit area is the same throughout the ice-aluminum combination. Thus, this must be the heat per second per square meter that is conducted through the ice-aluminum combination.
680 THE TRANSFER OF HEAT
b. Applying Equation (1) to the aluminum in the absence of any ice gives:
[240 J/(s m C )] ( 10.0 C) ( 25.0 C) Q 2.40 106 J/(s m2 ) At Al 0.0015 m ______________________________________________________________________________
10. REASONING The water in both pots is boiling away at the same rate. This means that the Q heat per second being delivered to the water through the bottom of the pot is the same in t each case. The heat passes through the bottom of either pot via conduction. Therefore, we Q kA T know that Equation 13.1 applies, so that , where k is the thermal conductivity of t L the material from which the pot bottom is made, A is the cross-sectional area of the bottom, T is the difference in temperature between the inner and outer surface of the pot bottom, and L is the thickness of the bottom. We will apply this relation to the aluminum and to the copper bottom. SOLUTION Applying Equation 13.1, we obtain
Q t Q t kcopper A T L
copper
or
aluminum
kaluminum A T L
aluminum
copper
Note that the area A and thickness L are the same for each pot. Algebraically eliminating these terms gives
kcopper T
copper
kaluminum
T
aluminum
Solving for
T
copper
, we find that
T
copper
Theating element Twater
kaluminum
T
aluminum
kcopper
Thus, the heating element on which the copper bottom rests has a temperature of
Theating element Twater kaluminum T
aluminum
kcopper 240 J/ s m C 155.0 C 100.0 C 134 C
100.0 C
390 J/ s m C
Chapter 13 Problems
681
11. SSM REASONING The heat lost per second due to conduction through the glass is given by Equation 13.1 as Q/t = (kA T)/L. In this expression, we have no information for the thermal conductivity k, the cross-sectional area A, or the length L. Nevertheless, we can apply the equation to the initial situation and again to the situation where the outside temperature has fallen. This will allow us to eliminate the unknown variables from the calculation. SOLUTION Applying Equation 13.1 to the initial situation and to the situation after the outside temperature has fallen, we obtain
Q t
kA TIn TOut, initial
Initial
L
and
Q t
kA TIn TOut, colder
Colder
L
Dividing these two equations to eliminate the common variables gives
kA TIn TOut, colder Q/t Q/t
Colder Initial
kA TIn
L TOut, initial L
TIn TOut, colder TIn TOut, initial
Remembering that twice as much heat is lost per second when the outside is colder, we find
2 Q/t Q/t
Initial Initial
2
TIn TOut, colder TIn TOut, initial
Solving for the colder outside temperature gives
TOut, colder 2TOut, initial TIn 2 5.0 C 25 C 15 C
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12. REASONING
Heat Q flows along the length L of the bar via conduction, so that kA T t Equation 13.1 applies: Q , where k is the thermal conductivity of the material L from which the bar is made, A is the cross-sectional area of the bar, T is the difference in temperature between the ends of the bar, and t is the time during which the heat flows. We will apply this expression twice in determining the length of the bar. SOLUTION Solving Equation 13.1 for the length L of the bar gives
L kA T t Q k A TW TC t Q
(1)
682 THE TRANSFER OF HEAT
where TW and TC, respectively are the temperatures at the warmer and cooler ends of the bar. In this result, we do not know the terms k, A, t, or Q. However, we can evaluate the heat Q by recognizing that it flows through the entire length of the bar. This means that we can also apply Equation 13.1 to the 0.13 m of the bar at its cooler end and thereby obtain an expression for Q: k A T TC t Q D where the length of the bar through which the heat flows is D = 0.13 m and the temperature at the 0.13-m point is T = 23 C, so that T T TC . Substituting this result into Equation (1) and noting that the terms k, A, and t can be eliminated algebraically, we find
L
k A TW TC t Q
k A TW TC t k A T TC t D
k A TW TC t D k A T TC t
TW TC D T TC
48 C 11 C 0.13 m 23 C 11 C
0.40 m
