fisica para ingenieria y ciencias bauer vol i - capitulo 18 soluciones

Bauer/Westfall: University Physics, 1E

732

Chapter 18: Heat and the First Law of Thermodynamics

In-Class Exercises

18.1. a 18.2. c 18.3. b 18.4. (1) False (2) False (3) False (4) True

Multiple Choice

18.1. d 18.2. b 18.3. b 18.4. d 18.5. c 18.6. b 18.7. b 18.8. d 18.9. f 18.10. e

Questions

18.11. The average human, approximated as a cylinder, has a radius of about 10.0 cmR = to about (upper and lower limit) with a minimum height of 150 cmh = to maximum 200 cm.h = The surface area of a cylinder of radius R and height h is A 2 .Rhπ= The temperature of a healthy human is 37.0 CT = ° or

310. K.T = Since we are assuming the person is black body, they will have an emissivity of 1.ε = The power radiated is given by the Stefan-Boltzmann equation, P = σεAT 4 ; therefore, the range of power emitted by an average person is given by:

( )( )( )( )( )( )( )( )( )( )( )( )( )( )

48 2 4min

48 2 4max

5.67 10 W/ m K 1 2 0.160 m 1.5 m 310 K 789.6 W

5.67 10 W/ m K 1 2 0.300 m 2.00 m 310 K 1974.1 W

P

P

−

−

= ⋅ =

= ⋅ =

π

π

The average person radiates between 800 W and 2000 W.

18.12. The house that has snow on the roof is releasing less heat to the atmosphere; hence less snow is melting and therefore, the house would be better insulated.

18.13. A bathmat (most likely cotton or wool) would have a much lower thermal conductivity than the tile. Assuming the bathmat and tile are the same temperature (room temperature) then the tile will feel colder since its higher thermal conductivity takes thermal energy from your feet faster than a mat would, and hence feels colder. Even though the thermal energy flow is proportional to the temperature difference, when your feet are colder, the thermal energy flow would be more noticeable to you and hence seem colder.

18.14. A blackbody can be constructed by making a cavity out of this material designed such that any electromagnetic radiation that enters the cavity cannot escape (at least not until after many bounces off the walls of the cavity). After every bounce, half the energy is deposited. So after many bounces nearly all of the original energy has been deposited, giving the impression that it is an ideal black body since virtually no light is reflected.

18.15. After the eruption, the atmosphere was filled with increased amounts of ash and dust. This blocks out some sunlight and prevents the Earth from heating. As a result the temperature decreases until after many years the particles settle and the normal amount of sunlight makes it through to return the temperature to normal.

18.16. Even though the hot coals have a large temperature, they also have a very small thermal conductivity. This means that as long as the contact time between the coals and the person’s feet is small (which is true when walking a steady pace) then the coals will not transfer enough heat and the person will not feel it.

18.17. When dry, the coat can be considered to be soaked in air. Air has a much lower thermal conductivity than water, so when it is dry, the air will insulate better than water. Also, when the coat is dry (and fluffy) it will


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be thicker; as opposed to when it is wet the fluffiness will disappear and flatten out. When the overall thickness decreases, it allows more heat to escape.

18.18. With an increased amount of dust in the atmosphere, more sunlight would be reflected back to space and this would prevent the heating of the Earth’s atmosphere. This would overall drop the Earth’s temperature.

18.19. (a) In order to push the piston down, the person must apple a force. This force applied to the area of the piston produces a pressure. Therefore, as the piston is pushed down, the pressure of the gas increases. (b) Since the gas is thermally insulted from the environment, there is no thermal energy flow ( )0 .Q = The increase in pressure is a result of work done on the system. Since there is work done on the system and the volume decreases, the internal energy of the system increases ΔEint = −W = −PΔV( ). If the energy increases in a thermally insulated region, its temperature must increase. (c) Other than pressure and temperature increasing and the volume decreasing, no other changes occur.

18.20. The thermal energy transferred per unit time is given by the equation P = Q / t = kA Th −Tc( )/ L. Consider a 10 cm long glass rod and a 10 m long aluminum rod where the cross sectional area and temperature difference are the same. The ratio of the thermal energy transferred is

PAl

Pglass

=kAl A Th −Tc( )/ LAl

kglass A Th −Tc( )/ Lglass

=kAlLglass

kglassLAl

=220 W/(m K)( ) 0.10 m( )

0.8 W/(m K)( ) 10 m( ) = 2.75.

Therefore, despite the length difference, the aluminum transfers heat better than the glass rod by a factor of ≈ 3.

18.21. If the two canteens have a similar thickness, then the plastic bottle, which has a much lower thermal conductivity, will insulate the water better than the aluminum can.

18.22. The mass of the penny (copper) is about 3.0 g, while that of the silver dollar is about 8.0 g. The temperature of the penny should be about ( )0 C 273 K° since it was outside in the cold while the silver dollar should be ( )37 C 310 K° since it was in the girl’s hand. The specific heat of the penny and the dollar are ( )p 0.386 KJ / kg Kc = and ( )s 0.235 KJ / kg K .c = Since the two coins are on wood (an insulator) they only exchange heat between each other. The final temperature T is then calculated using

( ) ( )p p p s s s ,m c T T m c T T− = − which can be rewritten as mpcpTp +mscsTs ( )p p s s ;m c m c T= + therefore,

( ) ( )p p p s s s p p s s/ .T m c T m c T m c m c= + + Then,

( ) ( )( )( ) ( ) ( )( )( )( ) ( )( ) ( ) ( )( )

0.0030 kg 0.386 kJ / kg K 273 K 0.0080 kg 0.235 kJ / kg K 310 K296 K.

0.0030 kg 0.386 kJ / kg K 0.0080 kg 0.235 kJ / kg KT

+= =

+

The final temperature of both coins is about 23 C.° Problems

18.23. (a) The work to lift the elephant is W mgh= or ( )( )( )3 2 45.0 10 kg 2.0 m 9.81m/s 9.8 10 J.W = ⋅ = ⋅

(b) A food calorie is equal to 34.1868 10 J.⋅ The task of lifting the elephant consumes

( )4 39.8 10 J 1 cal / 4.1868 10 J 23.4307W = ⋅ ⋅ = food calories. Assuming that the body converts 100% of

food energy into mechanical energy then the number of doughnuts needed is ( )23.4307 cal / 250 cal / doughnut 0.093627.= It takes less than one doughnut to power its consumer to

lift an elephant. The body usually converts only 30 % of the energy consumed. This corresponds to 0.31 of a doughnut.


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18.24. The work done to expand the gas is given by W = Vi

Vf pdV . The pressure is related to the volume by

P = αV 3 . The work is then W = Vi

Vf αV 3 dV = α 1/ 4( )V 4 | Vi

Vf = α / 4( ) Vf4 −Vi

4( ). If the final volume is 3

times larger, then the work done is:

W = α / 4( ) 3Vi( )4 −Vi

4( )= α / 4( ) 34 −1( )Vi4 = 4.00 N/m11 / 4( ) 80( ) 2.00 m3( )4

=1280 J.

18.25. The work per cycle is given by the area enclosed by the pressure vs. volume graph. For this case the area is given by ( )( )( )2 4 31/ 2 2.0 10 kPa 4.0 10 m 0.040 KJ 40. J.W −= ⋅ ⋅ = =

18.26. The process is adiabatic; thus, the change in internal energy is equal to the work done on the gas. The final internal energy is int int, f int, i ,E W E E PdVΔ = − = − = − or

( ) ( )( )33int, f int, i 500. J 3.00 atm 101325 Pa / atm 100. cm m /100. cm 470. J.E E PdV = − = − =

18.27. The temperature of the material will be ( )f iQ cm T cm T T= Δ = − or ( )f i i/ / .T T Q cm T Q cV ρ= + = + Since the final temperature is inversely proportional to the specific heat, ,c and the density. The material with the largest final temperature will be steam. A large specific heat will give a lower final temperature. The material with the largest specific heat and density, in this case, water, has the smallest final temperature. An example of the calculation for aluminum:

( ) ( )( )Al 3 3 3 3f 22.0 C 1.00 J/ 1.00 cm 2.375 10 kg/cm 0.900 10 J/ kg K 22.4678 C.T − = ° + ⋅ ⋅ = °

Note that we need the density of the material.

Material Specific Heat ( )KJ/kg K

Density

( )3g/cm Final Temperature C°

Lead 0.129 11.34 22.684Copper 0.386 8.94 22.290Steel 0.448 7.85 22.284Aluminum 0.900 2.375 22.468Glass 0.840 2.5 22.476Ice 2.22 0.9167 22.491Water 4.19 1.00 22.239Steam 2.01 45.974 10−⋅ 8350

18.28. The energy would only be transferred between the two bodies of water. This implies 1 1 2 2m c T m c TΔ = Δ where 1 2,T TΔ Δ indicate the water that starts at 20.0 C° and 32.0 C,° respectively. Further manipulation gives 1 1 1 1 2 2 2 2m c T m c T m c T V c Tρ ρΔ = Δ = Δ = Δ or 1 1 2 2 .V T V TΔ = Δ The first volume of water will have a temperature increase but the temperature of the second volume of water decreases. Solving for the final temperature:

( ) ( )( ) ( ) ( ) ( )( ) ( )

1 f 1 2 f 2 1 f 2 f 1 1 2 2

f 1 1 2 2 1 2

/ 7.00 L 20.0 C 3.00 L 32.0 C / 7.00 L 3.00 L 23.6 C.

V T T V T T V T V T V T V TT V T V T V V

− = − − + = +

= + + = ° + ° + = °

The final temperature of the water is 23.6 C.°

18.29. The heat transfer is equal for the aluminum and the water. The aluminum’s temperature will decrease, but the water’s temperature will increase. The temperature of the water is 10. C 283 K,wT = ° = and the temperature of the aluminum is Al 85 C 358 K.T = ° = ( ) ( )Al Al f Al w w f w Q m c T T m c T T= − − = −

w w f Al Al f Al Al Al w w w .m c T m c T m c T m c T+ = + The mass of one liter of water is one kilogram.


