Cnx Physics Ism Ch 14 Heat and Heat Transfer

College Physics Instructor Solutions Manual Chapter 14

366

CHAPTER 14: HEAT AND HEAT TRANSFER METHODS 14.2 TEMPERATURE CHANGE AND HEAT CAPACITY

1. On a hot day, the temperature of an 80,000-‐L swimming pool increases by . What is the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation.

Solution

Therefore,

2. Show that .

Solution

3. To sterilize a 50.0-‐g glass baby bottle, we must raise its temperature from to. How much heat transfer is required?

Solution

4. The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of the following, originally at : (a) water; (b) concrete; (c) steel; and (d) mercury.

Solution

C1.50°

. kg 108.00L 1000

m 1L) 00080)(kg/m 10(1.00 43

33 ×=××=ρ= ,Vm

J 1002.5C)501C)(J/kg 4186()kg 108.00( 84 ×=°°⋅×=Δ= .TmcQ

Ckcal/kg1Ccal/g1 °⋅=°⋅

Ckcal/kg 1kg 1

g 1000cal 1000

kcal1Cg

cal1°⋅=××

°⋅

C22.0°C95.0°

J 103.07J 3066C)3.07C)(J/kg 840kg)( 10(50.0 3-3 ×==°°⋅×=Δ= TmcQ

C20.0°

mcQTTmcQ =Δ⇒Δ=


367

(a)

(b)

(c)

(d)

5. Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 7.50 cm per rub, and with an average frictional force of 40.0 N, what is the temperature increase? The mass of tissues warmed is only 0.100 kg, mostly in the palms and fingers.

Solution Let be the number of hand rubs and be the average frictional force of a hand

rub:

6. A 0.250-‐kg block of a pure material is heated from to by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is most likely composed.

Solution

It is copper.

7. Suppose identical amounts of heat transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water?

Solution

C 021C 001C 020C 001C)kcal/kg 001kg)( 001(

kcal 001°=°+°=⇒°=

°⋅=Δ ...T.

...T

C25.0C0.5C20.0C0.5C)kcal/kg kg)(0.20 (1.00

kcal 1.00°=°+°=⇒°=

°⋅=Δ TT

C3.92C26.9C20.0C26.9C)kcal/kg kg)(0.108 (1.00

kcal 1.00°=°+°=⇒°=

°⋅=Δ TT

C0.50C0.30C20.0C0.30C)kcal/kg kg)(0.0333 (1.00

kcal 1.00°=°+°=⇒°=

°⋅=Δ TT

N F

C0.171C)J/kg kg)(3500 (0.100m) 10N)(7.50 20(40.0 2

°=°⋅

×==Δ⇒Δ==

−

mcNFdTTmcNFdQ

C20.0° C65.0°

Ckcal/kg 09240C)045kg)( 2500(

kcal 041°⋅=

°=

Δ=⇒Δ= .

...

TmQcTmcQ

10.8Ckcal/kg 09240

Ckcal/kg 1

c

w

w

c

ccww

=°⋅

°⋅==

Δ==Δ

.cc

mm

TcmQTcm


368

8. (a) The number of kilocalories in food is determined by calorimetry techniques in which the food is burned and the amount of heat transfer is measured. How many kilocalories per gram are there in a 5.00-‐g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-‐kg aluminum cup, causing a temperature increase? (b) Compare your answer to labeling information found on a package of peanuts and comment on whether the values are consistent.

Solution (a)

(b) A label for unsalted dry roasted peanuts says that 33 g contains 200 calories (kcal),

which is which is consistent with our results to part (a),

to one significant figure.

9. Following vigorous exercise, the body temperature of an 80.0-‐kg person is . At what rate in watts must the person transfer thermal energy to reduce the body temperature to in 30.0 min, assuming the body continues to produce energy at the rate of 150 W?

Solution

Thus, .

C54.9°

( )

kcal/g 5.73g 005

kcal 6328

kcal 6328C)954(C)kcal/kg 2150kg)( 1000(

C)kcal/kg 001kg)( 5000(

p

AlAlwwAlAlww

==

=°⎥⎦

⎤⎢⎣

⎡

°⋅

+°⋅=

Δ+=Δ+Δ=

..

mQ

......

Q

TcmcmTcmTcmQ

,kcal/g 6g 33

kcal 200

p

==mQ

C40.0°

C37.0°( )J/s1=W1orndjoule/seco1=watt1

W104.67min) s/1 min)(60 (30J 108.40

J 1040.8C)37-CC)(40J/kg kg)(3500 (80.0

25

cooling

5bodyhuman

×=×

==

×=°°°⋅=Δ=

tQP

TmcQ

W617 W150 W 467bodycoolingrequired =+=+= PPP


369

10. Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails . (a) Calculate the rate of temperature increase in degrees Celsius per second if the mass of the reactor core is and it has an average specific heat of

. (b) How long would it take to obtain a temperature increase of , which could cause some metals holding the radioactive materials to melt?

(The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the steel containment vessel would also begin to heat up.)

Solution (a)

Recall that 1 W = 1 J/s. Thus for 1 s is given by

(b)

14.3 PHASE CHANGE AND LATENT HEAT

11. How much heat transfer (in kilocalories) is required to thaw a 0.450-‐kg package of frozen vegetables originally at if their heat of fusion is the same as that of water?

Solution

12. A bag containing ice is much more effective in absorbing energy than one containing the same amount of water. (a) How much heat transfer is necessary to raise the temperature of 0.800 kg of water from to ? (b) How much heat transfer is required to first melt 0.800 kg of ice and then raise its temperature? (c) Explain how your answer supports the contention that the ice is more effective.

megawatt)1=MW1andJ/s1=W1orndjoule/seco1=watt(1)C/s(°

kg1060.1 5×CkJ/kg0.3349 °⋅

C2000°

kg-105 5×

mcQTTmcQ =Δ⇒Δ=

TΔ

C/s2.80RateC2.80C)kcal/kg kg)(0.0800 10(1.60

)kcal/4186J J)(1 10(1505

6

°=⇒°=°⋅×

×=ΔT

min 11.9s 714C/s2.80C2000

==°

°=t

C0°

kcal 35.9kcal/kg) kg)(79.8 (0.450f ===mLQ

C0°C0°

C0° C30.0°C0°


370

Solution (a)

(b)

(c) The ice is much more effective in absorbing heat because it first must be melted, which requires a lot of energy, then it gains the same amount of heat as the bag that started with water. The first of heat is used to melt the ice, then it absorbs the of heat as water.

13. (a) How much heat transfer is required to raise the temperature of a 0.750-‐kg aluminum pot containing 2.50 kg of water from to the boiling point and then boil away 0.750 kg of water? (b) How long does this take if the rate of heat transfer is 500 W [ ]?

Solution (a)

(b)

14. The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.00 g of condensation forms on a glass containing both water and 200 g of ice, how many grams of the ice will melt as a result? Assume no other heat transfer occurs.

Solution

(Note that for water at is used here as a better approximation than for

water.)

J 101.00C)C)(30.0J/kg kg)(4186 (0.800 5×=°°⋅=Δ= TmcQ

J 1068.3J 101.005J/kg) 10kg)(334 (0.800 553f ×=×+×=Δ+= TmcmLQ

J 102.67 5×J 101.00 5×

C30.0°

J/s)1=W(1ndjoule/seco1=watt1

kcal 591kcal 590.5Qkcal/kg) kg)(539 (0.750

C)C)(70.0kcal/kg kg)(0.215 (0.750C)C)(70.0kcal/kg kg)(1.00 (2.50v

'wAlAlww

==

+

°°⋅+°°⋅=

+Δ+Δ=

QLmTcmTcmQ

s 104.94 W500J/kcal 4186kcal) (590.5

.time andpower where,

3×=⎟⎠

⎞⎜⎝

⎛=

===⇒=

t

tPPQtPtQ

g 58.1kcal/kg 79.8kcal/kg 580g) (8.00

f

vwiceficevw =⎟⎟

⎠

⎞⎜⎜⎝

⎛==⇒=

LL

mmLmLm

vL C37° vLC001 °


371

15. On a trip, you notice that a 3.50-‐kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at and completely melts to water in exactly one day [

]?