13. REASONING The heat Q required to change liquid water at 100.0 C into steam at 100.0 C is given by the relation Q = mLv (Equation 12.5), where m is the mass of the water and Lv is the latent heat of vaporization. The heat required to vaporize the water is conducted through the bottom of the pot and the stainless steel plate. The amount of heat kA T t conducted in a time t is given by Q (Equation 13.1), where k is the thermal L conductivity, A and L are the cross-sectional area and length, and T is the temperature difference. We will use these two relations to find the temperatures at the aluminum-steel interface and at the steel surface in contact with the heating element. SOLUTION a. Substituting Equation 12.5 into Equation 13.1 and solving for T, we have
T
QL k At
mLv L k At
The thermal conductivity kAl of aluminum can be found in Table 13.1, and the latent heat of vaporization for water can be found in Table 12.3. The temperature difference TAl between the aluminum surfaces is
TAl
mLv L kAl At
0.15 kg 22.6 105 J/kg 3.1 10 3 m 240 J/ s m C 0.015 m2 240 s 1.2 C
Chapter 13 Problems
683
The temperature at the aluminum-steel interface is TAl-Steel = 100.0 C + TAl = 101.2 C . b. Using the thermal conductivity kss of stainless steel from Table 13.1, we find that the temperature difference Tss between the stainless steel surfaces is
Tss
mLv L kss At
0.15 kg 22.6 105 J/kg 1.4 10 3 m 14 J/ s m C 0.015 m 2 240 s 9.4 C
The temperature at the steel-burner interface is T = 101.2 C + Tss = 110.6 C . ______________________________________________________________________________ 14. REASONING Heat flows along the rods via conduction, so that Equation 13.1 applies: kA T t , where Q is the amount of heat that flows in a time t, k is the thermal Q L conductivity of the material from which a rod is made, A is the cross-sectional area of the rod, and T is the difference in temperature between the ends of a rod. In arrangement a, this expression applies to each rod and T has the same value of T TW TC . The total heat Q is the sum of the heats through each rod. In arrangement b, the situation is more complicated. We will use the fact that the same heat flows through each rod to determine the temperature at the interface between the rods and then use this temperature to determine T and the heat flow through either rod. SOLUTION For arrangement a, we apply Equation 13.1 to each rod and obtain for the total heat that k1 A TW TC t k2 A TW TC t k1 k2 A TW TC t Q Q1 Q2 (1) L L L For arrangement b, we use T to denote the temperature at the interface between the rods and note that the same heat flows through each rod. Thus, using Equation 13.1 to express the heat flowing in each rod, we have
k1 A TW T t L
Heat flowing through rod 1
k2 A T TC t L
Heat flowing through rod 2
or
k1 TW T
k2 T TC
Solving this expression for the temperature T gives
T
k1TW
k2TC
k1 k2
(2)
Applying Equation 13.1 to either rod in arrangement b and using Equation (2) for the interface temperature, we can determine the heat Q that is flowing. Choosing rod 2, we find that
684 THE TRANSFER OF HEAT
Q
k2 A T TC t L k2 A
k2 A
k1TW k2TC TC t k1 k2 L
k1TW k1TC t k1 k2 L
k2 Ak1 TW TC t L k1 k2
(3)
Using Equations (1) and (3), we obtain for the desired ratio that
k1 k2 A TW TC t Q Q L k2 Ak1 TW TC t L k1 k2
Using the fact k2
k1 that k2 A TW TC t L k1 k2 L k2 A k1 TW TC t
k1 k2 k2 k1
2
2k1 , we obtain
Q Q k1 k2 k2 k1
2
k1 2k1 2k1 k1
2
4.5
15. SSM WWW REASONING If the cylindrical rod were made of solid copper, the amount of heat it would conduct in a time t is, according to Equation 13.1, Qcopper (kcopper A2 T / L)t . Similarly, the amount of heat conducted by the lead-copper combination is the sum of the heat conducted through the copper portion of the rod and the heat conducted through the lead portion:
Qcombination
kcopper ( A2
A1 ) T / L klead A1 T / L t .
Since the lead-copper combination conducts one-half the amount of heat than does the solid copper rod, Qcombination 1 Qcopper , or 2
kcopper ( A2 L A1 ) T klead A1 T L 1 kcopper A2 T 2 L
This expression can be solved for A1 / A2 , the ratio of the cross-sectional areas. Since the cross-sectional area of a cylinder is circular, A known, the ratio of the radii can be determined.
r 2 . Thus, once the ratio of the areas is
SOLUTION Solving for the ratio of the areas, we have
Chapter 13 Problems
685
A1 A2
kcopper 2 kcopper klead
2 ( r 2 ) /( r2 ) (r / r2 )2 ; therefore, 1 1
The cross-sectional areas are circular so that A / A2 1
r1 r2 kcopper 2(kcopper klead )
390 J/(s m C ) 2[390 J/(s m C ) 35 J/(s m C )]
0.74
where we have taken the thermal conductivities of copper and lead from Table 13.1.