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( ) ( )( )( ) ( ) ( )( )( )( )( ) ( )( ) ( ) ( )( )

Al Al Al w w wf

Al Al w w3 33 3

3 33

0.900 10 J/ kg K 4.19 10 J/ kg K1.0 kg25 10 kg 25 10 kg358K 283K0.900 10 J/ kg K 4.19 10 J/ kg K1.0 kg25 10 kg

280 K

m c T m c TT

m c m c− −

−

+=

++⋅ ⋅⋅ ⋅

=+⋅ ⋅⋅

=

The equilibrium temperature is 280 K.

18.30. The kinetic energy of the bullet is ( ) 21/2 .K mv= If 75% of this energy is converted to heat then

( )( ) 20.75 3/4 1/2 .Q K mv mc T= = = Δ The change in temperature is:

( )( )

22

3

3 250 m/s3 8 181.69 C.8 0.129 10 J/ kg K

vTc

Δ = = = ° ⋅

Assuming the bullet is initially at room temperature of 20. C,° the final temperature of the bullet is

f i220. C 181.67 C 201.69 C 2.0 10 C.T T T= + Δ = ° + ° = ° = ⋅ °

18.31. THINK: The problem calls for calculating the change in energy for the copper and the water. The volume of the materials does not change, so the process is isochoric. An isochoric process implies the change in internal energy is equal to the heat transferred. Therefore, to calculate the energies of the materials the final temperatures of the samples are needed. The magnitude of the heat transferred will be equal for both materials. Knowing this and the mass, initial temperatures and the specific heat it is possible to calculate the final temperature. Copper has a mass of c 1.00 kg,m = an initial temperature of c 80.0 C,T = ° and a specific heat of ( )c 386 J/ kg K .c = The volume of water is 2.00 L which is equal to a mass of

w 2.00 kg,m = an initial temperature of w 10.0 C,T = ° and a specific heat of ( )w 4190 J/ kg K .c = SKETCH:

RESEARCH: The heat transferred is equal to .Q mc T= Δ SIMPLIFY: The temperature change for the water and the copper will be positive and negative, respectively. The heat lost by the copper plus the heat gained by the water equals zero, so

( ) ( )C W c c f c w f w0 .wQ Q m c T T m c T T+ = = − + − Solving this equation for fT yields

( ) ( )f c c c w w w w c c/ .wT m c T m c T m c m c= + + The magnitude of the change in energy is given by

int .E Q mc TΔ = = Δ

CALCULATE: ( )( )( ) ( )( )( )

( )( ) ( )( )f

1.00 kg 386 J/kgK 80.0 C 2.00 kg 4190 J/kgK 10.0 C13.0824 C

1.00 kg 386 J/kgK 2.00 kg 4190 J/kgKT

° + °= = °

+

( )( )( )( )( )( )

int,w

int,Cu

2.00 kg 4190 J/kgK 13.0824 C 10.0 C 25830 J1.00 kg 386 J/kgK 13.0824 C 80.0 C 25830 J

EE

Δ = ° − ° =Δ = ° − ° = −

ROUND: The energy should be rounded to three significant figures: int 25800 J.EΔ =


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DOUBLE-CHECK: The magnitude of the change in energy for the water and copper must be equal since there are no other sources of change in energy. The signs must be opposite so energy is conserved. This is a reasonable amount of heat for a system of this size. Because copper has a much lower specific heat than water, it is expected that the copper will undergo a larger change in temperature.

18.32. THINK: The question asks how much heat must be added to bring an aluminum pot and water to 95.0 C.° The masses of the aluminum and water are 1.19 kg and 2.31 kg, respectively. Both materials start at a temperature of 19.7 C° and the temperature is kept uniform during the process. SKETCH:

RESEARCH: The heat is given by .Q mc T= Δ SIMPLIFY: The total heat needed is the sum of the heat needed for each material respectively.

( ) ( )( )tot Al w Al Al w w Al Al w w Al Al w w f iQ Q Q m c T m c T m c m c T m c m c T T= + = Δ + Δ = + Δ = + − CALCULATE:

( )( ) ( )( ) ( )tot 1.19 kg 0.900 kJ/ kg K 2.31 kg 4.19 kJ/ kg K 95.0 C 19.7 C 729.9 kJ.Q = + ° − ° =

ROUND: The heat is reported to 3 significant figures. The heat needed to bring the kettle and water is 730. kJ. DOUBLE-CHECK: One calorie is defined as the energy required to heat one gram of water by one degree.

( )( )32.31 10 g 95 C 19.7 C 173943 cal 730 kJ.⋅ ° − ° = ≈ The low specific heat of aluminum ensures that the bulk of the energy goes toward heating the water. This is a reasonable answer.

18.33. THINK: Using a calorimeter, the type of material can be determined by calculating its specific heat. The heat transferred from brick is equal to the heat increase of the copper and water. The material has a mass of 3.0 kgm = and an initial temperature of 300.0 C.T = ° The copper and water have masses of 1.5 kg and 2.0 kg respectively and initial temperature of 20.0 C.° The equilibrium temperature of 31.7 C.T = ° SKETCH:

RESEARCH: The heat is given by .Q mc T= Δ SIMPLIFY: ( )? w Cu w w Cu Cu Cu w w Cu Cu .w wQ Q Q mc T m c T m c T m c m c T= + = Δ = Δ + Δ = + Δ Solving for ,c the specific heat of the unknown material:

( ) ( )( ) ( )w w Cu Cu w w Cu Cu/ / .w eq w eqc m c m c T m T m c m c T T m T T = + Δ Δ = + − − −


737

CALCULATE:

( ) ( )( ) ( ) ( )( ) ( )( ) ( )

2.0 kg 4190 J/ kg K 1.5 kg 386 J/ kg K 31.7 C 20.0 C130.227 J/ kg K

3 kg 31.7 C 300.0 Cc

+ ° − ° = =− ° − °

ROUND: The specific heat of the unknown material is reported to 2 significant figures: ( )130 J/ kg K ,c = which is very close to the specific heat of lead. The conclusion is that the material is lead. DOUBLE-CHECK: The mass of the brick is similar to the combined mass of the calorimeter and water. The large change in temperature of the object compared to the small change in temperature of the calorimeter and water points to a material of low specific heat. The calculated value and conclusion are reasonable.

18.34. THINK: The equilibrium temperature of the three materials is desired after they are placed in thermal contact. The copper has a mass of 2

Cu 2.0 10 gm = ⋅ and an initial temperature of Cu 450 K.T = The

aluminum has a mass of 2Al 1.0 10 gm = ⋅ and an initial temperature of 2

Al 2.0 10 K.T = ⋅ The water has a

mass of 2w 5.0 10 gm = ⋅ and an initial temperature of w 280 K.T = The values are given to two significant

figures. SKETCH:

RESEARCH: The heat is given by .Q mc T= Δ The copper will most likely decrease its temperature and the other two materials will increase their temperatures. SIMPLIFY: ( ) ( ) ( )Cu w Al Cu Cu Cu eq w w eq w Al Al eq Al Q Q Q m c T T m c T T m c T T= + − = − + −

Solving for eq :T ( )eq w w Al Al Cu Cu Cu Cu Cu w w w Al Al Al ,T m c m c m c m c T m c T m c T+ + = + + or,

Cu Cu Cu w w w Al Al Aleq

w w Al Al Cu Cu

.m c T m c T m c T

Tm c m c m c

+ +=

+ +

CALCULATE: Without units,

( )( )( ) ( )( )( ) ( )( )( )( )( ) ( )( ) ( )( )

2

eq

2 2

2 2 2

2.0 10 0.386 450 5.0 10 4.19 280 1.0 10 0.900 200

2.0 10 0.386 5.0 10 4.19 1.0 10 0.900282.619.

T⋅ + ⋅ + ⋅

=⋅ + ⋅ + ⋅

=

The units are: ( )( )( ) ( )( )( ) ( )( )( )

( )( ) ( )( ) ( )( )eq

g kJ/kg K K g kJ/kg K K g kJ/kg K Kg kJ/kg K g kJ/kg K g kJ/kg K

K

T+ +

= + +=

Altogether, eq 282.619 K.T =

ROUND: The equilibrium temperature is reported to 2 significant figures, eq 280 K.T =

DOUBLE-CHECK: This value is between the lowest and highest temperature. This value is also below the melting point of the metals and below the boiling point of water, so there is no need to worry about phase changes.


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18.35. THINK: I want to find the original temperature of the water before it was measured. Initially there is 11.0 g of Pyrex glass and 4.00 g of mercury at a temperature of 20.0 C.° The thermometer reads 29.0 C° when the water is added. The specific heat of Pyrex glass is given as 800. J/(kg K), and the specific heat of mercury is given as 140. J/(kg K), both at room temperature. SKETCH:

RESEARCH: The heat is given by .Q mc T= Δ SIMPLIFY: The water will decrease its temperature while the other materials increase their temperature.

w p m .Q Q Q= + ( ) ( ) ( )w w w eq p p eq m m eq .m c T T m c T T m c T T− = − + − Solving for eq :T

( )( ) ( ) ( )w w w p p m m eq w w eq w p p m m w w eq eq / .m c T m c m c T T m c T T m c m c m c T T T = + − + = + − +

CALCULATE:

( ) ( )( ) ( ) ( )( )( )( ) ( )w

11.0 g 800. J/ kgK 4.00 g 140. J/ kg K29.0 C 20.0 C 29.0 C 32.35 C

6.00 g 4190 J/kgKT

+= ° − ° + ° = °

ROUND: The temperature is reported to three significant figures. The initial temperature of the water is 32.4 C.° DOUBLE-CHECK: This is a reasonable answer. Note that if the water had a mass of 200. g the measure would have been 29.1 C,° much closer to the thermometer as would be expected.