Solution

16. On a certain dry sunny day, a swimming pool’s temperature would rise by if not for evaporation. What fraction of the water must evaporate to carry away precisely enough energy to keep the temperature constant?

Solution Let be the mass of pool water and be the mass of pool water that evaporates.

(Note that for water at is used here as a better approximation than for water.)

17. (a) How much heat transfer is necessary to raise the temperature of a 0.200-‐kg piece of ice from to , including the energy needed for phase changes? (b) How much time is required for each stage, assuming a constant 20.0 kJ/s rate of heat transfer? (c) Make a graph of temperature versus time for this process.

Solution (a) (i) Heat needed to warm ice to

(ii) Heat needed to melt ice at

(iii) Heat required to warm water to

(iv) Heat required to vaporize water at

(v) Heat required to warm vapor to

C0°C0°

J/s)1=W(1ndjoule/seco1=watt1

W 13.5 W3.531s 64008

C)kJ/kg kg)(334 (3.50f ==°⋅

===tmL

tQP

C1.50°

M m

3

C)v(37C)v(37 102.59

kcal/kg 580C)C)(1.50kcal/kg (1.00 −

°° ×=

°°⋅=

Δ=⇒=ΔLTc

MmmLTMc

vL C37° vLC001 °

C20.0°− C130°

C0°kJ8.36C)C)(20kJ/kgkg)(2.090(0.200ii1 =°°⋅=Δ= TcmQ

C0°

kJ8.66kJ/kg)kg)(334(0.200fi2 === LmQ

C0° C100°

kJ83.7kJ83.73C)C)(100kJ/kgkg)(4.186(0.200 oowi3 ==⋅=Δ= TcmQ

C100°

kJ451kJ2.451kJ/kg)kg)(2256(0.200vi4 ==== LmQ

C100° C130°


372

Total heat required .

(b)

(i)

(ii)

(iii)

(iv)

(v)

Total time

(c)

18. In 1986, a gargantuan iceberg broke away from the Ross Ice Shelf in Antarctica. It was approximately a rectangle 160 km long, 40.0 km wide, and 250 m thick. (a) What is the mass of this iceberg, given that the density of ice is ? (b) How much heat transfer (in joules) is needed to melt it? (c) How many years would it take sunlight alone to melt ice this thick, if the ice absorbs an average of , 12.00 h per day?

kJ12.9C)C)(30kJ/kgkg)(1.520(0.200vi5 =°°⋅=Δ= TcmQ

kcal148kcal9.147kJ619.054321 ===++++= QQQQQQ

PQt

tQP =⇒=

s4180kJ/s20kJ8.361

1 .PQt ===

s343kJ/s20kJ66.82

2 .PQt ===

s4.19s1854kJ/s20kJ83.73

3 ==== .PQ

t

s622kJ/s20kJ5144

4 .PQt ===

s456.0kJ/s20kJ9.125

5 ===PQt

s31s00.3154321 ==++++= tttttt

3kg/m917

2W/m100


373

Solution (a)

(b)

(c)

19. How many grams of coffee must evaporate from 350 g of coffee in a 100-‐g glass cup to cool the coffee from to ? You may assume the coffee has the same thermal properties as water and that the average heat of vaporization is 2340 kJ/kg (560 cal/g). (You may neglect the change in mass of the coffee as it cools, which will give you an answer that is slightly larger than correct.)

Solution The heat gained in evaporating the coffee equals the heat leaving the coffee and glass to lower its temperature, so that where is the mass of coffee that evaporates. Solving for the evaporated coffee gives:

20. (a) It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases of energy when burned. To illustrate this difficulty, calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water has its temperature raised from to , it boils, and the resulting steam is raised to . (b) Discuss additional complications caused by the fact that crude oil has a smaller density than water.

Solution (a)

( )kg 101.47kg 101.467

m) m)(250 10m)(40.0 10)(160kg/m (9171515

333

×=×=

××=== lwhVm ρρ

J 104.90J/kcal) 186kcal/kg)(4 kg)(79.8 10(1.467 2015f ×=×== mLQ

y 48.5d 365.25

y 1d 101.773J 102.765

J/kcal) )(4186 kcal 10(1.171

J 102.765h 1

s 3600h) m)(12 10m)(40.0 10)(160 W/m(100

416

17

day

16332day

=××=×

×==

×=⎟⎠

⎞⎜⎝

⎛××=

QQn

Q

C95.0° C45.0°

,ΔΔ ggccv TcmTcmML += M

[ ] g 33.0C)cal/g g)(0.20 (100C)cal/g g)(1.00 (350cal/g 560

C)45.0C(95.0

)(

v

ggcc

=°⋅+°⋅⋅°−°

=

+Δ=

LcmcmT

M

J1080.2 7×

C20.0° C100°C300°

ssvww TmcmLTmcQ Δ++Δ=


374

(b) Crude oil is less dense than water, so it floats on top of the water, thereby exposing it to the oxygen in the air, which it uses to burn. Also, if the water is under the oil, it is less able to absorb the heat generated by the oil.

21. The energy released from condensation in thunderstorms can be very large. Calculate the energy released into the atmosphere for a small storm of radius 1 km, assuming that 1.0 cm of rain is precipitated uniformly over this area.

Solution We have a phase change . We need to find mass of rain in a cloud of radius

1 km. . With and

, we find – about the energy released in the first atomic bomb explosion.

22. To help prevent frost damage, 4.00 kg of water is sprayed onto a fruit tree. (a) How much heat transfer occurs as the water freezes? (b) How much would the temperature of the 200-‐kg tree decrease if this amount of heat transferred from the tree? Take the specific heat to be , and assume that no phase change occurs.

Solution (a)

(b)

23. A 0.250-‐kg aluminum bowl holding 0.800 kg of soup at is placed in a freezer. What is the final temperature if 377 kJ of energy is transferred from the bowl and soup, assuming the soup’s thermal properties are the same as that of water? Explicitly show how you follow the steps in the Problem-‐Solving Strategies for the Effects of Heat Transfer.

Solution To bring the system to requires heat, , of:

L 9.67m 10

L 1kg 101.00

m 1kg 9.67

kg 9.67C)C)(200J/kg (1520J/kg 102256C)C)(80.0J/kg (4186

J 102.80

333

3

3

7ssvww

=××

×=⇒

=°°⋅+×+°°⋅

×=

Δ++Δ=

−V

TcLTcQm

vmLQ =

kg 10)m 10m)( 01.0)(kg/m 1000( 7263 ×=×== ππρVm vmLQ =

kJ/kg 2256v =L J 107 13×=Q

C0°

CkJ/kg3.35 °⋅

kcal 319kcal 319.2C)kcal/kg kg)(79.8 (4.00f ==°⋅==mLQ

C2.00C)kcal/kg kg)(0.800 (200

kcal 319.2°=

°⋅==Δ⇒Δ=

mcQTTmcQ

C25.0°

C0° Q


375

This leaves to freeze all the soup, leaving

to be removed. So, we can now determine the final temperature of the frozen soup:

24. A 0.0500-‐kg ice cube at is placed in 0.400 kg of water in a very well-‐insulated container. What is the final temperature?