____________________________________________________________________________________________
16. REASONING The radiant energy Q absorbed by the person's head is given by Q e T 4 At (Equation 13.2), where e is the emissivity, is the Stefan-Boltzmann constant, T is the Kelvin temperature of the environment surrounding the person (T = 28 C + 273 = 301 K), A is the area of the head that is absorbing the energy, and t is the time. The radiant energy absorbed per second is Q/t = e T 4 A. SOLUTION a. The radiant energy absorbed per second by the person's head when it is covered with hair (e = 0.85) is
Q t
e T4A
0.85 5.67 10
8
J/ s m2 K 4
301 K
4
160 10
4
m2
6.3 J/s
b. The radiant energy absorbed per second by a bald person's head (e = 0.65) is
Q 4 e T 4 A 0.65 5.67 10 8 J/ s m2 K 4 301 K 160 10 4 m 2 4.8 J/s t ______________________________________________________________________________
17. SSM WWW REASONING AND SOLUTION Solving the Stefan-Boltzmann law, Equation 13.2, for the time t, and using the fact that Qblackbody Qbulb , we have
tblackbody
Qblackbody T A
4
Qbulb T A
4
Pbulb tbulb T 4A
where Pbulb is the power rating of the light bulb. Therefore,
tblackbody
(100.0 J/s) (3600 s) 5.67 10 8 J/(s m 2 K 4 ) (303 K)4 (6 sides)(0.0100 m)2 / side
1h 3600 s
1d 24 h
= 14.5 d
686 THE TRANSFER OF HEAT
______________________________________________________________________________ 18. REASONING According to the Stefan-Boltzmann law, the radiant power emitted by the Q "radiator" is e T 4 A (Equation 13.2), where Q is the energy radiated in a time t, e is the t emissivity of the surface, is the Stefan-Boltzmann constant, T is the temperature in Kelvins, and A is the area of the surface from which the radiant energy is emitted. We will apply this law to the "radiator" before and after it is painted. In either case, the same radiant power is emitted. SOLUTION Applying the Stefan-Boltzmann law, we obtain the following:
Q t
4 eafter Tafter A after
and
Q t
4 ebefore Tbefore A before
Since the same radiant power is emitted before and after the "radiator" is painted, we have
Q t Q t or
before 4 4 eafter Tafter A ebefore Tbefore A
after
The terms
and A can be eliminated algebraically, so this result becomes
eafter
4 Tafter A
ebefore
4 Tbefore A
or
4 eafter Tafter
4 ebefore Tbefore
Remembering that the temperature in the Stefan-Boltzmann law must be expressed in Kelvins, so that Tbefore = 62 C +273 = 335 K (see Section 12.2), we find that
4 Tafter 4 ebefore Tbefore
eafter
or
Tafter
4
ebefore eafter
Tbefore
4
0.75 335 K 0.50
371 K
On the Celsius scale, this temperature is 371 K
273 = 98 C .
19. REASONING AND SOLUTION We know from Equation 13.2 that
A
Q/t e T4
6.0 101 W (0.36) 5.67 10
8
J/(s m K ) (3273 K)
2
4
4
2.6 10
5
m2
______________________________________________________________________________ 20. REASONING The radiant energy Q radiated by the sun is given by Q e T 4 At (Equation 13.2), where e is the emissivity, is the Stefan-Boltzmann constant, T is its temperature (in Kelvins), A is the surface area of the sun, and t is the time. The radiant energy emitted per second is Q/t = e T 4 A. Solving this equation for T gives the surface temperature of the sun.
Chapter 13 Problems
687
SOLUTION The radiant power produced by the sun is Q/t = 3.9 1026 W. The surface area of a sphere of radius r is A = 4 r2. Since the sun is a perfect blackbody, e = 1. Solving Equation 13.2 for the surface temperature of the sun gives
T
4
Q/t e 4 r2
3.9 1026 W
4
1 5.67 10
8
J/ s m K
2
4
4
6.96 10 m
8
2
5800 K
______________________________________________________________________________ 21. SSM REASONING AND SOLUTION The net power generated by the stove is given by Equation 13.3, Pnet e A ( T 4 T04 ) . Solving for T gives
T
7300 W (302 K) 4 532 K 8 2 4 2 10 J / (s m K )](2.00 m ) ______________________________________________________________________________
FP T I GA J e H K R S (0.900)[5.67 T
net 4 0
1/ 4
U V W
1/4
22. REASONING The net rate at which energy is being lost via radiation can not exceed the production rate of 115 J/s, if the body temperature is to remain constant. The net rate at which an object at temperature T radiates energy in a room where the temperature is T0 is given by Equation 13.3 as Pnet = e A(T4 T04). Pnet is the net energy per second radiated. We need only set Pnet equal to 115 J/s and solve for T0. We note that the temperatures in this equation must be expressed in Kelvins, not degrees Celsius. SOLUTION According to Equation 13.3, we have
Pnet e A T 4 T04 or T04 T4 Pnet e A
Using Equation 12.1 to convert from degrees Celsius to Kelvins, we have T = 34 + 273 = 307 K. Using this value, it follows that
T0
4 T4
Pnet e A
4
4
307 K
115 J/s 0.700 5.67 10
8
J/ s m 2 K 4
1.40 m 2