18.36. THINK: When the water is poured on the ice, the heat transferred to the ice will warm it until it reaches a temperature of 0 C° (the melting point of ice). If there is still heat left for transfer, it will start to convert the ice to water. Once all the ice melts into water, if there remains heat to be transferred than the melted water will increase its temperature from m 0 C.T = ° The water has a mass of 1 400. gm = and an initial temperature of 1 30.0 C.T = ° The ice has a mass if 2 60.0 gm = and an initial temperature of 2 5.00 C.T = ° SKETCH:

RESEARCH: First, check the heat required to bring water from 30.0 C° to 0 C,° 1 ,Q and compare it to the heat required to bring ice from 5.00 C− ° to 0 C° and melt the ice completely, 2 .Q If 1Q is less than

2 ,Q the ice will only partially melt. If 1Q is greater than or equal to 2Q , then the water will totally melt the


739

ice. ( )1 1 w 1 mQ m c T T= − and ( )2 2 2 m 2 2 .mQ m c T T m L= − + If 1 2 ,Q Q< the ice left, ,ζ must be calculated.

Setting the transfer of heat of the water and ice equal gives ( ) ( ) ( )1 w 1 m 2 2 m 2 2 m .m c T T m c T T m Lζ− = − + − The heating is given by Q mc T= Δ or .Q mL= If 1 2Q Q≥ then the temperature of the water is desired. SIMPLIFY: Solving the equation for the transfer of heat for ζ gives

( ) ( )m 2 2 m 2 2 1 w 1 m 2 1 2 1 m / .mL m c T T m L m c T T Q Q Q Q Lζ ζ= − + − − = − = − Solve the following equation for the equilibrium temperature eq :T

( ) ( ) ( )( )

( )( ) ( )

1 w 1 eq 2 i m 2 2 m 2 w eq m

2 w eq 1 w eq 1 w 1 2 i m 2 2 m 2 w m

eq 1 w 1 2 i m 2 2 m 2 w m 2 1 w

/ .

m c T T m c T T m L m c T T

m c T m c T m c T m c T T m L m c T

T m c T m c T T m L m c T m m c

− = − + + −

+ = − − − +

= − − − + +

CALCULATE: ( ) ( )( )( )1 0.400 kg 4190 J/ kg K 30.0 C 0 C 50280 JQ = ° − ° =

( ) ( )( ) ( )( ) ( )( )32 0.0600 kg 2060 J/ kg K 0 C 5.00 C 0.0600 kg 334 10 J 26220 JQ = ° − − ° + ⋅ =

Since 1 2Q Q> the equilibrium temperature is calculated as follows. Without units,

( )( )( ) ( )( ) ( )( ) ( )( )( ) ( )( )( )( )( )eq

400. 4.19 30.0 60.0 2.22 0 5.00 60.0 1.00 334 60.0 4.19 015.343.

400. 60.0 4.19T

− − − − + = =+

The units for eqT are:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )( )eq

kJ kJ kJ kJg C g C C g kJ g Ckg K kg K kg K kg K

Cg g kJ/ kg K

T

° − ° − ° − + ° = = ° +

Altogether, eq 15.343 C.T = °

ROUND: The temperature is reported to three significant figures. The water reaches 15.3 C.° DOUBLE-CHECK: This temperature is in between 5.00 C− ° and 30.0 C° which makes sense.

18.37. The heat given off by the person is 180 kcal. This energy is consumed by converting water into steam. The amount of water is given by ( ) ( )vap/ 180 kcal / 539 cal/g 0.33395 kg 334 g.m Q L= = = = The amount of water converted to steam is 330 g.

18.38. A block of aluminum of mass Al 1.3 kgm = and temperature of 21 C.° The aluminum must be brought to a temperature of 932 K or 659 C° before it will melt. The heat required to bring the aluminum to this point is ( )1 m i .Q mc T T= − At this temperature, more heat is needed to melt the aluminum equal to

2 fus .Q mL= The total energy is then:

( ) ( )( )( )( )

tot 1 2 m i fus m i fus

1.3 kg 0.900 kJ/kgK 932 K 294 K 396 kJ/kg K1300 kJ.

Q Q Q mc T T mL m c T T L= + = − + = − +

= − + =

The heat required to melt1.3 kg at a temperature of 21 C° is 1300 kJ.

18.39. The time needed to vaporize the liquid is given by vap/ / .t Q P mL P= = For liquid nitrogen the time of

vaporization is calculated to be ( )( ) ( )5 41.00 kg 2.00 10 J/kg / 10.0 W 2.00 10 s.t = ⋅ = ⋅ The time helium

takes to vaporize is given by ( )( ) ( )41.00 kg 2.09 10 J/kg / 10.0 W 2090 s.t = ⋅ = Thus it takes about 10 times longer to vaporize liquid nitrogen than liquid helium.

18.40. THINK: The question asks for the equilibrium temperature inside the calorimeter after the steam is added. The steam will be converted to water and then the temperature will change to the equilibrium


740

temperature. The steam has a mass of s 10. gm = at a temperature of s 100.00 C,T = ° and has the same mass after it has been converted to water at w 19.0 C.T = ° The water in the cup has a mass of w 100. g,m = and the aluminum has a mass of 35 g , both at w 19.0 C.T = ° SKETCH:

RESEARCH: The heat from the change of temperature is .Q cm T= Δ The heat from the change of state is

vap .Q mL=

SIMPLIFY: The heat lost by the steam is equal to the heat gained by the water and the aluminum. s w Al .Q Q Q− = +

( )( ) ( ) ( ) ( )( )

( )

s vap s w eq s w w eq w Al Al eq w w w Al Al eq w

s vap s w eq s w s w w eq w w w Al Al eq Al Al w

s w eq w w eq Al Al eq s vap s w s w w w Al Al w

eq s w w w Al Al s vap s

m L m c T T m c T T m c T T m c m c T T

m L m c T m c T m c T m c T m c T m c Tm c T m c T m c T m L m c T m c T m c T

T m c m c m c m L m

− − + − = − + − = + −

− + = − + −+ + = + + +

+ + = + w s w w w Al Al w

s vap s w s w w w Al Al weq

s w w w Al Al

c T m c T m c Tm L m c T m c T m c T

Tm c m c m c

+ ++ + +

=+ +

CALCULATE: ( )( ) ( )( )( ) ( )( )( ) ( )( )( )

( )( ) ( )( ) ( )( )eq

10. g 539 10. 1.00 100.00 273 100. 1.00 19.0 273 35 .215 19.0 27310. 1.00 100. 1.00 35 .215

344.75

T+ + + + + +

=+ +

=( )( ) ( )( )( ) ( )( )( ) ( )( )( )

( )( ) ( )( ) ( )( )eq

g cal/g g cal/g K K g cal/g K K g cal/g K Kg cal/g K g cal/g K g cal/g K

K

T+ + +

= + +=

eq 344.75 K 71.75 CT∴ = = °

ROUND: The equilibrium temperature is reported to 2 significant figures since the masses are given to the least precise values (having two significant figures): eq 72 C.T = °

DOUBLE-CHECK: The condensation of the steam alone provides 5390 calories of heat energy to the system. By definition, this is sufficient to raise the temperature of 100 g of water by about 54 C.° The low specific heat of aluminum ensures that the bulk of the energy goes toward raising the temperature of the water. The result is reasonable.

18.41. THINK: 0.10 kg of molten aluminum is dropped into 1.00 L of water. The temperatures of the aluminum and water are 932 K and 295 K, respectively. I want to determine how much water will evaporate, how much aluminum will solidify, and what the equilibrium temperature will be. I also want to consider how the result would change if the temperature of the aluminum is increased to 1150 K. In order to determine the final state, the heat to solidify the aluminum, Al, sQ the heat to bring the water to its boiling point,

w,b ,Q and the heat to vaporize the water, w,vQ , are required. The aluminum has a mass of Al 0.100 kgm = at

Al 932 K 659 C.T = = ° The water has mass of w 1.00 kgm = at w 22 C.T = °


741

SKETCH:

RESEARCH: The heat is given by Q mc T= Δ and .Q mL= SIMPLIFY: The heat required to solidify the aluminum is ( )Al, s Al fusion,Al .Q m L= − The heat required to

bring the water to its boiling point ( )w, b w w boiling w .Q m c T T= − To find the equilibrium temperature use the

following equation ( ) ( )Al Al, m Al Al Al eq w w eq w .m L m c T T m c T T+ − = − The equilibrium temperature is then:

( ) ( )eq Al Al, m Al Al Al w w w Al Al w w/ .T m L m c T m c T m c m c= + + +

CALCULATE: (a,b) ( )( )Al, s 0.100 kg 396 kJ/kg 39.6 kJ.Q = − = −

( )( )( )w, b 4.19 kJ/kgK 1.00 kg 373 K 295 K 326.82 kJ.Q = − = Because Al,s w,bQ Q< , the molten aluminum does not supply enough heat to boil the water.

(c)( ) ( ) ( ) ( ) ( )

( ) ( )eq

kJ kJ kJ.10 kg 396 .10 kg 0.900 932 K 1.0 kg 4.19 295 Kkg kgK kgK

kJ kJ.10 kg 0.900 1.0 kg 4.19 kgK kgK

317.65 K 44.65 C

T

+ +

=

+

= = °

ROUND: The equilibrium temperature is reported to 2 significant figures. (a) None of the water boils away. (b) The aluminum will completely solidify. (c) The final temperature is 45 C.° (d) No. It is not possible to complete without knowing the specific heat of aluminum in its liquid phase. DOUBLE-CHECK: The large specific heat value of water makes it a very efficient coolant. This is a reasonable answer.