Solution First bring the ice up to and melt it with heat

This lowers the temperature of water by

Now, the heat lost by the hot water equals that gained by the cold water ( is the final temperature):

25. If you pour 0.0100 kg of water onto a 1.20-‐kg block of ice (which is initially at ), what is the final temperature? You may assume that the water cools so

rapidly that effects of the surroundings are negligible.

[ ] kcal 21.34C)(25.0C)kcal/kg kg)(1.00 (0.800C)kcal/kg kg)(0.215 (0.250ssAlAl

=°°⋅+°⋅=

Δ+Δ= TcmTcmQ

kcal 68.66kcal )34.210.90( =−

kcal 4.82kcal )84.6366.68('' =−=Q

C10.6C)kcal/kg kg)(0.500 (0.800C)kcal/kg kg)(0.215 (0.250

kcal 4.82

".C0)()("

ssAlAlf

fssAlAlssAlAl

°−=°⋅+°⋅

−=

+−

=

°+=Δ+=

cmcmQT

)-T(cmcmTcmcmQ

C30.0°− C35.0°

C0° :1Q

( )[ ]kcal 4.74

kcal/kg) (79.8C)(30.0Ckcal/kg 0.500kg) (0.0500f11

=

+°°⋅=+Δ= mLTmcQ

:2TΔ

C23.15C11.85C35.0 New

C11.85C)kcal/kg kg)(1.00 (0.400

kcal 4.74

w

1221

°=°−°=

°=°⋅

==Δ⇒Δ=

TmcQTTmcQ

fT

C20.6K 293.7kg 0.450

K) kg)(273.15 (0.0500K) kg)(296.3 (0.400)()(

hc

cchhf

fhwhcfwc

°==+

=+

+=

−=−

mmTmTm

T

TTcmTTcm

C20.0°C15.0°−


376

Solution First, we need to calculate how much heat would be required to raise the temperature of the ice to :

Now, we need to calculate how much heat is given off to lower the water to :

Since this is less than the heat required to heat the ice, we need to calculate how much heat is given off to convert the water to ice:

Thus, the total amount of heat given off to turn the water to ice at :

Since , we have determined that the final state of the water/ice is ice at some temperature below . Now, we need to calculate the final temperature. We set the heat lost from the water equal to the heat gained by the ice, where we now know that the final state is ice at :

Substituting for the change in temperatures (being careful that is always positive) and simplifying gives:

Solving for the final temperature gives

and so finally,

26. Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of rock must be placed in 4.00 kg of

water to bring its temperature to , if 0.0250 kg of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings and take the average specific heat of the rocks to be that of granite.

C0°J 10762.3C)C)(15J/kg kg)(2090 (1.20 4

ice ×=°°⋅=Δ= TmcQ

C0°J 837.2C)C)(20.0J/kg kg)(4186 (0.010011 =°°⋅=Δ= TmcQ

J 103.340J/kg) 10kg)(334 (0.0100 33f2 ×=×==mLQ

C0° . J 10177.4 3

water ×=Q

waterice QQ >C0°

C0f °<T

?15iceice?0icewaterfwater020waterwater ice,by gainedby waterlost or →−→→ Δ=Δ++Δ= TcmTcmLmTcmQQ

TΔ

C)].15([)]0)((C)20([ ficeiceficefwaterwater °−−=−++° TcmTcLcm

iceicewater

iceicefwaterwaterf )(

)C15(])C20([cmm

cmLcmT+

°−+°=

C13.2C)J/kg kg)(2090 1.20kg (0.0100

C)C)(15J/kg kg)(2090 (1.20C)J/kg kg)(2090 1.20kg (0.0100

J/kg] 10334C)C)(20J/kg kg)[(4186 (0.0100 3

f

°−=

°⋅+°°⋅

−

°⋅+×+°°⋅

=T

C500°C15.0° C100°


377

Solution Let the subscripts r, e, v, and w represent rock, equilibrium, vapor, and water, respectively.

27. What would be the final temperature of the pan and water in Calculating the Final Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a Hot Pan if 0.260 kg of water was placed in the pan and 0.0100 kg of the water evaporated immediately, leaving the remainder to come to a common temperature with the pan?

Solution Let the subscripts Al, e, v, and w represent aluminum pan, equilibrium, vapor, and water, respectively.

28. In some countries, liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A 3.00-‐hour delivery trip requires 200 L of liquid nitrogen, which has a density of . (a) Calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to . (Use and assume it is constant over the temperature range.) This value is the amount of cooling the liquid nitrogen supplies. (b) What is this heat transfer rate in kilowatt-‐hours? (c) Compare the amount of cooling obtained from melting an identical mass of ice with that from evaporating the liquid nitrogen.

kg 4.38C)100CC)(500J/kg (840

C)15CC)(100J/kg kg)(4186 (3.975J/kg) 10kg)(2256 (0.0250

)()()()(

3e1r

2ewwvvr

2ewwvve1rr

=

°−°°⋅°−°°⋅+×

=

−

−+=

−+=−

TTcTTcmLmm

TTcmLmTTcm

⇒−+=− )()( 2ewwvve1AlAl TTcmLmTTcm

C44.0

C)J/kg kg)(900 (0.500C)J/kg kg)(4186 (0.250J/kg) 10kg)(2256 (0.0100

C)C)(20.0J/kg kg)(4186 (0.250C)C)(150J/kg kg)(900 (0.5003

e

AlAlww

vv2ww1AlAle

°=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

°⋅+°⋅

×−

°°⋅+°°⋅

=

+

−+=

T

cmcmLmTcmTcm

T

3kg/m808C3.00° pc

C0°


378

Solution (a)

(b)

(c)

29. Some gun fanciers make their own bullets, which involves melting and casting the lead slugs. How much heat transfer is needed to raise the temperature and melt 0.500 kg of lead, starting from ?

Solution

14.5 CONDUCTION

30. (a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and that have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors. The surface area of the walls is and their inside surface is at , while their outside surface is at . (b) How many 1-‐kW room heaters would be needed to balance the heat transfer due to conduction?

Solution (a)

(b) 1 one-‐kilowatt room heater is needed.

31. The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly the windows transfer heat by conduction, calculate the rate of conduction in watts through a window that is

thick (1/4 in) if the temperatures of the inner and outer surfaces are and , respectively. This rapid rate will not be maintained—the inner

surface will cool, and even result in frost formation.

( )

[ ]kcal 101.57

C)195.8(C3.00C)kcal/kg (0.248kcal/kg 48.0

)kg/m )(808m 10(200

4

333

pvpv

×=

⎭⎬⎫

⎩⎨⎧

°−−°

⋅°⋅+×=

Δ+=Δ+=

−Q

TcLmTmcmLQ

hkW 18.3hkW 18.28J 103.60

hkW 1J/kcal) kcal)(4186 10(1.572 64 ⋅=⋅=⎟

⎠

⎞⎜⎝

⎛×

⋅×

kcal 101.29kcal 12,895kcal/kg) kg)(79.8 (161.6 4fice ×==== mLQ

C25.0°

( )( )( )[ ] kcal 7.53kcal/kg 5.85C25.0C327Ckcal/kg 0.0305kg) (0.500

ff

=+°−°°⋅⋅=

+Δ=+Δ=

QLTcmmLTmcQ

2m120C18.0° C5.00°

( ) [ ]

W101.01 W101.008m 0.130

C)5.00C)(18.0m (120C)mJ/s 2(0.042

33

212

×=×=

°−°°⋅⋅=

−=

dTTkA

tQ

2m-3.00cm0.635C5.00° C10.0°−


379

Solution

32. Calculate the rate of heat conduction out of the human body, assuming that the core internal temperature is , the skin temperature is , the thickness of the tissues between averages , and the surface area is .