287 K (14 C)
____________________________________________________________________________________________
688 THE TRANSFER OF HEAT
23. REASONING AND SOLUTION The heat Q conducted during a time t through a wall of thickness L and cross sectional area A is given by Equation 13.1:
Q
kA T t L
The radiant energy Q, emitted in a time t by a wall that has a Kelvin temperature T, surface area A, and emissivity e is given by Equation (13.2):
Q e T 4 At
If the amount of radiant energy emitted per second per square meter at 0 C is the same as the heat lost per second per square meter due to conduction, then
Q tA
conduction
Q tA
radiation
Making use of Equations 13.1 and 13.2, the equation above becomes
k T L
Solving for the emissivity e gives:
e k T L T4
e T4
[1.1 J/(s m K)](293.0 K 273.0 K) = 0.70 (0.10 m)[5.67 10 8 J/(s m 2 K 4 )] (273.0 K)4
Remark on units: Notice that the units for the thermal conductivity were expressed as J/(s.m.K) even though they are given in Table 13.1 as J/(s.m.C). The two units are equivalent since the "size" of a Celsius degree is the same as the "size" of a Kelvin; that is, 1 C = 1 K. Kelvins were used, rather than Celsius degrees, to ensure consistency of units. However, Kelvins must be used in Equation 13.2 or any equation that is derived from it. ______________________________________________________________________________ 24. REASONING The heat Q necessary to vaporize a mass m of any substance at its boiling point is Q mLv where Lv is the latent heat of vaporization. Therefore, the mass vaporized by an amount of heat Q is m Q / Lv . For the liquid helium system in question, it continually absorbs heat through radiation. The net power absorbed is given by Equation 13.3, Pnet e A(T 4 T04 ) where T0 is the temperature of the liquid helium, and T is the temperature maintained by the shield. Since the
Chapter 13 Problems
689
container is a perfect blackbody radiator, e 1 . Thus, the rate at which the mass of liquid helium boils away through the venting value is
m t
(Q / t ) Lv
Pnet Lv
e A(T 4 T04 ) Lv
This expression can be multiplied by the time t to determine the mass vaporized during that time. SOLUTION The rate at which liquid helium mass boils away is
m t (1)[5.67 10 8 J/(s m 2 K 4 )]4 (0.30 m)2 [(77 K)4 (4.2 K)4 ] 1.07 10 2.1 104 J/kg
4
kg/s
The mass of liquid helium that boils away in one hour is, therefore,
kg 3600 s (1.0 h) 0.39 kg s 1.0 h ______________________________________________________________________________ 1.07 10
4
25. SSM REASONING The total radiant power emitted by an object that has a Kelvin temperature T, surface area A, and emissivity e can be found by rearranging Equation 13.2, the Stefan-Boltzmann law: Q e T 4 At . The emitted power is P Q / t e T 4 A . Therefore, when the original cylinder is cut perpendicular to its axis into N smaller cylinders, the ratio of the power radiated by the pieces to that radiated by the original cylinder is Ppieces e T 4 A 2 (1) 4 Poriginal e T A1 where A1 is the surface area of the original cylinder, and A2 is the sum of the surface areas of all N smaller cylinders. The surface area of the original cylinder is the sum of the surface area of the ends and the surface area of the cylinder body; therefore, if L and r represent the length and cross-sectional radius of the original cylinder, with L 10 r ,
A1 (area of ends) (area of cylinder body) 2( r 2 ) (2 r ) L 2( r 2 ) (2 r )(10r ) 22 r 2
When the original cylinder is cut perpendicular to its axis into N smaller cylinders, the total surface area A2 is
A2
N 2( r 2 ) (2 r ) L
N 2( r 2 ) (2 r )(10r )
2 N 20
r2
690 THE TRANSFER OF HEAT
Substituting the expressions for A1 and A2 into Equation (1), we obtain the following expression for the ratio of the power radiated by the N pieces to that radiated by the original cylinder Ppieces e T 4 A 2 N 20 r 2 N 10 2
Poriginal
e T 4 A1
22 r 2
11
SOLUTION Since the total radiant power emitted by the N pieces is twice that emitted by the original cylinder, Ppieces / P original 2 , we have (N + 10)/11 = 2. Solving this expression for N gives N
12 . Therefore, there are 12 smaller cylinders .
______________________________________________________________________________ 26. REASONING The drawing shows a crosssectional view of the small sphere inside the larger spherical asbestos shell. The small sphere produces a net radiant energy, because its temperature (800.0 C) is greater than that of its environment (600.0 C). This energy is then conducted through the thin asbestos shell (thickness = L). By setting the net radiant energy produced by the small sphere equal to the energy conducted through the asbestos shell, we will be able to obtain the temperature T2 of the outer surface of the shell.