18.42. THINK: The question asks for the loss of internal energy during a rigorous work out. The question also asks for the amount of nutritional calories required to replace the loss of internal energy. The total work done is 51.80 10 J.W = ⋅ The heat required to evaporate 150 g of water can be found using the latent heat

of vaporization 6vap 2.42 10 J/kg.L = ⋅ A nutritional calorie is equal to 4186 J.

SKETCH: A sketch is not needed to solve the problem. RESEARCH: The first law of thermodynamics states int .Q E W= Δ + The heat loss is vap .Q mL= −

SIMPLIFY: int vap .E Q W mL WΔ = − = − −

CALCULATE: The change in internal energy is:

( )( ) ( )6 5 5int 0.150 kg 2.42 10 J/kg 1.80 10 J 5.4300 10 J.E Δ = − ⋅ − ⋅ = − ⋅

This energy is equivalent to the number of nutritional calories ( ) ( )55.4300 10 J / 4186 J/kcal 129.72 kcal.⋅ =

ROUND: The values will be reported to 2 significant figures. (a) The internal energy loses 55.4 10 J.⋅ (b) You should consume 130. nutritional calories to compensate for this loss.

DOUBLE-CHECK: This is a reasonable number of calories to burn in a light workout.


742

18.43. THINK: The question asks for the amount of water necessary to cool the carbon steel. The steel has a mass of 0.500 kg and must go from a temperature of 1346 F° to 500.0 F.° These temperatures in Celsius are ( )( )h 5/9 1346 F 32 F 730. CT = ° − ° = ° and ( )( )c 5/9 500. F 32 F 260. C.T = ° − ° = ° The blade will be surrounded by an unknown quantity of water and 2.000 kg of copper both at room temperature 20.0 C.° The specific heat of copper is 386 J/kgK. The table associated with the problem gives the specific heat of carbon steel at various temperature ranges. SKETCH:

RESEARCH: The heat loss by the carbon steel is equal to the heat gained by the copper and water. The heat is given by Q mc T= Δ or .Q mL= SIMPLIFY: The heat loss by the carbon steel is:

cs cs i ii

,Q m c T= Δ

where the summation goes over the different temperature ranges. The copper doesn’t reach its boiling point before 260. C.° The heat then from the copper is Cu Cu Cu Cu .Q m c T= Δ The water will be brought to its boiling point w w w b .Q m c T= Δ The water and the copper will reach a temperature of 100. C.° Equating the heat loss of the carbon steel to the heat gain of the water and copper:

cs cs Cu w Cu Cu Cu w w b .i ii

Q m c T Q Q m c T m c T= Δ = + = Δ + Δ Solving for the mass of the water:

( )w cs i i Cu Cu Cu w bi

.m m c T m c T c T = Δ − Δ Δ

CALCULATE: ( )( ) ( ) ( ) ( )( )

( )w

0.500 849 730 650 754 662 595 100 553 350. 260. 2.000 386 100. 20.04190 100. 20.0

0.290916,

m − + + + + − − − =

−=

with units of ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )

( )( )( )w

kg J/ kg K C J/ kg K C J/ kg K C kg J/ kg K Ckg.

J/ kg K Cm

° + ° + ° − ° = = °

Hence, w 0.290916 g. km = ROUND: The least precise value is given to three significant figures. If the water does not convert to steam, it takes 291 g of water to cool off the carbon steel. DOUBLE-CHECK: This is a reasonable answer. It takes relatively little water to accomplish the cooling due to the high specific heat of water. Note that the strength of the carbon steel also depends on the speed at which the steel hardens.

18.44. THINK: The question asks for the power dissipated in one Earth day by ethanol snowfall on Uranus. The polar regions are north of latitude 75.0° and south of latitude 75.0 .− ° In the polar regions, the amount of snow fall is 1.00 ft 0.3048 m.d = = The thermodynamic variables of the ethanol are ( )v 1.30 J/ g K ,c =

( )l 2.44 J/ g K ,c = and ( )sc 1.20 J/ g K .=


743

SKETCH:

RESEARCH: Since the minimum amount of energy that is lost to the atmosphere is required, assume that the ethanol starts at its boiling point, condenses to a liquid and then the liquid cools to the melting point and just freezes to become snow. The heat for a change in temperature is given by Q cm T= Δ and the heat for a change in phase is given by .Q mL= The area is given by 2 ,A r d= Ω where dΩ is the solid angle.

SIMPLIFY: The minimum amount of energy lost to the atmosphere is ( )v l f v l f .Q mL mc T mL m L c T L= + Δ + = + Δ +

In terms of the volume and density of the ethanol snow, ( ) ( )( ) ( )v l f v l f1 0.9 .Q V L c T L Ad L c T L= + Δ + = − + Δ +ρ ρ

The area covered by one pole is given by

( )2 2 2 0 2 o252 sin 2 cos | 2 1 cos25 .A r d r d r r°

°= Ω = = = − π θ θ π θ π The minimum amount of energy lost to the atmosphere for both poles is

( ) ( ) ( )2v l f4 1 cos25.0 1 0.9 .Q r d L c T Lπ ρ= − ° − + Δ +

The power dissipated is / .P Q t= CALCULATE:

( ) ( )( )( )( )( )( )( )( )

23 3 3

3 3 3

22

4 25559 10 m 1 cos25.0 0.100 0.3048 m 1.00 10 kg/m

858 10 J/kg 2.44 10 J/ kg K 351 K 156 K 104 10 J/kg

3.3707 10 J

Q π= ⋅ − ° ⋅

⋅ ⋅ + ⋅ − + ⋅

= ⋅

( )( )

2217

3.3707 10 J3.901 10 W

86400 sP

⋅= = ⋅

ROUND: To three significant figures, the minimum amount of energy lost to the atmosphere is 223.37 10 J,Q = ⋅ and the power dissipated in one Earth day is 173.90 10 W.P = ⋅

DOUBLE-CHECK: These large answers are reasonable for a planet.

18.45. By finding the thermal conductivity of the metal, it is possible to identify it. The ice has a volume of ( )( )100. mm 100. mm 5.00 mm ,V = a density of 3920. kg/m , starts at a temperature of 0 C° and melts in

0.400 s. The metal disk is 10.0 mm thick. The temperature on the other side of the disk is 100. C,° the boiling point of water. The power of heat transfer is given by ( )/ / .P Q t kA T d= = Δ The heat required to

melt the ice is ,Q mL= or in terms of volume and density .Q mL VLρ= = ( )/ / / .P Q t VL t kA T dρ= = = Δ Solving for the thermal conductivity gives: / .k VLd tA Tρ= Δ

( )( ) ( )( )( ) ( ) ( )

3 3 3920. kg/m 0.100 m 0.100 m 5.00 10 m 334 10 J/kg 0.0100 m384.1 W/ m K .

0.400 s 0.100 m 0.100 m 100. C 0 Ck

− ⋅ ⋅ = = ° − °

Rounding to three significant figures, the thermal conductivity of the metal is ( )384 W/ m K which is close to copper.


744

18.46. (a) The temperature at the copper-steel aluminum interface and the power flowing through the materials is desired. If the areas is a meter cubed, the copper has a thickness of 2.00 mm and one side is kept at a temperature of Cu 100.0 C.T = ° The steel on the other hand is 1.00 mm thick and is kept at S 25.0 C.T = ° Let S 1.00 mmL = be the width of the steel, and Cu 2.00 mmL = be the width of the copper.

The power of heat transfer is given by ( )/ / .P Q t kA T L= = Δ At the interface, the power must be the

same for both surfaces. Thus, ( ) ( )Cu Cu int Cu S int S S/ / .P k A T T L k A T T L= − = − Solving for the interface temperature, intT gives:

( ) ( )( )( )( ) ( )( )( )

( )( ) ( )( )

int Cu S Cu S Cu S S Cu Cu S/386 W/m K 0.00100 m 100.0 C 220. W/m K 0.00200 m 25.0 C

386 W/m K 0.00100 m 220. W/m K 0.00200 m60.048 C 60.0 C

T k L T k L T k L k L= + +° + °

=+

= ° ≈ °

(b) The result of part (a) can be used to calculate the power, ( )( )Cu Cu int Cu/ .P k A T T L= −

( )( ) ( ) ( )2 6386 W/mK 1.00 m 100. C 60.048 C / 0.00200 m 7710736 W 7.71 10 WP = ° − ° = ≈ ⋅

18.47. The question asks for the surface temperature of the Sun. This temperature can be determined by equating the power of a black body to the power reaching the Earth. 4

s Earth's orbitA .P A T Id I Aσε= = = The sun is

modeled as a black body, so 1.ε = The area of a sphere is 24 .rπ Using these facts 2 4 2s ES4 4 .r T I rσ π π=

Solving for the temperature gives ( )1/42 2ES s/ .T I r rπ σπ= Inputting the given values yields:

( )( ) ( )( )1/42 22 8 -8 4 2 51370 W / m 1.496 10 km / 5.67 10 W / K m 6.963·10 km 5778.99 KT = ⋅ ⋅ =

or 5506 C.°

The surface temperature of the Sun is 5780 K, or 5510 C.°

18.48. THINK: What is the equilibrium temperature of black and shiny engines? Then engine generates 11 kWP = and has an area of 2A 0.50 m= and a temperature of 0 27 CT = ° or 300. K. The emissivity of

the shiny engine is s 0.050ε = and the black engine has b 0.95.ε = SKETCH: A sketch is not needed to solve the problem. RESEARCH: The power of the radiation is 4 .P ATσε= SIMPLIFY: If the engines can only dissipate heat via thermal radiation, the power is given by

( )4 40AP T Tσε= − or ( )4 4

0/ .P A T Tσε = − Solving for T gives ( )1/440/ + .T P A Tσε=

CALCULATE: The shiny engine has a equilibrium temperature of:

( )( )( )( ) ( )

1/434

-8 4 2 2

11 10 W300 K 1669.478 K.

