Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 19 - Heat Engines and Refrigerators - Exercises and Problems - Page 550: 16

Answer

(a) $13\%$ (b) $320J$

Work Step by Step

(a) We can determine the required efficiency as follows: $W_{out}=\frac{1}{2}(300-100)\times 10^3(600-200)\times 10^{-6}=40J$ and $Q_c=100+180=280J$ Now $\eta=\frac{W_{out}}{W_{out}+Q_c}$ We plug in the known values to obtain: $\eta=\frac{40}{40+280}=13\%$ (b) We can determine the required heat as follows: $Q_c=W_{out}+Q_c$ We plug in the known values to obtain: $Q_c=40+280=320J$
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