Solution

33. Suppose you stand with one foot on ceramic flooring and one foot on a wool carpet, making contact over an area of with each foot. Both the ceramic and the carpet are 2.00 cm thick and are on their bottom sides. At what rate must heat transfer occur from each foot to keep the top of the ceramic and carpet at

?

Solution

For the wool carpet:

For the ceramic tile:

34. A man consumes 3000 kcal of food in one day, converting most of it to maintain body temperature. If he loses half this energy by evaporating water (through breathing and sweating), how many kilograms of water evaporate?

Solution

( )

[ ] W106.0 W5953m 100.635

C)10.0(C5.00)m C)(3.00mJ/s (0.84 32-

2

12

×==×

°−−°°⋅⋅=

−=

dTTkA

tQ

C37.0° C34.0°cm1.00 2m40.1

( ) W0.48m .01000

C)34.0C)(37.0m C)(1.40mJ/s (0.2 212 =

°−°°⋅⋅=

−=

dTTkA

tQ

2cm80.0C10.0°

C33.0°

( )dTTkA

tQ 12 −=

W0.368m .02000

C)10.0-C)(33.0m 10C)(80.0mJ/s (0.04 2-4w =

°°×°⋅⋅=

tQ

W73.7m .02000

C)32)(m 10C)(80.0mJ/s (0.84 2-4c =

°×°⋅⋅=

tQ

kg 2.59kcal/kg 580

kcal 1500

C)v(37C)v(37 ===⇒=

°° L

QmmLQ


380

35. (a) A firewalker runs across a bed of hot coals without sustaining burns. Calculate the heat transferred by conduction into the sole of one foot of a firewalker given that the bottom of the foot is a 3.00-‐mm-‐thick callus with a conductivity at the low end of the range for wood and its density is . The area of contact is , the temperature of the coals is , and the time in contact is 1.00 s. (b) What temperature increase is produced in the of tissue affected? (c) What effect do you think this will have on the tissue, keeping in mind that a callus is made of dead cells?

Solution (a)

(b) Taking the density of the callus to be , the change in temperature can be found from:

(c) At a temperature change of , the heat probably won’t do much damage, since a callus is made of dead cells.

36. (a) What is the rate of heat conduction through the 3.00-‐cm-‐thick fur of a large animal having a surface area? Assume that the animal’s skin temperature is

, that the air temperature is , and that fur has the same thermal conductivity as air. (b) What food intake will the animal need in one day to replace this heat transfer?

Solution (a)

(b)

3kg/m300 2cm25.0C700°

3cm25.0

( )

J 44.2m .003000

C)(1.00s)37.0-C)(700m 10C)(25.0mJ/s (0.0800 24-

12

=°°×°⋅⋅

=

−=

dtTTkAQ

3kg/m 300=ρ

C1.68C)J/kg )(3500m 10)(25.0kg/m (300

J 41.7363 °=

°⋅×==Δ

⇒Δ=Δ=

−VcQT

TVcTmcQ

ρ

ρ

C2°<

2m-1.40C32.0° C5.00°−

( ) W39.7 W39.71

m .03000C))(37.0m C)(1.40mJ/s (0.023 2

12 ==°°⋅⋅

=−

=dTTkA

tQ

kcal 028kcal 819.7J 4186

kcal 1s) 10J/s)(8.64 (39.71 4 ==⎟⎠

⎞⎜⎝

⎛×== PtW


381

37. A walrus transfers energy by conduction through its blubber at the rate of 150 W when immersed in water. The walrus’s internal core temperature is , and it has a surface area of . What is the average thickness of its blubber, which has the conductivity of fatty tissues without blood?

Solution

38. Compare the rate of heat conduction through a 13.0-‐cm-‐thick wall that has an area of and a thermal conductivity twice that of glass wool with the rate of heat

conduction through a window that is 0.750 cm thick and that has an area of , assuming the same temperature difference across each.

Solution

, so that

39. Suppose a person is covered head to foot by wool clothing with average thickness of 2.00 cm and is transferring energy by conduction through the clothing at the rate of 50.0 W. What is the temperature difference across the clothing, given the surface area is ?

Solution

40. Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.600 cm thick and heat conduction occurs through the same area and at the same rate as computed in Example 14.6, what is the temperature difference across it? Ceramic has the same thermal conductivity as glass and brick.

C1.00°− C37.0°2m2.00

( )

( )cm 10.1m 0.101

W150C))(38.0m C)(2.00mJ/s (0.2

/

212

12

==°°⋅⋅

=−

=

⇒−

=

tQTTkAd

dTTkA

tQ

2m10.02m2.00

dTTkA

tQ )( 12 −=

wall: window1 : 35or window,: wall0288.0m) 10)(13.0m C)(2.00mJ/s (0.84

m) 10)(0.750m C)(10.0mJ/s 0.042(2)/()/(

22

22

wallwindowwindow

windowwallwall

window

wall

=

×°⋅⋅

×°⋅⋅×==

−

−

dAkdAk

tQtQ

2m1.40

C17.9C17.86)m C)(1.40mJ/s (0.04

W)m)(50.0 10(2.00)/(

)(

2

2-

12

°=°=°⋅⋅

×==Δ

⇒Δ

=−

=

kAtQdT

dTkA

dTTkA

tQ


382

Solution

∆𝑇 = !(!/!)!"

= (!.!!×!"!!m)(!!"# W)(!.!" J/s∙m∙℃)(!.!"×!"!!m!)

= 1046℃ = 1.32×10! K

41. One easy way to reduce heating (and cooling) costs is to add extra insulation in the attic of a house. Suppose the house already had 15 cm of fiberglass insulation in the attic and in all the exterior surfaces. If you added an extra 8.0 cm of fiberglass to the attic, then by what percentage would the heating cost of the house drop? Take the single story house to be of dimensions 10 m by 15 m by 3.0 m. Ignore air infiltration and heat loss through windows and doors.

Solution The original heat loss by conduction is given by . We need to consider all

6 sides that contribute to the heat loss. We will put the loss through the attic in a separate part.

If we add 8 cm to the attic, the new addition is

So the percentage of savings in heat transfer = .

42. (a) Calculate the rate of heat conduction through a double-‐paned window that has a area and is made of two panes of 0.800-‐cm-‐thick glass separated by a 1.00-‐

cm air gap. The inside surface temperature is , while that on the outside is . (Hint: There are identical temperature drops across the two glass panes.

First find these and then the temperature drop across the air gap. This problem ignores the increased heat transfer in the air gap due to convection.) (b) Calculate the rate of heat conduction through a 1.60-‐cm-‐thick window of the same area and with the same temperatures. Compare your answer with that for part (a).

K101.05C 1046)m 10C)(1.54mJ/s (0.84

W)m)(2256 10(6.00)/(

)(

322-

3-

12

×=°=×°⋅⋅

×==Δ

⇒Δ

=−

=

kAtQdT

dTkA

dTTkA

tQ

dTkA

tQ Δ=

[ ]

TC)J/s/ (126 TC)J/s/ 42 CJ/s/ 84(m 0.15

Tm) 15m 10C(J/s/m/042.0m 0.15

Tm) 10m (152)m 3m 15()2m 3m (10C)J/s/m/ 042.0(

oo

o

o

Δ=Δ+°=

Δ×+

Δ⋅+×⋅+×⋅=

tQ

( ) TTT

Δ=

Δ+=Δ×+

CJ/s/ 111C)J/s/ 27CJ/s/ (84m] )/0.23m 150CJ/s/m/ (0.042CJ/s/ 84[

o

oo2oo

%12126/)111126( =−

2m-50.1C15.0°

C10.0°−


383

Solution

In equilibrium, the heat flows across each “slab” are equal.