5.67 10 W / K m 0.05 0.50 mT

⋅ = + = ⋅

The black engine’s temperature is:


745

( )( )( )( )

( )1/43

4

-8 4 2 2

11 10 W300 K 803.362 K.

5.67 10 W / K m 0.50 m 0.95T

⋅ = + = ⋅

ROUND: The values will be reported to 2 significant figures. The shiny and black engine have temperature of 1400 C° and 530 C° respectively. DOUBLE-CHECK: These temperatures are very high, which is expected. Other sources of heat dissipation will cool these engines further.

18.49. THINK: I want to know how long it will take to freeze the Popsicle. The values are not given in SI units. The volume of the juice is 8.00 oz = ( )8.00 oz 0.0295735296 L/oz 0.2366 L.= The temperature of the

juice is j 71.0 FT = ° or ( )( )5 / 9 71.0 F 32 22.0 C.° − = ° The cooling power is:

( ) ( )( )4000 BTU/hr 1055.06 J/BTU / 3600 s/hr 1172.3 W.P = = I assume that the juice is similar to water.

SKETCH: A sketch is not needed to solve the problem. RESEARCH: The power is given by / .P Q t= The heat is given by Q mc T= Δ + fusion .mL The mass is equal to the volume. SIMPLIFY: The time is given by / .t Q P= The juice must first reach 0 C° before it can freeze.

( ) ( )( )j f fusion j f fusion/ / .t mc T T mL P m c T T L P = − + = − + The mass is given by the density times volume

( )( )j f / .t V c T T L Pρ = − +

CALCULATE: ( )( ) ( )( ) 34190 J/kg K 22.0 C 0 C 334 10 J/kg1.00 kg/L 0.2366 L 86.01 s.

1172.3 Wt

° − ° + ⋅= =

ROUND: The time is reported to 3 significant figures. It will take the juice 86.0 s to freeze. DOUBLE-CHECK: This is an unusually brief time to freeze an 8.00 oz Popsicle. The reasoning for this short time may be part of a classroom discussion: How efficient are freezers?

18.50. THINK: How long does it take for the ice to melt? The copper rod will transmit heat from the 90.0 C° water to the ice cube, melting it. The ice cube has dimensions 10.0 cm,s = with a density of

3 30.917 g/cm 917 kg/mρ = = and its temperature is i 0 CT = ° , its melting point. The area of the copper rod had a cross section of 10 cm and has a length of 20.0 cm. The other side of the copper rod is at

w 90.0 C.T = ° SKETCH:

RESEARCH: The heat required to melt the ice cube is fusionQ mL= or fusion .VLρ The rate of heat transfer of the rod is ( )W I / / .P kA T T L Q t= − =

SIMPLIFY: Solving for the time gives ( ) ( ) ( )fusion fusion

w i w i w i

.mL L VL LQLt

kA T T kA T T kA T Tρ

= = =− − −

CALCULATE: Using the heat of melting the time becomes:

( )( )( )( )( )( )( )

3 3 3 3

2

917 kg/m 1.00 10 m 334 10 J/kg 0.200 m17.633 s.

386 W/m K 0.100 m 363 K 273 Kt

−⋅ ⋅= =

−

ROUND: The values are reported to three significant figures. It takes about 17.6 seconds to melt the ice.


746

DOUBLE-CHECK: Because the copper rod is immersed in a large pool of water, it has an effectively infinite heat reservoir to draw on, so the ice melts very quickly.

18.51. THINK: I want to find the rate of heat flow through a window which is 0.32 cm thick, and has an area of 1.2 m by 1.4 m. The inside and outside temperatures of the window are 8.5 C° and 4.1 C° , respectively. SKETCH:

RESEARCH: The power is ( )h c / .P kA T T L= − SIMPLIFY: Not required.

CALCULATE: ( )( )( )( )

2

0.8 W/m K 1.2 m 1.4 m 281.5 K 277.1 K1848 W

0.32 10 mP −

−= =

⋅= 1.848 kW

ROUND: The heat loss rate is reported to 2 significant figures, so 1.8 kW.P = DOUBLE-CHECK: This is a reasonable answer. This is why heating costs are one of the most expensive energy wise.

18.52. THINK: The rate of heat loss due to radiation and the boil-off rate are desired. The temperatures are 2

h 3.0 10 KT = ⋅ and c 4.2 K.T = The area is 20.50 m .A = The latent heat of vaporization of liquid helium is v 20.9 kJ/kg,L = and its density is 0.125 kg/L.ρ = SKETCH: A sketch is not needed to solve the problem. RESEARCH: The heat loss due to radiation is given by 4 ,P ATσε= where 1,ε = since the dewar is treated as a black body. The net rate of heat transfer is the difference between the heat radiated by the black-body dewar, and the heat absorbed by it. The heat radiated is given by 4

rad ,DP ATσε= and the heat

absorbed is given by 4abs .HP ATσε= The boil-off rate will be given by the equation / /P Q t dQ dt= =

( )vap vap/ / .dmL dt dm dt L= = The volume is related to the mass by V.m ρ=

SIMPLIFY: (a) ( )4 4 4 4

h cD HP AT AT A T Tσε σε σε= − = −

(b) The boil-off rate is ( )/P dm dt L= ( )/d V dt Lρ= ( )/dV dt Lρ= . Solving for the rate of volume boil-off, /dV dt , gives / / .dV dt P Lρ= CALCULATE:

(a) ( )( )( ) ( ) ( )4 4-8 4 2 25.67 10 W / K m 1 0.50 m 300 K 4.2 K 229.635 WP = ⋅ − =

(b) ( )( )229.635 W/ 0.087899 L/s

0.125 kg/L 20.9 kJ/kgdV dt = =

ROUND: The values will be reported to 2 significant figures. (a) The rate of heat loss due to radiation is 0.23 kW. (b) The volume boil-off rate of the dewar is 0.088 L/s. DOUBLE-CHECK: The dewar loses heat at a rapid rate. This heat is transferred to the liquid helium contained in the dewar, causing a rapid boil-off rate. It is for this reason that cryogenic dewars are manufactured of reflective materials.


747

18.53. THINK: The solar irradiance on the surface of Mars and its temperature are desired. Mars is 1.52 times farther from the Sun than the Earth. Mars has a diameter of 0.532 that of Earth. SKETCH:

RESEARCH: The solar irradiance on the surface of Earth is approximately 1400 W/m2 . The average surface temperature on Earth is approximately 288 K. The intensity of the light reaching the planets is given by the equation

P = I dA = I 4πr 2 . The temperature can be found using the rate of heat loss for

radiation 4/ .P Q t ATσε= = SIMPLIFY: To find the irradiance note that the power of the Sun is equal for both Earth and Mars:

2 2M MS E ES4 4I r I rπ π= or 2 2

M E ES MS/ .I I r r= Using this irradiance, the temperature of Mars can be calculated assuming it is a blackbody, IMπrM

2 = σε4πrM2TM

4 . Solving for the temperature of the surface of Mars:

TM = IM

4σ4 .

CALCULATE: ( )( ) ( ) ( ) ( )2 22 2 2 2

M ES ES1368 W/m / 1.52 1368 W/m / 1.52 592.105 W/mI r r= = =

TM = 592.105 W/m2

4 5.67 ⋅10−8 W/K4m2( )4 = 226.04 K.

ROUND: The values are given to 3 significant figures. (a) The solar irradiance is 2592 W / m at the surface of Mars. (b) The temperature on the surface of Mars is 226 K. DOUBLE-CHECK: The published value for the solar irradiance on the surface of Mars is 2590 W / m . The temperature on the surface of Mars is around 210 K . Our answers are reasonable.

18.54. THINK: (a) The question asks for the rate of heat flow if the copper bar has a length of 2.00 m,L = a cross section of 0.100 ms = and is bordered by reservoirs 80.0 C° and 20.0 C.° (b) If the bar from part (a) has an area that varies as ( ) ( )( )20.0100 m 1 / 2.00 mA x= + where x is the distance along the bar from the cold end, what is the rate of heat flow? What is the rate of change of the temperature with distance at cold end, the middle and the hot end? SKETCH: (a)

(b)


748

RESEARCH: (a) The rate is given by ( )h c/ / .P Q t kA T T L= = − (b) The rate of heat flow will not change, since the contact area at the cold end is the same as in part a). The power is constant throughout the bar. The change in the temperature with distance can be found using ( )/ x .P kA T= Δ Δ SIMPLIFY:

(a) The area of the bar is 2A s= so ( )2

h c .ks T TQP

t L−

= =

(b) Solving the rate equation for / xTΔ Δ gives ( ) ( )( )2.

0.010 m 1 / 2.0 mT P Px kA k x

Δ = = Δ +

CALCULATE: (a) ( )( ) ( )2386 W / m K 0.100 m 80.0 C 20.0 C / 2.00 m 115.8 WP = ° − ° =

(b) At the cold end 0 m,x = thus ( )( )( )2

115.8 W= 30.0 C/m.386 W/m K 0.0100 m 1 0 / 2 m

Tx

Δ = ° Δ + The

middle of the bar is located at 1.00 mx = giving:

( )( )( )2

115.8 W= 20.0 C/m.386 W/m K 0.0100 m 1 1.00 m / 2 m

Tx

Δ = ° Δ +

At the hot end, 2.0 m.x = The rate of temperature change with the distance at the hot end is:

( )( )( )2

115.8 W= 15.0 C/m.386 W/mK 0.0100 m 1 2.00 m / 2.00 m

Tx

Δ = ° Δ +

ROUND: (a) The rate is reported to 3 significant figures. The rate of heat transfer is 0.116 kW.P = (b) The values are reported to 3 significant figures. The rate of heat flow is the same throughout the bar and is 0.116 kW. At the cold end, the middle and the hot end the rate of temperature change with distance is 30.0 C/m,° 20.0 C/m,° and 15.0 C/m,° respectively. DOUBLE-CHECK: (a) This value of a reasonable order of magnitude. (b) Since the rate of heat flow is constant, the temperature change must be less for a larger area. These are reasonable results.

