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: 23

Answer

a) $ \eta = 0.25 =25\%$ b) $T_C = 231.75^oC$

Work Step by Step

a) We have $W = Q_H -Q_C $ $ 200J = Q_H - 600J$ $Q_H = 800J$ Now, $ \eta = \frac{W}{Q_H} = \frac{200}{800}$ $ \eta = 0.25 =25\%$ b) Now, $ \eta = 1- \frac{T_C}{T_H}$ $ 0.25 =1- \frac{T_C}{(400+273)K}$ $ T_C = 0.75* 673K = 504.75K$ $T_C = 231.75^oC$
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