(a)

Adding two equations, we obtain:

Now, we have

Q 2

Q 3Q 1

TL T1 T2 TR

d1 d1d2

Q

2RL1321 and TTTTtQ

tQ

tQ

−=−==

21R11222

1222L111

31

2R12R312232

122L1121

13

33

2

22

1

11

) since()()()(unit timeper

)()(unit timeper

,,

TaTaTaTaTaTaTaTa

aaTTAaTTAaTTAaQQ

TTAaTTAaQQ

adK

adKa

dKa

−=−

−=−

=

−=−=−⇒=

−=−⇒=

====

12

LR21122

2

LR12

112122211LR1

2)()(

)(2

)(2)()(

aaTTAaaTTAa

tQ

TTaa

aTTTTaTTaTTa

+

−=−=

−⎟⎟⎠

⎞⎜⎜⎝

⎛

+=−⇒−=−+−

Thus . C10.0;C15.0

CmJ/s 2.3m 101.00

CmJ/s 0.023

CmJ/s 105m 100.800CmJ/s 0.84

LR

22-

2

22

22-

1

11

°−=°=

°⋅⋅=×

°⋅⋅==

°⋅⋅=×

°⋅⋅==

TTdKa

dKa


384

Since

(b)

The single-‐pane window has a rate of heat conduction equal to 1969/83, or 24 times that of a double pane window.

43. Many decisions are made on the basis of the payback period: the time it will take through savings to equal the capital cost of an investment. Acceptable payback times depend upon the business or philosophy one has. (For some industries, a payback period is as small as two years.) Suppose you wish to install the extra insulation in Problem 14.41. If energy cost $1.00 per million joules and the insulation was $4.00 per square meter, then calculate the simple payback time. Take the average for the 120 day heating season to be .

Solution We found in Problem 14.41 that as baseline energy use. So the

total heat loss during this period is .

At the cost of $1/MJ, the cost is $1960. From Problem 14.41, the savings is 12% or . We need of insulation in the attic. At this is a $600 cost. So

the payback period is .

44. For the human body, what is the rate of heat transfer by conduction through the body’s tissue with the following conditions: the tissue thickness is 3.00 cm, the change in temperature is , and the skin area is . How does this compare with the average heat transfer rate to the body resulting from an energy intake of about 2400 kcal per day? (No exercise is included.)

Solution The rate of heat transfer by conduction is

( )( ) ( )

W82.6CmJ/s 105C)mJ/s 2(2.3

C25.0)m (1.50CmJ/s 2.3CmJ/s 1052

)(22

222

12

LR212

=

°⋅⋅+°⋅⋅

°°⋅⋅°⋅⋅=

+

−=

aaTTAaa

tQ

W83 W6.82312 ==⇒===tQ

tQ

tQ

tQ

tQ

W1097.1 W9691m 101.60

C))(25.0m C)(1.50mJ/s (0.84)( 32-

2LR ×==

×

°°⋅⋅=

−=

dTTkA

tQ

TΔC15.0°

CJ/s 126 °⋅Δ= TtQ

J 101960s/day) 1086.4(days) 120)(C0.15(C)J/s126( 63 ×=×°°⋅=Q

y /235$ 2m 150 2m/4$cost)labor (excluding years 2.6 year /235$/600$ =

C2.00° 2m1.50


385

On a daily basis, this is 1,728 kJ/day.

Daily food intake is

So only of energy intake goes as heat transfer by conduction to the environment at this .

14.6 CONVECTION

45. At what wind speed does air cause the same chill factor as still air at ?

Solution 10 m/s (from Table 14.4)

46. At what temperature does still air cause the same chill factor as air moving at 15 m/s?

Solution (from Table 14.4)

47. The “steam” above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot coffee. What is the final temperature of 250 g of hot coffee initially at if 2.00 g evaporates from it? The coffee is in a Styrofoam cup, so other methods of heat transfer can be neglected.

Solution Let be the mass of coffee that is left after evaporation and be the mass of coffee that evaporates.

48. (a) How many kilograms of water must evaporate from a 60.0-‐kg woman to lower her body temperature by ? (b) Is this a reasonable amount of water to evaporate in the form of perspiration, assuming the relative humidity of the surrounding air is low?

( )( ) . W0.20m 1000.3

C 00.2m 50.1C)mJ/s2.0(2

2o

=×

°⋅⋅=

Δ=

−dTkA

tQ

kJ/day. 10,050J/kcal4186kcal/d 2400 =×

%2.17TΔ

C10°− C29°−

C5°−

C26°−

90.0°C

M m

C85.7C4.35C90.0

C4.35C)kcal/kg g)(1.00 (248

kcal/kg) g)(539 (2.00

if

vv

°=°−°=Δ−=

°=°⋅

==Δ⇒=Δ

TTTMcmLTmLTMc

0.750°C


386

Solution (a) is the mass of the woman and is the mass of water that evaporates:

(b) Yes, 64.4 g of water is reasonable. If the air is very dry, the sweat may evaporate without even being noticed.

49. On a hot dry day, evaporation from a lake has just enough heat transfer to balance the of incoming heat from the Sun. What mass of water evaporates in 1.00 h from each square meter? Explicitly show how you follow the steps in the Problem-‐Solving Strategies for the Effects of Heat Transfer.

Solution

(Note that we can use the value at as a closer approximation of the temperature on a hot day than .)

50. One winter day, the climate control system of a large university classroom building malfunctions. As a result, of excess cold air is brought in each minute. At what rate in kilowatts must heat transfer occur to warm this air by (that is, to bring the air to room temperature)?

Solution

51. The Kilauea volcano in Hawaii is the world’s most active, disgorging about of lava per day. What is the rate of heat transfer out of Earth by convection if this lava has a density of and eventually cools to ? Assume that the specific heat of lava is the same as that of granite.

M m

kg 106.44kcal/kg 580

C)C)(0.750kcal/kg kg)(0.83 (60.0 2

C)v(37

C)v(37

−

°

°

×=°°⋅

=Δ

=

⇒=Δ

LTMcm

mLTMc

2kW/m1.00

kg 1.48J/kg 102430

s) W)(360010(1.00so , W 101.00kW 1.00,m 1

3

3

C)v(37C)v(37

32

=×

×==⇒==

×==⇒=

°° L

PtmmLQPt

PA

vL C37°C100°

3m50010.0°C

( )( )( )( )

.kW 77.5 W107.75s 60.0

C10.0CJ/kg 721m 500kg/m 1.29

4

33

=×=

°°⋅=

Δ=

Δ==

tTVc

tTmc

tQP ρ

35 m10 5×1200°C

3kg/m2700 30°C


387

Solution

For c, we use the specific heat of granite, which is formerly molten rock.

52. During heavy exercise, the body pumps 2.00 L of blood per minute to the surface, where it is cooled by . What is the rate of heat transfer from this forced convection alone, assuming blood has the same specific heat as water and its density is ?

Solution

53. A person inhales and exhales 2.00 L of air, evaporating of water from the lungs and breathing passages with each breath. (a) How much heat transfer occurs due to evaporation in each breath? (b) What is the rate of heat transfer in watts if the person is breathing at a moderate rate of 18.0 breaths per minute? (c) If the inhaled air had a temperature of , what is the rate of heat transfer for warming the air? (d) Discuss the total rate of heat transfer as it relates to typical metabolic rates. Will this breathing be a major form of heat transfer for this person?