T2 600.0 C r1 800.0 C r2 L
SOLUTION The heat Q conducted during a time through the thin asbestos shell is given by kasbestos A2 T t Equation 13.1 as Q , where kasbestos is the thermal conductivity of L 2 asbestos (see Table 13.1), A2 is the area of the spherical shell A2 r2 , T is the temperature difference between the inner and outer surfaces of the shell ( T = 600.0 C T2), and L is the thickness of the shell. Solving this equation for the T2 yields
T2 QL kasbestos r22 t 600.0 C
The heat Q is produced by the net radiant energy generated by the small sphere inside the asbestos shell. According to Equation 13.3, the net radiant energy is Q Pnet t e A1 T 4 T04 t , where e is the emissivity, is the Stefan-Boltzmann constant, A1 is the spherical area of the sphere
A1
r12 , T is the temperature of the sphere
(T = 800.0 C = 1073.2 K) and T0 is the temperature of the environment that surrounds the
Chapter 13 Problems
691
sphere (T0 = 600.0 C = 873.2 K). Substituting this expression for Q into the expression above for T2, and algebraically eliminating the time t and the factors of , gives
e T2 600.0 C
T4
T04 r2 r1
2
L
kasbestos
0.90 5.67 10 600.0 C
8
J s m
2
K
4
1073.2 K J s m C
4
873.2 K
2
4
1.00 10 2 m
0.090
10
557.7 C ______________________________________________________________________________
27. SSM REASONING AND SOLUTION According to Equation 13.1, the heat per second lost is
Q t
kA T L
[0.040 J/(s m C o )] (1.6 m 2 )(25 C o ) 2.0 10
3
m
8.0 10 2 J/s
where the value for the thermal conductivity k of wool has been taken from Table 13.1. ______________________________________________________________________________ 28. REASONING AND SOLUTION a. The heat lost by the oven is
Q
( kA T ) t L 8.6 10 6 J
0.045 J / (s m C ) 1.6 m 2 160 C 50 C 6.0 h 0.020 m
c
b h
b F1 h J g g3600 s I G K H
b. As indicated on the page facing the inside of the front cover, 3.600 106 J = 1 kWh, so that 1 J = 2.78 107 kWh. Therefore, Q = 2.4 kWh. At $ 0.10 per kWh, the cost is $ 0.24 . ______________________________________________________________________________ 29. REASONING According to the discussion in Section 13.3, the net power Pnet radiated by the person is Pnet
e A T 4 T04 , where e is the emissivity,
is the Stefan-Boltzmann
constant, A is the surface area, and T and T0 are the temperatures of the person and the
692 THE TRANSFER OF HEAT
environment, respectively. Since power is the change in energy per unit time (see Equation 6.10b), the time t required for the person to emit the energy Q contained in the dessert is t = Q/Pnet. SOLUTION The time required to emit the energy from the dessert is
t Q Pnet Q e A T 4 T04
4186 J , and the Kelvin temperatures are 1 Calorie T = 36 C + 273 = 309 K and T0 = 21 C + 273 = 294 K. The time is
The energy is Q
260 Calories
260 Calories t 0.75 5.67 10
8
4186 J 1 Calorie 1.3 m
2
J/ s m K
2
4
309 K
4
K
4
1.2 104 s
______________________________________________________________________________ 30. REASONING AND SOLUTION a. The radiant power lost by the body is PL = e T 4A = (0.80)[5.67 108 J/(s m2 K4)](307 K)4(1.5 m2) = 604 W
The radiant power gained by the body from the room is Pg = (0.80)[5.67 108 J/(s m2 K4)](298 K)4(1.5 m2) = 537 W Pg =
67 W
The net loss of radiant power is P = PL b. The net energy lost by the body is
1 Calorie 58 Calories 4186 J ______________________________________________________________________________ Q Pt (67 W)(3600 s)
Chapter 13 Problems
693
31. SSM REASONING AND SOLUTION The power radiated per square meter by the car when it has reached a temperature T is given by the Stefan-Boltzmann law, Equation 13.2, Pradiated / A e T 4 , where Pradiated Q / t . Solving for T we have
T
( Pradiated / A) e
1/ 4
1/4
560 W/m2 (1.00) 5.67 10 8 J/(s m2 K 4 )
320 K
______________________________________________________________________________ 32. REASONING AND SOLUTION According to Equation 13.2, for the sphere we have Q/t = e AsTs4, and for the cube Q/t = e AcTc4. Equating and solving we get Tc4 = (As/Ac)Ts4 Now As/Ac = (4 R2)/(6L2) The volume of the sphere and the cube are the same, (4/3)
2/3
R = L , so R
3
3
3 4
1/ 3
L.