18.55. THINK: The Planck spectrum distribution is given by ( ) ( )( )B/2 3T 2 / / 1 ,hf k Tf h c f eε π= − where h is

Planck’s constant and c is the speed of light. The frequency of the peak of this distribution is needed. The Boltzmann constant is 23 2 -2 -1

B 1.38 10 m kg s .K k −= ⋅ SKETCH:

RESEARCH: The frequency of the peak of the Planck distribution is determined from solving / 0.d dfε = The derivative of the Planck distribution is given by:


749

( )B

B

12 2/32 /

B

32 1 .1

hf k Thf k T

fd h hf edf k Tc eε π

−− = − − −

SIMPLIFY: Solving / 0,d dfε = it is found that:

( ) ( )( ) ( ) ( ) ( )B B B B B1 2 2/ / / / /2 3 2

B B3 1 / 1 0 1 3 1 / 0.hf k T hf k T hf k T hf k T hf k Tf e f h k T e f e e hf k T e− − − − − − = − − − =

This leads to 3 3 0,x xe xe− − = where Bf / .x h k T= Simplifying yields 3 3 xx e−− = or

( )3 3 3 1 .x xx e e− −= − = − Solving x iteratively with a starting value 0 3,x = it is found that

( ) ( ) ( ) ( )0 13 2.85060 1 23, 3 1 3 1 2.8506, 3 1 3 1 2.8266x xx x e e x e e− −− −= = − = − = = − = − = and after few

iterations, the most is 2.8215.x = This means that B/ 2.8215hf k T = or ( )B2.8215 / .f k h T= CALCULATE: (a) Substituting 23 2 -2 -1

B 1.38 10 kg sm Kk −= ⋅ and 346.626 10 J sh −= ⋅ gives ( )105.8792 10 Hz/K.f T= ⋅

(b) At 36.00 10 K,T = ⋅ the frequency of the peak is:

( )( )10 3 145.8792 10 Hz/K 6.00 10 K 3.5275 10 Hz.f = ⋅ ⋅ = ⋅ (c) At 2.735 K,T = , the frequency of the peak is:

( )( )10 115.8792 10 Hz/K 2.735 K 1.60796 10 Hz.f = ⋅ = ⋅ (d) At 300 K,T = the frequency of the peak is:

( )( )10 13 135.8792 10 Hz/K 300. K 1.7638 10 Hz 1.76 10 Hz.f = ⋅ = ⋅ ≈ ⋅

ROUND: (a) ( )105.88 10 Hz/Kf T= ⋅

(b) The temperature of the Sun is given to three significant figures in the question: 143.53 10 Hz.f = ⋅

(c) The temperature of the background radiation is given to four significant figures: 111.608 10 Hz.f = ⋅

(d) The Earth’s temperature is given only to three significant figures in the question: 131.76 10 Hz..f = ⋅ DOUBLE-CHECK: The result in (a), where ( )constantf T= is known as Wien’s displacement law. The rest of the calculated frequencies have Hertz as their units, which is appropriate. As one might expect, the frequencies increase with the temperatures.

18.56. Energy required to raise the temperature of an object by TΔ is given by .Q mc T= Δ Substituting the specific heat of aluminum, ( )910 J/ kg K ,c = the mass 0.300 kgm = and

( )( )100.0 273 20.0 273 K 80. KTΔ = + − − = , hence, ( )( )( )910 J/kg K 0.300 kg 80. K 21840 J 22 kJ.Q = = ≈

18.57. The R value of an object is equal to / ,R L k= where L is the thickness of the object and k is its thermal conductivity. Inserting the thermal conductivity of fiberglass batting ( )68.0 10 BTU/ ft F sk −= ⋅ ° and its thickness 4.0 inL = or 0.33 ftL = yields:

( )2 2

6

0.33 ft 1 hr 11.458 ft F hr/BTU 11 ft F hr/BTU.3600 s8.0 10 BTU/ ft F s

R −

= = ° ≈ ° ⋅ °

18.58. The amount of heat required to change the temperature of 10.0 kg of water by 10.0 K is

( )( )( ) 54190 J/kg K 10.0 kg 10.0 K 4.19 10 J.Q cm T= Δ = = ⋅ The kinetic energy of a car with 31.00 10 kgm = ⋅ and a speed of 27.0 m/s is:

( ) ( )( )( )22 3 5 51/2 1/2 1.00 10 kg 27.0 m/s 3.645 10 J 3.65 10 J.K mv= = ⋅ = ⋅ ≈ ⋅ It should be noted that Q and K are about the same.


750

18.59. The conduction rate through a spherical glass is given by ( )cond /P kA T L= Δ where TΔ is the temperature difference, k is the thermal conductivity of the glass and A and L are the area and thickness of the glass. Simplifying gives cond / .T P L kAΔ = Using the area of sphere, it is found that ( )2

cond / 4 .T P L k rπΔ =

Substituting ( )cond 0.95 100.0 W 95 W,P = = 30.50 10 m,L −= ⋅ 23.0 10 mr −= ⋅ and 1.0 W/mKk = gives

( )( ) ( )( )( )23 295 W 0.50 10 m / 1.0 W/mK 4 3.0 10 m 4.2 K.T π− −Δ = ⋅ ⋅ =

18.60. “Calories” here refers to food calories. The amount of heat transferred to the soft drink is .Q cm T= Δ Inserting ( )4190 J/ kg K ,c = 0.355 kgm = and ( )37 10.0 K 27 KTΔ = − = yields the expression

( )( )( )( ) 34190 J/ kg K 0.355 kg 27 K 40.16 10 J 9594 cal.Q = = ⋅ = Recall that the energy content of food

given in calories on the label is actually in food calories. The net energy content is 0 150 kcal 9594 cal 159744 cal.E E= = + = Rounding the value of E to 2 significant figures gives

51.6 10 food calories.E = ⋅

18.61. The conduction rate of the skin is given by ( )cond / .P kA T L= Δ After rearrangement, cond / .k P L A T= Δ

Substituting 33.0 10 m,L −= ⋅ cond 100. W,P = 21.5 mA = and ( )37.0 27.0 K 10.0 KTΔ = − = gives the

thermal conductivity, ( )( )

( )( )( )

32

2

100. W 3.0 10 m2.0 10 W/ m K .

1.5 m 10.0 Kk

−−

⋅= = ⋅

18.62. THINK: A lead bullet is fired at a wall. It is assumed that the bullet receives 75% of the work done on it by the wall as it stops. It is assumed the bullet has an initial temperature of 293 K (room temperature). Before the bullet starts to melt, the bullet’s temperature needs to be increased to the melting point which is 601 K. The heat of fusion for lead is 23.2 kJ/kg,L = and the specific heat of lead is ( )Pb 0.129 kJ/ kg K .c = SKETCH:

RESEARCH: The amount of energy absorbed by the bullet is given by the work done by the wall times 0.75. Therefore, ( )( ) ( )( )( ) ( )( )2 2

absorbed 0 00.75 0.75 1/ 2 0 0.375 .Q K mv mv= Δ = − = Before the bullet begins to melt, its temperature must be raised to the melting point of 601 K. The heat needed for this is given by

( )1 f i .Q cm T T= − After the melting point has been reached, the heat needed to completely melt the bullet is 2 fusion .Q mL= SIMPLIFY: (a) Thus, the minimum speed needed to start melting the bullet is found by equating the heat absorbed with the heat required to raise the temperature of the bullet 1.Q

( )( ) ( )2absorbed 1 0 f i 0.375 ,Q Q mv cm T T= = − which implies that ( )0 f i / 0.375.v c T T= −

(b) In order to completely melt the bullet, the heat absorbed must be equal to the heat required to raise the temperature and to melt the bullet, that is ( )( ) ( )2

absorbed 1 2 0 f i fusion 0.375 .Q Q Q mv cm T T mL= + = − +

Thus, the speed required is ( )0 f i fusion / 0.375.v c T T L= − +

CALCULATE: (a) ( )( )0 129 J/ kg K 601 K 293 K / 0.375 325.5 m/sv = − =


751

(b) ( )( )0 129 J/ kg K 601 K 293 K 23200 J/kg / 0.375 409.7 m/sv = − + =

ROUND: Rounding to two significant figures, (a) 0 330m/sv = and (b) 0 410 m/s.v = DOUBLE-CHECK: The results in (a) and (b) are typical speeds of bullets.