Solution (a)

(b)

(c)

tTVc

tTmc

tQ Δ

=Δ

=ρ

( )( )( )( )

MW 102 W102 W101.54s 108.64

C1701CJ/kg 840m 105kg/m 2700

41010

4

353

×=×=×=

×

°°⋅×=

tQ

2.00°C

3kg/m1050

( )( )( )( ) W293s 06

C00.2CJ/kg 4186m 102.00kg/m 1050 33-3

=°°⋅×

=

Δ=

Δ=

tTVc

tTmc

tQ ρ

37.0°C g1000.4 2−×

C20.0°

J 97.2J/kg) 10kg)(2430 10(4.00 3-5C)v(37 =××== °mLQ

W29.2s 60.0J) 18(97.2===

tNQP

W49.9s 60.0J) 18(31.6

:is lossheat of rate theso ,J/breath 31.6C)20.0CC)(37.0J/kg )(721m 10)(2.00kg/m (1.29 333

===

=

°−°°⋅×=

Δ=Δ=−

tNQP

TVcTmcQ ρ


388

(d) The total rate of heat loss would be 29.2 W + 9.49 W = 38.7 W. While sleeping, our body consumes 83 W of power, while sitting it ranges 120-‐210 W. Therefore, the total rate of heat loss from breathing will not be a major form of heat loss for this person.

54. A glass coffee pot has a circular bottom with a 9.00-‐cm diameter in contact with a heating element that keeps the coffee warm with a continuous heat transfer rate of 50.0 W. (a) What is the temperature of the bottom of the pot, if it is 3.00 mm thick and the inside temperature is ? (b) If the temperature of the coffee remains constant and all of the heat transfer is removed by evaporation, how many grams per minute evaporate? Take the heat of vaporization to be 2340 kJ/kg.

Solution (a)

(b)

14.7 RADIATION

55. At what net rate does heat radiate from a black roof on a night when the roof’s temperature is and the surrounding temperature is ? The emissivity of the roof is 0.900.

Solution

Note that the negative answer implies heat loss to the surroundings.

56. (a) Cherry-‐red embers in a fireplace are at and have an exposed area of and an emissivity of 0.980. The surrounding room has a temperature of

. If 50% of the radiant energy enters the room, what is the net rate of radiant heat transfer in kilowatts? (b) Does your answer support the contention that most of the heat transfer into a room by a fireplace comes from infrared radiation?

60.0°C

C88C88.07

C28.07m) 104.50()CmJ/s 48.0(

)m 10 W)(3.00.050()/(

i

22

3

2

°=°=⇒Δ+=

°=×°⋅⋅

×==Δ⇒

Δ=

−

−

TTTTrkdtQT

dTkA

tQ

ππ

g 1.282340J/g

s) W)(60.0(50.0

vv ===⇒=

LPtmmLPt

2m-275C0.30 ° C0.15 °

( ) ( )[ ]kW 7.12

K 303K 288)m 75)(0.900)(2KmJ/s 10(5.67)( 44242841

42

−=

−⋅⋅×=−= −TTeAtQ

σ

C850°2m200.0

C0.18 °


389

Solution (a)

Note that the negative answer implies heat loss to the surroundings.

(b) This answer is quite large, so it does indeed suggest that the heat put into a room by a fireplace comes mainly from infrared radiation (which is hotter than red embers).

57. Radiation makes it impossible to stand close to a hot lava flow. Calculate the rate of heat transfer by radiation from of fresh lava into surroundings, assuming lava’s emissivity is 1.00.

Solution

58. (a) Calculate the rate of heat transfer by radiation from a car radiator at into a environment, if the radiator has an emissivity of 0.750 and a

surface area. (b) Is this a significant fraction of the heat transfer by an automobile engine? To answer this, assume a horsepower of 200 hp (1.5 kW) and the efficiency of automobile engines as .

Solution (a)

(b) Assuming an automobile engine is 200 horsepower and the efficiency of a gasoline

engine is 25%, the engine consumes

Therefore, 600 horsepower is lost due to heating. The radiator transfers

from radiation, which is not a significant fraction

because the heat is primarily transferred from the radiator by other means.

( ) ( )[ ]kW 80.8

K 1123K 291)m .200)(0.980)(0KmJ/s 10(5.6721

)(21

442428

41

42

−=

−⋅⋅×=

−=

−

TTeAtQ

σ

2m00.1 C1200° C0.30 °

( ) ( )[ ]kW 266

K 1473K 303)m 00)(1.00)(1.KmJ/s 10(5.67)( 44242841

42

−=

−⋅⋅×=−= −TTeAtQ

σ

110°C 50.0°C 2m-20.1

25%

( ) ( )[ ] W543K 383K 323)m .20)(0.750)(1KmJ/s 10(5.67

)(

442428

41

42

−=−⋅⋅×=

−=

−

TTeAtQ

σ

horsepower 00825%

horsepower 200=

hp 0.728 W746

hp 1 W543 =×


390

59. Find the net rate of heat transfer by radiation from a skier standing in the shade, given the following. She is completely clothed in white (head to foot, including a ski mask), the clothes have an emissivity of 0.200 and a surface temperature of , the surroundings are at , and her surface area is .

Solution

60. Suppose you walk into a sauna that has an ambient temperature of . (a) Calculate the rate of heat transfer to you by radiation given your skin temperature is

, the emissivity of skin is 0.98, and the surface area of your body is . (b) If all other forms of heat transfer are balanced (the net heat transfer is zero), at what rate will your body temperature increase if your mass is 75.0 kg?

Solution (a)

(b)

61. Thermography is a technique for measuring radiant heat and detecting variations in surface temperatures that may be medically, environmentally, or militarily meaningful.(a) What is the percent increase in the rate of heat transfer by radiation from a given area at a temperature of compared with that at , such as on a person’s skin? (b) What is the percent increase in the rate of heat transfer by radiation from a given area at a temperature of compared with that at

, such as for warm and cool automobile hoods?

Solution (a)

(b)

C0.10 °15.0°C− 2m1.60

( ) ( )[ ] W0.63K 283K 258)m .60)(0.200)(1KmJ/s 10(5.67

)(

442428

41

42

−=−⋅⋅×=

−=

−

TTeAtQ

σ

50.0°C

37.0°C 2m1.50

( ) ( )[ ] W371K 310K 323)m 50)(0.98)(1.KmJ/s 10(5.67

)(

442428

41

42

=−⋅⋅×=

−=

−

TTeAtQ

σ

C/min 0.0314C/s 105.24C)J/kg kg)(3500 (75.0

W)(1371 4 °=°×=°⋅

==Δ

⇒Δ=

−

mctQ

tT

TmcQ

34.0°C 33.0°C

34.0°C20.0°C

1.31%100%1K 306K 307 4

=×⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟

⎠

⎞⎜⎝

⎛

20.5%100%1K 293K 307 4

=×⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟

⎠

⎞⎜⎝

⎛


391

62. The Sun radiates like a perfect black body with an emissivity of exactly 1. (a) Calculate the surface temperature of the Sun, given that it is a sphere with a radius that radiates into 3-‐K space. (b) How much power does the Sun radiate per square meter of its surface? (c) How much power in watts per square meter is that value at the distance of Earth, away? (This number is called the solar constant.)

Solution (a)

(b)

(c) Let be the radius of a sphere with the sun at the center and the earth at a point on the surface of the sphere.

63. A large body of lava from a volcano has stopped flowing and is slowly cooling. The interior of the lava is at , its surface is at , and the surroundings are at

. (a) Calculate the rate at which energy is transferred by radiation from of surface lava into the surroundings, assuming the emissivity is 1.00. (b)

Suppose heat conduction to the surface occurs at the same rate. What is the thickness of the lava between the surface and the interior, assuming that the lava’s conductivity is the same as that of brick?