The ratio of the areas is then
As Ac
4 R2 6 L2
1/4
4 6
3 4
0.806 . The temperature of the cube is,
Tc
As Ac
Ts
0.806
1/ 4
773 K
732 K
______________________________________________________________________________ 33. REASONING The heat Q required to melt ice at 0 C into water at 0 C is given by the relation Q = mLf (Equation 12.5), where m is the mass of the ice and Lf is the latent heat of fusion. We divide both sides of this equation by the time t and solve for the mass of ice per second (m/t) that melts: Q (1 m t ) t Lf The heat needed to melt the ice is conducted through the copper bar, from the hot end to the kA T t cool end. The amount of heat conducted in a time t is given by Q (Equation L 13.1), where k is the thermal conductivity of the bar, A and L are its cross-sectional area and
694 THE TRANSFER OF HEAT
length, and T is the temperature difference between the ends. We will use these two relations to find the mass of ice per second that melts. SOLUTION Solving Equation 13.1 for Q/t and substituting the result into Equation (1) gives kA T m kA T L t Lf L Lf The thermal conductivity of copper can be found in Table 13.1, and the latent heat of fusion for water can be found in Table 12.3. The temperature difference between the ends of the rod is T = 100 C , since the hot end is in boiling water (100 C) and the cool end is in ice (0 C). Thus,
m t
kA T L Lf
390 J/ s m C
4.0 10 4 m 2 100 C
4
3.1 10
5
kg/s
1.5 m 33.5 10 J/kg
______________________________________________________________________________ 34. REASONING AND SOLUTION The rate of heat transfer is the same for all three materials so Q/t = kpA Tp/L = kbA Tb/L = kwA Tw/L Let Ti be the inside temperature, T1 be the temperature at the plasterboard-brick interface, T2 be the temperature at the brick-wood interface, and To be the outside temperature. Then kpTi and kbT1 Solving (1) for T2 gives T2 = (kp + kb)T1/kb a. Substituting this into (2) and solving for T1 yields (kp/kb)Ti kbT2 = kwT2 kwTo (2) kpT1 = kbT1 kbT2 (1)
T1
kp /kb 1 kw /kb Ti 1 kw /kb 1 kp /kb
kw /kb T0 1
21 C
b. Using this value in (1) yields
T2 18 C
______________________________________________________________________________
Chapter 13 Problems
695
35. SSM WWW REASONING The rate at which heat is conducted along either rod is given by Equation 13.1, Q / t k A T / L . Since both rods conduct the same amount of heat per second, we have
ks As T Ls
=
ki Ai T Li
(1)
Since the same temperature difference is maintained across both rods, we can algebraically cancel the T terms. Because both rods have the same mass, ms mi ; in terms of the densities of silver and iron, the statement about the equality of the masses becomes s Ls As i Li Ai , or
As Ai
i Li s Ls
(2)
Equations (1) and (2) may be combined to find the ratio of the lengths of the rods. Once the ratio of the lengths is known, Equation (2) can be used to find the ratio of the cross-sectional areas of the rods. If we assume that the rods have circular cross sections, then each has an area of A r 2 . Hence, the ratio of the cross-sectional areas can be used to find the ratio of the radii of the rods. SOLUTION a. Solving Equation (1) for the ratio of the lengths and substituting the right hand side of Equation (2) for the ratio of the areas, we have
Ls Li
ks As ki Ai
ks ki
i Li s Ls
or
Ls Li
2
ks ki
i s
Solving for the ratio of the lengths, we have
Ls Li ks ki
i s
[420 J/(s m C )](7860 kg/m3 ) [79 J/(s m C )](10 500 kg/m3 )
2.0
b. From Equation (2) we have
rs2 ri2
i Li s Ls
or
rs ri
2
i Li s Ls
Solving for the ratio of the radii, we have
696 THE TRANSFER OF HEAT
rs ri
i s
Li Ls
7860 kg/m3 1 3 2.0 10 500 kg/m
0.61
______________________________________________________________________________ 36. REASONING AND SOLUTION ice is The heat which must be removed to form a volume V of Q = mLf = VLf = AhLf The heat is conducted through the ice to the air, so Q is Q = kA( T)t/L. Thus, we have
h
k Tt Lf L
[2.2 J/(s m C )] 15 C 917 kg/m
3 5
3.0 102 s
1.1 10 4 m
0.11 mm
3.35 10 J/kg 0.30 m
where the values for the thermal conductivity k, the density , and the heat of fusion Lf have been taken from Table 13.1, Table 12.3, and Table 11.1, respectively. ______________________________________________________________________________ 37. CONCEPT QUESTIONS a. The temperature is 100.0 C, because water boils at 100.0 C under one atmosphere of pressure. The temperature remains at 100.0 C until all the water is gone. b. When water boils, it changes from the liquid to the vapor phase. The heat needed to make the water change phase is Q = mLv, according to Equation 12.5, where m is the mass and Lv is the latent heat of vaporization of water. c. The temperature of the heating element must be greater than 100.0 C. This is because heat flows via conduction from a higher to a lower temperature and the temperature of the boiling water is 100.0 C. SOLUTION Applying Equation 13.1 to the heat conduction and using Equation 12.5 to express the heat needed to boil away the water, we have
Q
kcopper A T t L
mLv
,
The thermal conductivity of copper can be found in Table 13.1 kcopper = 390 J/ s m C
and the latent heat of vaporization for water can be found in Table 12.4 (Lv = 22.6 105 J/kg). The area A is the area of a circle or A = R2. Finally, the temperature difference is T = TE 100.0 C. Using this expression for conduction equation and solving for TE gives T in the heat-
Chapter 13 Problems
697
kcopper A TE 100.0 C t L
3
mLv
or
TE
100.0 C+
LmLv kCopper At
2.0 10 TE 100.0 C+
m 0.45 kg 22.6 105 J/kg 0.065 m
2
103.3 C 120 s
390 J/ s m C
______________________________________________________________________________ 38. CONCEPT QUESTIONS a. According to Equation 13.1 less heat is lost when the area through which the heat flows is smaller. Since the window has the smaller area, it would lose less heat than the wall, other things being equal. b. According to Equation 13.1 more heat is lost when the thickness through which the heat flows is smaller. Since the window has the smaller thickness, it would lose more heat than the wall, other things being equal. c. According to Equation 13.1 more heat is lost when the thermal conductivity of the material through which the heat flows is greater. According to Table 13.1 the thermal conductivity of glass is kG = 0.80 J/ s m C , while the value for Styrofoam is
kS = 0.010 J/ s m C . Therefore, the window would lose more heat than the wall, other things being equal.