18.63. THINK: Solar radiation reaches the Earth’s surface at about 21.4 kW/m . It is assumed here that the Earth is a black body. This means that there is no reflection due to Earth’s atmosphere, and all solar radiation that reaches the Earth is absorbed by the Earth’s surface. SKETCH:

RESEARCH: The Sun emits radiation uniformly in all directions. This means that if the total power of solar radiation is ,P then the intensity of radiation at a distance r from the Sun is distributed uniformly

over a spherical surface area. Thus, 2 .Spherical Area 4

P PIrπ

= = Therefore, the intensity of radiation that

reaches the Earth is 2E SE/ 4 .I P rπ= Similarly for Mars, the intensity that reaches Mars is 2

M SM/ 4 .I P rπ= SIMPLIFY: Since EI is known, the power of radiation P can be eliminated from the above equations

giving the intensity on Mars as ( ) ( )2 2 2 2M SE SM E SE SM E4 / 4 / .I r r I r r Iπ π= =

CALCULATE: Substituting 2E 1.4 kW/m ,I = 11

SE 1.496 10 mR = ⋅ and 11M 2.28 10 mR = ⋅ yields:

2113 2 2 2

M 11

1.496 10 m 1.4 10 W/m 6.03 10 W/m .2.28 10 m

I ⋅= ⋅ = ⋅ ⋅

ROUND: Keeping only two significant figures gives 2 2M 6.0 10 W/m .I = ⋅

DOUBLE-CHECK: Since SMR is larger than SER it is expected that MI is less than E .I

18.64. THINK: The radiated power of a body depends on its surface area, temperature and emissivity. Assume that your body has an emissivity of 1.00,ε = that is, that you are a black body. SKETCH:

RESEARCH: The radiated power of an object is given by 4b .P ATσε= The power radiated by my body is

4rad b .P ATσε= If the surrounding environment has a temperature env ,T the power absorbed by my body is

4abs env .P ATσε= Since my body radiates energy to the environment and at the same time absorbs energy,

the net power is net rad abs+ .P P P=

SIMPLIFY: ( )4 4net b envP A T Tσε= −


752

CALCULATE: (a) The radiated power from my body is:

( )( )( ) ( )( )44 8 2 4 2rod b 5.6703 10 W/m K 1.00 2.00 m 273 33.0 K 994.3 W.P ATσε −= = ⋅ + =

(b) The net body radiated power is:

( )( )( ) ( ) ( )( )( )4 4 48 2 4 2net 5.6703 10 W/m K 1.00 2.00 m 273 33.0 273 20.0 K 158.5 W.P − = ⋅ + − + =

(c) The net body radiated power is:

( )( )( ) ( ) ( )( )( )4 4 48 2 4 2net 5.6703 10 W/m K 1.00 2.00 m 273 27.0 273 20.0 K 82.8 W.P − = ⋅ + − + =

ROUND: The final answers should be rounded to three significant figures. (a) rod 994 WP = (b) net 159 WP = (c) net 82.8 WP = DOUBLE-CHECK: The answers are reasonable. It is expected that the body radiates less heat when its surface temperature is lower.

18.65. THINK: A 10.0 g ice cube has an initial temperature of 10.0 C.− ° The ice cube is dropped into 40.0 g of water at 30.0 C.° The equilibrium temperature of the water must be calculated. Then, a second similar ice cube is added. The new equilibrium temperature is required. SKETCH:

RESEARCH: It is noted that the ice temperature is below the freezing point. Therefore, any heat transferred to the ice initially can only increase the temperature of the ice. The heat 1Q required to raise the temperature of the ice is ( )1 I I 1 I I I0 C .Q c m T c m T= Δ = ° − After the temperature of the ice reaches 0 C,° the heat 2Q needed to melt the ice is 2 F I .Q L m= Then, after all ice has melted, the heat 3Q needed to increase the temperature by 2TΔ is ( )3 W I 2 W I f 0 C .Q c m T c m T= Δ = − ° The amount of heat 4Q transferred

from the water to the ice is ( )4 W W 3 W W f W .Q c m T c m T T= Δ = − Since the system is isolated, the net heat must be zero. 1 2 3 4 0.Q Q Q Q+ + + = When a second ice cube is added to the system, the process must be repeated with the new starting temperature, and taking into account the added mass of the first ice cube. SIMPLIFY: Adding all heats, it is found that:

( ) ( ) ( )I I I F I W I f W W f W0 C T 0 C 0.c m L m c m T c m T T° − + + − ° + − = Solving for fT gives:

( )W W W I I I F I

fW W I

.c m T c m T L m

Tc m m

+ −=

+

CALCULATE: (a) The final temperature of the water is:


753

( )( ) ( )( ) ( )( )( )

3 2

f 2 2

J J J4190 40.0 g 30.0 C 2220 10.0 g 10.0 C 334 10 1.00 10 kgkg K Kg K kg

4190 J/kg K 4.00 10 kg 1.00 10 kg6.998 C

T

−

− −

° − ° − ⋅ ⋅

=⋅ + ⋅

= °

(b) The final temperature using w 50.0 gm = and w 6.998 CT = ° is:

( )( ) ( )( ) ( )

( )( )( )

3

f

J J J4190 50.0 k 6.998 C 2220 10.0 g 10.0 C 334 10 10.0 gkg K kg K kg

8.337 C4190 J/ kg K 50.0 g 10.0 g

T

° − ° − ⋅

= = − °+

Since fT is negative, this means that the ice has not melted completely or has not reached the melting point. Let us calculate the value of 1 ,Q 2Q and 4Q assuming f 0T = in order to determine the state of the ice. The heat is:

( )( )( ) ( )( )( )( )

( )( )( )( )

1

32

4

2220 J/ kg K 20 g 0 C 10 C 444 J

334 10 J/kg 20 g 6680 J

4190 J/ kg K 40 g 0 C 30 C 5028 J.

Q

Q

Q

= ° − − ° =

= ⋅ =

= ° − ° = −

Since 1 2Q Q+ is larger than the magnitude of 4Q and 1Q is less than the magnitude of 4 ,Q this means that the ice has not melted completely and the final temperature is therefore, f 0 C.T = ° ROUND: Rounding to three significant figures, (a) f 7.00 CT = ° . (b) f 0.00 CT = ° DOUBLE-CHECK: The largest part of the heat required for the process is that which causes the phase change from solid to liquid. Doubling the mass of the ice effectively doubles this heat requirement. It is not surprising that the system does not contain sufficient heat to complete the process.

18.66. THINK: The mass of water is 60.0 kg and the temperature is 35.0 C.° It is assumed the Sun gives out 3 21.00 10 W/m⋅ and the dimension of a mirror is 25.0 cm by 25.0 cm. The mirrors are held at an angle of

45.0 .° The total power received by the cylinder of water depends on the amount of power reflected by the mirrors. SKETCH:

RESEARCH: The amount of power reflected by a mirror is given by ( )1 Effective Area ,P I= I is intensity of solar radiation. 1 cos .P IA θ= The total power reflected is 1 cos .P NP NIA θ= = For an interval of time t, the energy absorbed by the water is cos .E Pt NIA tθ= = The amount of heat to raise the temperature of the water is ( )f i .Q cm T T= −

SIMPLIFY: Equating E with Q gives the time required is ( )f i .

coscm T T

tNIA θ

−=


754

CALCULATE: Substituting 3 21.00 10 W/m ,I = ⋅ 34.20 10 J/kg C,c = ⋅ ° 60.0 kg,m = f 100. C,T = °

iT 35.0 C,= °

( )( ) 2 2A 0.250 0.250 m 0.0625 m= = and 45.0θ = ° yields:

( )( )( )( )( )( ) ( )

3

3 2 2

4.20 10 J/kg C 60.0 kg 100. C 35.0 C7.41 s.

50000 1.00 10 W/m 0.0625 m cos 45.0t

⋅ ° ° − °= =

⋅ °

ROUND: Using three significant figures gives 7.41 s.t = DOUBLE-CHECK: 50,000 mirrors each with an area of 20.0625 m is equivalent to a single mirror with an area of more than more than 23 km . Considering the fact that all of the power reflected by this massive mirror is focused on a single point, the result is not surprising.

18.67. THINK: Assume the Gulf Stream is a box shaped object with a length of 38.00 10 km,⋅ a width of 100. km and a depth of 500. m. Assume also that the temperature inside the box is uniform. SKETCH:

RESEARCH: The power radiated by an object is 4 .P ATσε= The surface area of the Gulf Stream is equal to 2 2 2 .A LW Ld Wd= + + Here L, W and D are the length, width and depth of the Gulf Stream. The absorbed power is from the Sun, which corresponds to receiving 1400 W/m2 for half the day on the surface of the water,

Pabs = 1/ 2( ) 1400. W/m2( )(LW ) = 700.0 W/m2( )(LW ).

SIMPLIFY: Pnet = Prad − Pabs = σεAT 4 − 700.0 W/m2( )(LW ).

CALCULATE: To make the equation fit on the page, compute radP first without units:

( )( )( ) ( )( ) ( )( ) ( )( ) ( )48 6 5 6 5rad

14

5.6703 10 0.930 2 8.00 10 1.00 10 8.00 10 500. 1.00 10 500. 290

5.9978 10 .

P − = ⋅ ⋅ ⋅ + ⋅ + ⋅ = ⋅

Then, the units of radP are: ( )( ) ( )( ) ( )( ) ( )( ) ( )44 2rad W/ K m m m m m m m K W,P = + + = so altogether

14rad 5.9978 10 W.P = ⋅

Pabs = 700. W/m2( ) 8.00 ⋅106 m( )1.00 ⋅105 m( ) = 5.60 ⋅1014 W.

Pnet = 5.9978 ⋅1014 W − 5.60 ⋅1014 W=3.978 ⋅1014 W. ROUND: Rounding to three significant figures gives Pnet = 3.99 ⋅1013 W. DOUBLE-CHECK: It is well known that the Gulf Stream is responsible for significantly warming the waters in the North Atlantic, so one would expect its radiated power to be very large.

18.68. THINK: 1.00 kg of steam at 100.0 C° is poured over a 4.00 kg block of ice at 0.00 C.° After the system reaches equilibrium, the final temperature is needed.


