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TitleTippens capitulo 17
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Page 1

Chapter 17. Quantity of Heat Physics, 6th Edition

Chapter 17. Quantity of Heat

NOTE: Refer to Tables 18-1 and 18-2 for accepted values for specific heat, heat of vaporization,

and heat of fusion for the substances in the problems below

Quantity of heat and Specific Heat Capacity

17-1. What quantity of heat is required to change the temperature of 200 g of lead from 20 to

1000C? [Answer given in calories--also worked below for joules.]

0 0(0.20 kg)(130 J/kg K)(100 C - 20 C)Q mc t= ∆ = g ; Q = 2080 J

0 0 0(200 g)(0.031 cal/g C )(100 C - 20 C)Q mc t= ∆ = g ; Q = 496 cal

17-2. A certain process requires 500 J of heat. Express this energy in calories and in Btu.

1 cal
500 J

4.186 J
Q

 =  
 

Q = 119 cal

1 cal Btu
500 J

4.186 J 252 cal
Q

  =   
  

Q = 0.474 Btu

17-3. An oven applies 400 kJ of heat to a 4 kg of a substance causing its temperature to increase

by 80 C0. What is the specific heat capacity?

0

400,000 J

(4 kg)(80 C )

Q
c

m t
= =


; c = 1250 J/kg C0

17-4. What quantity of heat will be released when 40 lb of copper cools form 78 to 320F?

Q = (40 lb)(0.093 Btu/lb F0)(780F – 320F); Q = 171 Btu

17-5. A lawn mower engine does work at the rate of 3 kW. What equivalent amount of heat will

be given off in one hour?

J 3600 s
3000 3000

s 1 h
P W

 = =  
 

; Q = 10,800,000 J/h; Q = 10.8 MJ/h

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Page 8

Chapter 17. Quantity of Heat Physics, 6th Edition

17-31. A heating element supplies an output power of 12 kW. How much time is needed to melt

completely a 2-kg silver block? Assume no power is wasted.

Q = mLf = (2 kg)(88.3 x 103 J/kg) = 176,600 J; P = 12,000 J/s

176,600 J
;

12,000 J/s

Heat Heat
P t

t P
= = = ; t = 14.7 s


*17-32. How much ice at -100C must be added to 200 g of water at 500C to bring the equilibrium

temperature to 400C? (The ice must first be brought to 00C, then it must be melted, and

the resulting water brought to the equilibrium temperature.)

mici(10 C0) + miLf + micw(400C – 00C) = mwcw(500C – 400C)

mi(0.5 cal/g C0)(10 C0) + (80 cal/g)mi + (40 C0)(1 cal/g C0)mi = (200 g)(1 cal/g C0)(10 C0)

(125 cal/g)ms = 2000 cal; ms = 16.0 g

*17-33. Assume that 5 g of steam at 1000C are mixed with 20-g of ice at 00C. What will be the

equilibrium temperature? (Total heat lost by steam = Total heat gained by ice)

msLv+ mscw(1000C – te) = miLf + micw(te – 00C)

(5 g)(540 cal/g) + (5 g)(1 cal/g C0)(1000C – te) = (20 g)(80 cal/g) + (20 g)(1 cal/g C0)te

2700 cal + 500 cal – (5 cal/C0)te = 1600 cal + (20 cal/C0)te

(25 cal/C0)te = 1600 cal; te = 64.00C

*17-34. How much heat is developed by the brakes of a 4000-lb truck to bring it to a stop from a

speed of 60 mi/h? Work = ½mvf2 - ½mvo2 vo= 60 mi/h = 88 ft/s; vf = 0; m = W/g

2

4000 lb
125 slugs

32 ft/s
m = = ; Work = 0 - ½(125 sl)(88 ft/s)2 = -484,000 ft lb

1 Btu
484,000 ft lb

778 ft lb
Heat

 = −  
 

; Heat = 622 Btu

241

Page 14

Chapter 17. Quantity of Heat Physics, 6th Edition

*17-48. If equal masses of ice at 00C, water at 500C, and steam at 1000C are mixed and allowed to

reach equilibrium. Will all of the steam condense? What will be the temperature of the

final mixture? What percent of the final mixture will be water and what percent will be

steam? (Let m be the initial mass of water, ice, and steam.)

Because the heat of vaporization is the largest, lets assume that the equilibrium

temperature is 1000C and that only mass mx condenses. Then look at the results.

mxLv = mLf + mcw(100 C0) + mcw(1000C – 500C)


0 0(100 C ) (50 C )x v f w w

m
L L c c

m
= + + ; 0v f wL = L + (150 C )c

xm

m

0 0 0(150 C ) 80 cal/g (150 C )(1 cal/g C )

540 cal/g
f wx

v

L cm

m L

+ +
= = ; x

m

m
= 0.426

This answer is reasonable based on the assumptions. Thus, all the steam

does NOT condense and te = 1000C

The total mass is ms + mw + ms = 3m and mx = 0.426 m (condensed) , Thus

Percent steam =
0.426 0.574

0.191
3 3

m m m

m m


= = ; 19.1% steam

The remainder is water; 80.9% water

*17-49. If 100 g of water at 200C is mixed with 100 g of ice at 00C and 4 g of steam at 1000C, find

the equilibrium temperature and the composition of the mixture?.

First let’s assume that not all of the ice melts, then check to see if answer is reasonable.

Heat lost = Heat gained: msLv + mscw(1000C – 00C) + mwcw(200C – 00C) = mxLf

(4 g)(540 cal/g) + (4 g)(1 cal/g C0)(100 C0) + (100 g)(1 cal/g C0)(20 C0) = mx(80 cal/g)

mx = 57 g; This means that 100 g – 57 g or 43 g of ice remain unmelted

Total water = 4 g + 100 g + 57 g = 161 g; Total ice = 43 g

247

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