SOLUTION The percentage of the heat lost by the window is
Percentage
Qwindow Qwall Qwindow
100
kG AG T t LG kS AS T t kG AG T t LS LG
100
kG AG LG kS AS kG AG LS LG
100
Here, we algebraically eliminated the time t and the temperature difference T, since they are the same in each term. The percentage is
698 THE TRANSFER OF HEAT
Percentage
kG AG LG kS AS kG AG LS LG
100
0.80 J/ s m C 0.010 J/ s m C 0.10 m 97 %
0.16 m 2 0.16 m 2
3
2.0 10 3 m 18 m 2 0.80 J/ s m C 2.0 10
100
m
______________________________________________________________________________ 39. CONCEPT QUESTIONS a. The cross-sectional area A through which the heat flows is greater for arrangement b; the cross-sectional area in b is twice that in a. b. The thickness L of the material through which the heat flows is greater for arrangement a; the thickness in a is twice that in b. c. For two reasons, Qa is less than Qb. First, the area in arrangement a is smaller, and the heat flows in direct proportion to the area. A smaller area means less heat. Second, the thickness in arrangement a is greater, and the heat flows in inverse proportion to the thickness. A greater thickness means less heat. SOLUTION Applying Equation 13.1 for the conduction of heat to both arrangements gives
Qa kAa La T t and Qb kAb Lb T t
Note that the thermal conductivity k, the temperature difference T, and the time t are the same in both arrangements. Dividing Qa by Qb gives
kAa Qa Qb kAb Lb La
T t Aa Lb T t Ab La
Remember that Ab = 2Aa and that La = 2Lb. As expected then, we find that
Chapter 13 Problems
699
Qa Qb
Aa Lb Ab La
Aa Lb 2 Aa 2 Lb
1 4
______________________________________________________________________________ 40. CONCEPT QUESTIONS a. According to Equation 6.10b, power is the change in energy divided by the time during which the change occurs. In this case, then, the power is P = Q/t. b. According to the Stefan-Boltzmann law (Equation 13.2), the power radiated is Q/t = e T 4A. The power is proportional to the fourth power of the temperature T (in Kelvins). Thus, a higher temperature promotes more radiated power. c. According to the Stefan-Boltzmann law (Equation 13.2), the power radiated is Q/t = e T 4A. The power is proportional to the area A. Thus, a smaller area generates less radiated power. d. The higher temperature of bulb #1 promotes a greater radiated power. The only way for both bulbs to radiate the same power, then, is for the filament area of bulb #1 to be smaller than that of bulb #2, in order to offset the effect of the higher temperature. SOLUTION Using the Stefan-Boltzmann law (Equation 13.2) for both bulbs, we have
P Q1 / t1 e T14 A1 1
Since P1 = P2, we see that
and
P2
Q2 / t2
e T24 A2
e T14 A1 e T24 A2
or
T14 A1 T24 A2
As expected, solving for the ratio A1/A2 gives a value less than one:
A1 T24 2100 K
4
0.37 4 2700 K ______________________________________________________________________________ A2 T14
41. CONCEPT QUESTIONS a. The power radiated by the object is given by Q / t e T 4 A (Equation 13.2), where e is the emissivity of the object, is the Stefan-Boltzmann constant, T is the Kelvin temperature of the object, and A is the surface area of the object. b. The power that the object absorbs from the room is given by Q/t = e T04A. Except for the temperature T0 of the room, this expression has the same form as that for the power radiated by the object. Note especially that the area A is the radiation area for the object, not the
700 THE TRANSFER OF HEAT
room. Review part b of the solution to Example 8 in the text to understand this important point. c. The power emitted by the object is proportional to the fourth power of the its temperature, and the power absorbed is proportional to the fourth power of the room's temperature. Since the object emits more power than it absorbs, its temperature T must be greater than the room's temperature T0. SOLUTION The object emits three times more power than it absorbs from the room, so it follow that (Q/t)emit = 3(Q/t)absorb. Using the Stefan-Boltzmann law for each of the powers, we find e T 4 A 3e T04 A Solving for T gives
T
4
3T0
4
3 293 K
386 K
______________________________________________________________________________ 42. CONCEPT QUESTIONS a. According to the Stefan-Boltzmann law, the power radiated by an object is Q/t = e T 4A, where A is the area from which the radiation is emitted. The power radiated is proportional to the fourth power of the temperature T. Therefore, other things being equal, the greater surface temperature of Sirius B would imply that its radiated power is greater than that of our sun. b. The fact that Sirius B radiates less power than our sun, means that something is offsetting the effect of the greater surface temperature in the Stefan-Boltzmann law. This can only be the surface area A. The power radiated is proportional to A, according to the law. A smaller area means a smaller radiated power. Therefore, the surface area of Sirius B must be less than the surface area of our sun. c. The surface area of a sphere is 4 R2, where R is the radius. Therefore, having less surface area, Sirius B must also have a radius that is less than the radius of our sun. SOLUTION Writing the Stefan-Boltzmann law (Equation 13.2) for both stars, we have
QSirius / tSirius