Solution (a)

Note the negative answer implies heat lost to the surroundings.

m-107.00 8×W103.80 26×

m101.50 11×

K105.74m) 10)(7.00)(4KmJ/s 10(5.67

W103.80

)4 is sphere a of area surface (the4

31/4

2848

26

241

24

×=⎥⎦

⎤⎢⎣

⎡

×π⋅⋅×

×=

π⎥⎦

⎤⎢⎣

⎡

πσ=⇒σ=

−T

r)r(e

t/QTeATtQ

/

2728

26

2 W/m106.17m) 10(7.004

W103.804

×=×

×==

ππrP

AP

r

23211

26

2 W/m1034.1m) 10(1.504

W103.804

×=×

×=

ππrP

1200°C 450°C27.0°C

2m 1.00

450°C 1200°C

( ) ( )[ ]kW 15.0 W101.50 W101.503

K 723K 300)m 00)(1.00)(1.KmJ/s 10(5.67

)(

44

442428

41

42

−=×−=×−=

−⋅⋅×=

−=

−

tQ

TTeAtQ

σ


392

(b)

64. Calculate the temperature the entire sky would have to be in order to transfer energy by radiation at —about the rate at which the Sun radiates when it is directly overhead on a clear day. This value is the effective temperature of the sky, a kind of average that takes account of the fact that the Sun occupies only a small part of the sky but is much hotter than the rest. Assume that the body receiving the energy has a temperature of .

Solution

65. (a) A shirtless rider under a circus tent feels the heat radiating from the sunlit portion of the tent. Calculate the temperature of the tent canvas based on the following information: The shirtless rider’s skin temperature is and has an emissivity of 0.970. The exposed area of skin is . He receives radiation at the rate of 20.0 W—half what you would calculate if the entire region behind him was hot. The rest of the surroundings are at . (b) Discuss how this situation would change if the sunlit side of the tent was nearly pure white and if the rider was covered by a white tunic.

Solution (a)

cm 4.2m 104.19 W101.503

C))(750m C)(1.00mJ/s (0.84/

24

2

=×=×

°°⋅⋅=

Δ=

⇒Δ

=

−

tQTkAd

dTkA

tQ

2W/m1000

27.0°C

K 401)(1.00)KmJ/s 10(5.67

W/m1000K) (300

W/m1000/

/)(

1/4

428

24

1

2

4/14

214

14

2

=⎥⎦

⎤⎢⎣

⎡

⋅⋅×

−−=∴

−=

⎟⎠

⎞⎜⎝

⎛ −=⇒−=

−T

AtQ

eAtQTTTTeA

tQ

σσ

34.0°C2m0.400

34.0°C

C48.5K 321.63

K) (307)m .400)(0.970)(0KmJ/s 10(5.67

W)2(20.0

)/(2)2/(

/

, )(2

1/44

2428

4/14

1

4/14

12

41

42

°==

⎥⎦

⎤⎢⎣

⎡+

⋅⋅×=

⎥⎦

⎤⎢⎣

⎡ +=⎥⎦

⎤⎢⎣

⎡+=

−⎟⎠

⎞⎜⎝

⎛=

−

TeAtQT

AetQT

TTAetQ

σσ

σ


393

(b) A pure white object reflects more of the radiant energy that hits it, so the white tent would prevent more of the sunlight from heating up the inside of the tent, and the white tunic would prevent that radiant energy inside the tent from heating the rider. Therefore, with a white tent, the temperature would be lower than , and the rate of radiant heat transferred to the rider would be less than 20.0 W.

66. Integrated Concepts One day the relative humidity is , and that evening the temperature drops to , well below the dew point. (a) How many grams of water condense from each cubic meter of air? (b) How much heat transfer occurs by this condensation? (c) What temperature increase could this cause in dry air?

Solution (a) Let be the vapor density during the day. Percent relative humidity is equal to the vapor density divided by the saturation vapor density. Using the values for relative humidity and saturation vapor density, we have

(b)

(c)

67. Integrated Concepts Large meteors sometimes strike the Earth, converting most of their kinetic energy into thermal energy. (a) What is the kinetic energy of a meteor moving at 25.0 km/s? (b) If this meteor lands in a deep ocean and of its kinetic energy goes into heating water, how many kilograms of water could it raise by

(c) Discuss how the energy of the meteor is more likely to be deposited in the ocean and the likely effects of that energy.

Solution (a)

(b)

C48.5°

30.0°C 75.0%20.0°C

x

333

33

g/m 5.60g/m 17.2g/m 22.8

g/m 22.8750.0g/m 0.43

=−

=⇒= xx

kcal 3.02kcal/kg) kg)(539 10(5.60 3v =×== −mLQ

C6.13C)kcal/kg kg)(0.172 (1.29

kcal 3.02°=

°⋅==Δ⇒Δ=

mcQTTmcQ

kg109

80%

5.0°C?

J 103J 103.125m/s) 10kg)(25.0 0.5(1021 17172392 ×≈×=×== mvKE

kg 101kg 101.19C)C)(5.0J/kg (4186J) 1025(0.80)(3.1 1313

17

×≈×=°°⋅

×=

Δ=⇒Δ=

TcQmTmcQ


394

(c) When a large meteor hits the ocean, it causes great tidal waves, dissipating a large amount of its energy in the form of kinetic energy of the water.

68. Integrated Concepts Frozen waste from airplane toilets has sometimes been accidentally ejected at high altitude. Ordinarily it breaks up and disperses over a large area, but sometimes it holds together and strikes the ground. Calculate the mass of

ice that can be melted by the conversion of kinetic and gravitational potential energy when a piece of frozen waste is released at 12.0 km altitude while moving at 250 m/s and strikes the ground at 100 m/s (since less than 20.0 kg melts, a significant mess results).

Solution Let be the mass of the ice block and be the mass that melts before hitting the ground.

69. Integrated Concepts (a) A large electrical power facility produces 1600 MW of “waste heat,” which is dissipated to the environment in cooling towers by warming air flowing through the towers by . What is the necessary flow rate of air in ? (b) Is your result consistent with the large cooling towers used by many large electrical power plants?

Solution (a)

(b) This is equivalent to 12 million cubic feet of air per second. That is tremendous. This is too large to be dissipated by heating the air by only . Many of these cooling towers use the circulation of cooler air over warmer water to increase the rate of evaporation. This would allow there to be much smaller amounts of air

0°Ckg20.0

M m

2ff

2i

fi

21

21 MvmLMghMv

KEQPEKE

+=+

+=+

[ ]

[ ] kg 8.61J/kg 10334

m/s) 0.5(100m/s) 0.5(250m) 10)(12.0m/s (9.80kg) (20.0

)(5.0

3

2232f

2f

2i

=×

−+×=

−+=

LvvghMm

5.00°C /sm3

/sm 103.44/sm 103.44kg/m 1.29

kg/s 104.438

kg/s 104.438C)C)(5.00J/kg (721

J/s 101600

35353

5

56

×==⇒×=×

==

×=°°⋅

×=

Δ=⇒Δ=

tVFmV

TcQmTmcQ

ρ

C5°


395

necessary to remove such a large amount of heat, because evaporation removes larger quantities of heat than was considered in part (a).

70. Integrated Concepts (a) Suppose you start a workout on a Stairmaster, producing power at the same rate as climbing 116 stairs per minute. Assuming your mass is 76.0 kg and your efficiency is , how long will it take for your body temperature to rise if all other forms of heat transfer in and out of your body are balanced? (b) Is this consistent with your experience in getting warm while exercising?

Solution (a) You produce power at a rate of 685 W, and since you are 20% efficient, you must

have generated: .

If only 685 W of power was useful, the power available to heat the body is .

Now, so that

(b) This says that it takes about a minute and a half to generate enough heat to raise the temperature of your body by , which seems quite reasonable. Generally, within five minutes of working out on a Stairmaster, you definitely feel warm and probably are sweating to keep your body from overheating.

71. Integrated Concepts A 76.0-‐kg person suffering from hypothermia comes indoors and shivers vigorously. How long does it take the heat transfer to increase the person’s body temperature by if all other forms of heat transfer are balanced?