755

SKETCH:

RESEARCH: Since the ice is already at the freezing point, the heat needed to raise the ice to a final temperature fT is equal to the heat needed to melt the ice and the heat needed to increase the temperature of the ice, that is, ( )1 I I W I f 0 C .Q L m c m T= + − ° The heat released from the steam to decrease its temperature to the equilibrium temperature is equal to the heat released during condensation and the heat released when its temperature is reduced, that is, ( )2 S S W S f S .Q L m c m T T= − + − Since the system is isolated, this means there is no heat entering or leaving the system. Therefore, 1 2 0.Q Q+ = SIMPLIFY: ( ) ( )I I W I f S S W S f s0 C 0.L m c m T L m c m T T+ − ° − + − = Solving for fT yields:

( ) ( )f S S W I S I I W I S/ .T L m c m T L m c m m= + − + CALCULATE: The final temperature is:

( )( ) ( )( )( )( ) ( )( )( )( )( )

6 5

f

2.26 10 J/kg 1.00 kg 4186 J/ kg C 1.00 kg 100.0 C 3.33 10 J/kg 4.00 kg64.338 C.

4186 J/ kg C 1.00 kg 4.00 kgT

⋅ + ° ° − ⋅= = °

° +

ROUND: f 64.3 CT = ° DOUBLE-CHECK: Because S IL L> by a factor of 10, it is reasonable that the equilibrium temperature is slightly closer to the initial temperature of the steam.

18.69. THINK: The heat transfer through an object depends on its thermal conductivity, surface area and thickness, and the temperature difference between the heat reservoirs (in this case, the environment and the soda). SKETCH:

RESEARCH: First, the surface area of the soda cans and the plastic bottle are needed. The surface area of a cylindrical object is ( ) ( )2 2A 2 / 2 1/ 2 ,D Dh D Dhπ π π π= + = + where D is the diameter and h is the height. The heat transfer for the 6 soda cans is c c c c6 A / .cP k T L= Δ For the plastic bottle, the heat transfer is

B B B B BA / .P k T L= Δ SIMPLIFY: The ratio of the initial heat current into all six cans to the initial heat current into the bottle is:

( ) ( )B c c c c B B B Bratio / 6 A / / A / .cP P k T L k T L= = Δ Δ Since p cT TΔ = Δ and c B ,L L= the ratio becomes:

( ) ( )c c B Bratio 6 A / A .k k= Substituting ( ) 2c c c c1/ 2A D D hπ π= + and ( ) 2

B B B B1/ 2A D D hπ π= + gives:

( )( )

( )( )

2 2c c c c c c c

2 2B B B B B B B B

6 1/ 2 6 1/ 2ratio .

1/ 2 1/ 2

k D D h k D D h

k D D h k D D h

π π

π π

+ += =

+ +


756

CALCULATE: Putting in the numerical values yields:

( )( )( ) ( )( )( )( )( ) ( )( )

2

2

6 205 W/mK 1/ 2 0.0600 m 0.0600 m 0.120 mratio 87.11

0.100 W/mK 1/ 2 0.100 m 0.100 m 0.250 m

+= =

+

ROUND: Rounding to three significant figures gives ratio 87.1:1.= DOUBLE-CHECK: The cans have a larger combined surface area than the bottle. This, in combination with the larger thermal conductivity of aluminum leads to the expected result that the cans warm up more quickly than the bottle.

18.70. THINK: A 10.0 in. thick piece of fiberglass has an R-factor of ( )240.0 ft Fh/BTU .° I want to convert this value to SI units. I also want to determine the heat transfer through a wall that is insulated with this fiberglass, when the outdoor and indoor temperatures are 22.0 C and 23.0 C− ° ° , respectively. SKETCH:

RESEARCH: The conversion values for the units are 1 ft 0.3048 m,= 1 F 5/9 C,° = ° 31 h 3.60 10 s= ⋅ and

31 Btu 1.055 10 J.= ⋅ The heat transfer is given by : ( )cond i 0 / .P kA T T L= −

SIMPLIFY: Therefore, the heat transfer is ( ) ( )cond i 0 i 0/ / ,P kA T T L A T T R= − = − since / .R L k= The heat

transfer per 2m is ( )2cond i 0/ m / .P T T R= −

CALCULATE: (a) The thermal resistance R in SI unit is:

( ) ( )( )( ) ( )22 3 3

2

40.0 ft F h/BTU 0.3048 m/ft 5/9 K/ F 3.60 10 s/h / 1.055 10 J/BTU

7.045 m K/W.

R = ° ⋅ ° ⋅ ⋅

=

(b) The heat transfer per 2m is ( ) ( )2 2

cond 2

273 23.0 K 273 22.0 K/ m 6.388 W/m .

7.045 m K/WP

+ − −= =

ROUND: Rounding to three significant figures, 2 2 2 cond cond7.05 m K/W, / m 6.39 W/mP P= = .

DOUBLE-CHECK: This is within the range of R-values for commercial fiberglass insulation.

18.71. THINK: The conduction rate of insulation, condP , depends on the area and the R value of the material, and the temperature difference on either side of the insulation. Let H 294 K,T = and C 277 K.T = SKETCH:

RESEARCH: The conduction rate is given by ( )cond H C / .P A T T R= − The increase in the conduction rate

due to the change in the R value is ( ) ( )cond1 cond2 H C 1 H C 2/ / .P P P A T T R A T T R Δ = − = − − −


757

SIMPLIFY: ( ) ( ) ( )H C 1 21/ 1/P A T T R R Δ = − − (a) The change in heat that exits the room in an interval of time is:

( ) ( ) ( )H C 1 21/ 1/ .Q t P tA T T R R = Δ = − −

(b) In three months, the extra heat that exits the room is 90Q. CALCULATE: (a) Substituting ( )( ) 25.0 m 5.0 m 25 m ,A = =

( )2 21 19 0.176 m K/W 3.344 m K/W,R = = ( )2 2

2 30 0.176 m K/W 5.28 m K/WR = = and

( )24 3600 s 86400 st = = :

( )( ) ( )( ) ( )( )2 2 2 686400 s 25 m 294 K 277.0 K 1/ 3.344 m K/W 1/ 5.28 m K/W 4.03 10 J.Q = − − = ⋅

(b) ( )6 890 4.03 10 J 3.62 10 J.Q = ⋅ = ⋅ Since the electrical energy for heating costs 12 cents/kWh or 63.33 10 cents/J,−⋅ the increase in cost of electrical heating is ( )8 6cos 3.62 10 J 3.33 10 cents/Jt −= ⋅ ⋅

1206.7 cents.= ROUND: Keep only two significant figures. (a) 64.0 10 JQ = ⋅ (b) The extra cost is 1200 cents, or 12 dollars. DOUBLE-CHECK: The units of Q are Joules, and the cost is in cents. These units are expected units for these values. 4 million Joules is a reasonable amount of energy, corresponding to a cost of $12 dollars.

18.72. THINK: This is a problem of heat flow through three layers of material; two panes of glass and an air gap. Each glass pane has a thickness of 3.00 mm and the air gap is 1.00 cm thick. The thermal conductivities of glass and air are 1.00 W/m K and 0.0260 W/m K, respectively. SKETCH:

RESEARCH: The heat flow through all three layers must be the same. The conduction rate condP is given by ( )cond H C / .P kA T T L= − Therefore, the conduction rate through the layers is

( ) ( ) ( )cond G 0 1 G A 1 2 A G 2 3 G/ / / .P k A T T L k A T T L k A T T L= − = − = − The rate of heat energy lost through the

window per 2m is ( )cond G 0 1 G/ /P A k T T L= − . If the window had no air layer between the panes, the rate of

heat loss will become ( )cond G 0 1 G/ /P A k T T L= − SIMPLIFY: Therefore, two equations are obtained:

(1) ( ) ( )0 1 2 3T T T T− = − and (2) ( ) ( )( )1 2 G A A G 2 3/ .T T k L k L T T− = −

Adding (1) and (2) yields ( ) ( )( )0 2 G A A G 2 31 / .T T k L k L T T− = + − Solving for 2T gives

32

0 3

2A G A G G A

A G G A

k L T k L T k Lk

Tk L

TL

+ ++

=

The temperature 1T is calculated using 1 0 2 3 .T T T T= − +


758

CALCULATE: (a) Substituting the numerical values gives:

( )( )( ) ( )( )( ) ( )( )( )( )( ) ( )( )

3 3 2

2 3 2

293 0.0260 3.00 10 273 0.0260 3.00 10 273 1.00 1.00 10

0.0260 3.00 10 1.00 1.00273.154

10,

2T

− − −

− −

⋅ + ⋅ + ⋅

⋅ + ⋅= =

with units of ( )( )( ) ( )( )( ) ( )( )( )

( )( ) ( )( )2

K W/m K m K W/m K m K W/m K mW/m K m W/

m m

K.K

T = = + +

+ Therefore

2 273.154 K.T = 1 293.0 K 273.154 K 273.0 K 292.846 K 19.846 C.T = − + = = ° So, the temperatures of the four air-glass interfaces are 0 20.00 C,T = ° 1 19.846 C,T = ° 2 0.154 C,T = ° and 3 0.00 C.T = °

(b) ( ) 3 2cond / 1.00 W/mK 20.00 C 19.846 C / 3.00 10 m 51.33 W/mP A −= ° − ° ⋅ =

(c) ( ) 3 2cond / 1.00 W/mK 20.00 C 0.00 C / 6.00 10 m 3333 W/mP A −= ° − ° ⋅ =

(d) Window glass is not manufactured in this way because there is a pressure difference between the vacuum and the outside air. This pressure difference produces enough force to break the glass. If the glass can withstand the pressure difference, the glass must be very thick and the production of the glass with the vacuum gap would be very expensive ROUND: Keeping three significant figures; (a) 0 20.0 C,T = ° 1 19.8 C,T = ° 2 0.154 C,T = ° 3 0.00 CT = °

(b) 2cond / 51.3 W/mP A =

(c) 2cond / 3330 W/mP A =

DOUBLE-CHECK: The temperatures decrease with each interface, as expected.

fisica para ingenieria y ciencias bauer vol i - capitulo 18 soluciones

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