4 e TSirius ASirius
and
QSun / tSun
4 e TSun ASun
Dividing the equation for Sirius B by the equation for our sun and remembering that QSirius/tSirius = (0.040) QSun/tSun, we obtain
QSirius / tSirius QSun / tSun
4 e TSirius ASirius 4 e TSun ASun
or
0.040 QSun / tSun QSun / tSun
4 TSirius ASirius 4 TSun ASun
Simplifying this result and using the fact that the surface area of a sphere is 4 R gives
2
Chapter 13 Problems
701
0.040
4 2 TSirius RSirius 4 2 TSun RSun
Solving for the radius of Sirius B gives
RSirius
0.040
TSun TSirius
2
RSun
0.040
TSun 4TSun
2
6.96 108 m
8.7 106 m
As expected, the radius of Sirius B is less than that of our sun, so much so that it is called a white dwarf star. ______________________________________________________________________________ 43. CONCEPT QUESTIONS a. The heat Q conducted during a time t through a bar of length kA T t L and cross-sectional area A is Q (Equation 13.1). The heat depends on two L geometrical factors, the cross-sectional area and the length. Even though the cross-sectional area for heat conduction through the block in C is greater than that in A, it does not necessarily mean that more heat is conducted in C, because the lengths of the conduction paths are different. b. Even though the length of material through which heat is conducted in block A is greater than that in B, it does not necessarily follow that less heat is conducted in A, because the blocks have different cross-sectional areas. c. The heat Q conducted during a time t through a bar of length L and cross-sectional area A kA T t is Q (Equation 13.1). The cross-sectional area and length of each block are: L heat AA 2L2 and LA 3L0 , AB 3L2 and LB 2L0 , AC 6L2 and LC L0 . The 0 0 0 conducted through each block is
QA
2L 3 0
k Tt
QB
3L 2 0
k Tt
QC
6L0 k Tt
Therefore, the ranking of the heat conduction is (highest to lowest): C, B, A SOLUTION From the result of part c in the Concept Questions, the heat conducted in each case is: Case A
QA
2L 3 0
k Tt
2 3
0.30 m 250 J/ s m C
35 C 19 C 5.0 s
4.0 103 J
702 THE TRANSFER OF HEAT
Case B
QB
Case C
QC
3L 2 0
k Tt
3 2
0.30 m 250 J/ s m C
35 C 19 C 5.0 s
9.0 103 J
6 L0 k Tt
6 0.30 m 250 J/ s m C
35 C 19 C 5.0 s
3.6 10 4 J
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Chapter 13 Problems
703
44. CONCEPT QUESTIONS a. The net radiant power emitted by the bar in part (a) of the drawing is zero. The reason is that the temperature of the bar is the same as that of the room, and this temperature does not change. Therefore, the bar emits the same power into the room as it absorbs from the room, so the net radiant power emitted by the bar is zero. b. The two bars in part (b) of the drawing emit more power. According to Equation 13.2, the radiant power (or energy per unit time) emitted by an object is Q / t e T 4 A , which is directly proportional to its surface area A. The two bars in part (b) have a greater total surface area than the single bar in part (a). c. The two bars in part (b) of the drawing also absorb more power. From the results of Concept Question b, we know that the two bars emit more power because of their greater surface area. However, since their temperature does not change, the two bars must absorb as much power as they emit. Thus, they absorb more power from the room than the single bar in part (a). SOLUTION a. The power (or energy per unit time) absorbed by the two bars in part (b) of the drawing is given by Q / t e T 4 A2 , where A2 is the total surface area of the two bars: A2 28L2 . The 0 power absorbed by the single bar in A is Q / t
e T 4 A1 , where A1 is the total surface area
of the single bar: A 22L2 .The ratio of the power P2 absorbed by the two bars in part (b) 1 0 to the power P1absorbed by the single bar in part (a) is
P2 P 1
e T 4 28 L2 0 e T 4 22 L2 0
1.27
b. If the power absorbed by the two bars in part (b) of the drawing is the same as that absorbed by the single bar in part (a), then
e T14 A 1
Power absorbed by single bar in part (a )
e T24 A 2
Power absorbed by two bars in part (b )
Solving for the temperature of the room and the bars in part (b) gives
A1 A2 22 L2 0 28L2 0
T2
T1
4
450.0 K
4
424 K
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