Solution

20.0%C1.00º

W34250.20

W685efficiency

producedgenerated ===

PP

W102.740 W 685 W3425 3wasted ×=−=P

, wasted tTmc

tQP Δ==

s 97.1 W102.74

C)C)(1.00J/kg kg)(3500 (76.03

wasted

=×

°°⋅=

Δ=P

Tmct

C1.00°

C2.00º

min 20.9s 101.25 W425

C)C)(2.00J/kg kg)(3500 (76.0 3 =×=°°⋅

=Δ

=

⇒Δ==

PTmct

TmcPtQ


396

72. Integrated Concepts In certain large geographic regions, the underlying rock is hot. Wells can be drilled and water circulated through the rock for heat transfer for the generation of electricity. (a) Calculate the heat transfer that can be extracted by cooling of granite by . (b) How long will it take for heat transfer at the rate of 300 MW, assuming no heat transfers back into the of rock by its surroundings?

Solution (a)

(b)

73. Integrated Concepts Heat transfers from your lungs and breathing passages by evaporating water. (a) Calculate the maximum number of grams of water that can be evaporated when you inhale 1.50 L of air with an original relative humidity of 40.0%. (Assume that body temperature is also .) (b) How many joules of energy are required to evaporate this amount? (c) What is the rate of heat transfer in watts from this method, if you breathe at a normal resting rate of 10.0 breaths per minute?

Solution (a) To solve this, we calculate the mass of water initially in the breath and subtract this value from the mass of the water in an exhaled breath at 100% humidity. Using the saturation vapor density of water at ,

(b)

(Note that for water at is used here as a better approximation than for water at )

(c)

3km1.00 100°C3km1.00

J 102.27C)C)(100J/kg (840m) 10)(1.00kg/m (2700 17333 ×=°°⋅×=

Δρ=Δ= TVcTmcQ

y 24.0s 107.57 W10300

J 102.27 86

17

=×=×

×==⇒=

PQtQPt

37.0°C37.0°C

C37°

( )( )

( ) ( ) g 103.96g 102.64g 106.60

g 106.60)g/m (44.0m 101.50

g 102.64)(0.400)g/m (44.0m 101.50

222inex

2333ex

2333in

−−−

−−

−−

×=×−×=−=Δ

×=×=

×=×=

mmmmm

J 96.2J 96.23J/g) g)(2430 10(3.96 2C)v(37 ==×== −°mLQ

vL C37° vLC.001 °

W16.0s 60.0

J) 10(96.23===

tNQP


397

74. Integrated Concepts (a) What is the temperature increase of water falling 55.0 m over Niagara Falls? (b) What fraction must evaporate to keep the temperature constant?

Solution (a)

(b) Let be the mass of water that evaporates.

(Note that for water at is used here as a better approximation than for water at )

75. Integrated Concepts Hot air rises because it has expanded. It then displaces a greater volume of cold air, which increases the buoyant force on it. (a) Calculate the ratio of the buoyant force to the weight of air surrounded by air. (b) What energy is needed to cause of air to go from to (c) What gravitational potential energy is gained by this volume of air if it rises 1.00 m? Will this cause a significant cooling of the air?

Solution (a) The density of a given volume of air will be proportional to .

The buoyant force is equal to the weight of the displaced cold air (Archimedes’ principle.) Thus,

(b)

(c)

This will not cause a significant cooling of the air because it is much less than the

C0.117CJ/kg 4186

m) )(50.0m/s (9.80 2

°=°⋅

==Δ⇒Δ=cghTTmcmgh

M4

3

2

vv 1017.2

CJ/kg 102430m) )(50.0m/s (9.80 −×=°⋅×

==⇒=Lgh

mMMLmgh

vL C37° vLC.001 °

50.0°C 20.0°C3m1.00 20.0°C 50.0°C?

⇒= nRTPVT1

hh

1~T

ρc

c1~T

ρ

1.102K 293K 323

and

c

h

h

c

h

B

hhccB

====

===

TT

VgVg

wF

VgwVgwF

ρρ

ρρ

J 102.79C)C)(30.0J/kg )(721m )(1.00kg/m (1.29 433

pp

×=°°⋅=

Δ=Δ= TVcTmcQ ρ

J 12.6m) )(1.00m/s )(9.80m )(1.00kg/m (1.29 233 === mghPE


398

energy found in part (b), which is the energy required to warm the air from to .

76. Unreasonable Results (a) What is the temperature increase of an 80.0 kg person who consumes 2500 kcal of food in one day with of the energy transferred as heat to the body? (b) What is unreasonable about this result? (c) Which premise or assumption is responsible?

Solution (a)

This says that the temperature of the person is (b) Any temperature increase greater than about would be unreasonably large.

In this case the final temperature of the person would rise to .

(c) The assumption that the person retains 95% of the energy as body heat is unreasonable. Most of the food consumed on a day is converted to body heat, losing energy by sweating and breathing, etc.

77. Unreasonable Results A slightly deranged Arctic inventor surrounded by ice thinks it would be much less mechanically complex to cool a car engine by melting ice on it than by having a water-‐cooled system with a radiator, water pump, antifreeze, and so on. (a) If of the energy in 1.00 gal of gasoline is converted into “waste heat” in a car engine, how many kilograms of ice could it melt? (b) Is this a reasonable amount of ice to carry around to cool the engine for 1.00 gal of gasoline consumption? (c) What premises or assumptions are unreasonable?

Solution (a)

(b) No, the mass of ice is greater than 1/4 of a ton.

(c) Not all waste heat goes into the engine.

C0.20 ° C0.50 °

95.0%

.C36C)kcal/kg kg)(0.83 (80.0

kcal) 00(0.950)(25 that so , °=°⋅

==ΔΔ=mcQTTmcQ

! C73C36C37 °=°+°C3°

( )F163 C73 °°

80.0%0°C

( )( ) kg103.1kg4.311

J/kg 10334J 103.1800.0

Thus, J.103.1800.0

23

8

8gas

f

gasfgas

×==×

×=

×=

=⇒=

M

QLQ

MMLQ


399

78. Unreasonable Results (a) Calculate the rate of heat transfer by conduction through a window with an area of that is 0.750 cm thick, if its inner surface is at

and its outer surface is at . (b) What is unreasonable about this result? (c) Which premise or assumption is responsible?

Solution (a)

(b) This is very high power loss through a window. An electric heater of this power can keep an entire room warm.

(c) The surface temperatures of the window do not differ by as great an amount as assumed. The inner surface will be warmer, and the outer surface will be cooler.

79. Unreasonable Results A meteorite 1.20 cm in diameter is so hot immediately after penetrating the atmosphere that it radiates 20.0 kW of power. (a) What is its temperature, if the surroundings are at and it has an emissivity of 0.800? (b) What is unreasonable about this result? (c) Which premise or assumption is responsible?

Solution (a) Given (Note that the negative

sign indicates that the meteorite radiates heat to the surroundings.)

(b) The meteorite has too high a temperature. It would completely melt.

(c) The rate of radiation is probably too high.

2m1.0022.0°C 35.0°C

( )

kW46.1 W101.46 W1456m 10.7500

C)22.0-C)(35.0m C)(1.00mJ/s (0.84

3

2-

212

=×==

×

°°°⋅⋅=

−=

dTTkA

tQ

20.0°C

cm. 0.600 and W 100.02kW .020 3 =×−=−= RtQ

K 105.59

)m 100.600()(0.800)4KmJ/s 10(5.67 W100.02K) (293

/

3

1/4

22428

34

4/14

21

×=

⎥⎦

⎤⎢⎣

⎡

×⋅⋅×

×−−=

⎟⎠

⎞⎜⎝

⎛ −=

−− π

σeAtQTT

Documents

Cnx Physics Ism Ch 14 Heat and Heat Transfer