# Lavorem

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CHAPTER 11 HEAT
1. 2. 3. (d). (b), because 1 BTU = 1055 K and 1 cal = 4.186 J. 1 Cal = 1000 cal .

4.

No ; heat is the amount of energy added to or removed from the internal energy of an object, so a hot object might have more internal energy. However the mass of the object is also a factor.

5.

1500 Cal = 1500 × 103 cal = (1500 × 103 cal)(4.186 J/cal) = 6.279 × 106 J . 20 000 Btu (20000 Btu)(1055 J/Btu) Q = = = 5.86 × 103 W . 1h 3600 s ∆t (4 × 10 J/h)(24 h) = 9.6 × 10 J.
5 5 5

6.

1 BTU = 1055 J. P =

7.

In one day, the energy required is The number of Big Macs is

9.6 × 10 J (600 Cal)(4186 J.Cal) = 3.8 = 4 .
2

8 .

(a) Work done in each lift: Energy input:

W = Fd = (20 kg)(9.80 m/s )(1.0 m) = 196 J.
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E = 2800 Cal = (2800 J)(4186 J/Cal) = 1.17 × 10J. 1.17 × 107 J = 60 000 times . 196 J
5

So the number of lifts is

(b) t = 60 000(5.0 s) = 3.0 × 10 s = 83 h . 9. 10, (a), according to the definition of specific heat. (a). cp is always greater than cv for gases. At constant volume, no work is done when adding heat to the gases. Under constant pressure, more heat is needed because of the work done as the gas expands. 11. (b). From Q =cm∆T, if Q and m are the same, a greater ∆T results in a smaller c.

12 .

The answer is beach during summer day and also beach during winter night . Beach has a smaller specific heat than that of water, so it heats up and cools down faster than water.

© 2007 Pearson Prentice Hall, Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

166

Chapter 11

Heat

13.

Since Q = cm∆T = cm(Tf − Ti), specific heat and mass can cause the final temperature of the two objects to be different, if Q and Ti are the same.

14.

This is possible, because water has a high specific heat, so it canabsorb a lot of heat for a small temperature increase. Firewalkers often wet their feet before taking the plunge into the hot bed of burning coals, and the feet are largely made up of water. Other factors that are pertinent to this are that the wood coals, typically used in fire walking, are very poor conductors and contain relatively little thermal energy, little time is spent on the coals, and asmall part of the feet is in contact with the coals at any one time.

15.

(a) Copper requires (1) more heat , because it has a higher specific heat. ccopper = 390 J/(kg⋅C°) > clead = 130 J/(kg⋅C°). (b) Q = cm∆T, ∆Q = (∆c)m∆T = [390 J/(kg⋅C°) − 130 J/(kg⋅C°)](1.0 kg)(190 K − 110 K) = 2.1 × 104 J = copper requires 2.1 × 10 J more .
4

16 .

Q = mc∆T,

Q (200 J) ∆T = cm = = 43 C°. [920J/(kg·C°)](5.0 × 10−3 kg) Tf = Ti + ∆T = 20°C + 43 C° = 63°C .

So the final temperature is

17. 18.

Q = cm∆T = [4186 J/(kg·C°)](5.0 kg)(100°C − 20°C) = 1.7 × 106 J . Heat is lost, so Q is negative. Q = mc∆T, Q (−1500 J) ∆T = cm = = −1.43 C°. [4186 J/(kg·C°)](0.250 kg) 37.0°C − 1.43 C° = 35.6°C .

So the final temperature is

19 .

(a) Aluminum has a higher specific heat than copper. FromQ = mc∆T, if Q and ∆T are the same, a higher specific heat results in a lower mass. So the answer is (3) less than . (b) mAl cAl = mCu cCu mAl = mCu cCu (3.00 kg) [390 J/(kg·C°)] = 1.27 kg . cAl = 920 J/(kg·C°)

20.

Q = QAl + QFe = cAl mAl ∆T = cFe mFe ∆T = [920 J/(kg⋅C°)](25 kg)(120°C − 20°C) + [460 J/(kg⋅C°)](80 kg)(120°C − 20°C) = 6.0 × 10 J .
6

© 2007 Pearson Prentice Hall, UpperSaddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

167

Chapter 11

Heat

21.

Let the final temperature be T.

The heat lost by water is

Qlost = cm∆T = [4186 J/(kg·C°)](0.40 kg)(T − 90°C) =...