University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 20 - The Second Law of Thermodynamics - Problems - Exercises - Page 677: 20.13

Answer

(a) $W = 215 J $ (b) $T_C = 378K$ (c) $\epsilon = 39 \%$

Work Step by Step

(a) Work is done $W = 𝑄_𝐻 - 𝑄_𝐶 $ $W = 550 J- 335 J $ $W = 215 J $ (b) For Carnot cycle, $\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$ $T_C = \frac{Q_C}{Q_H} T_H $ $T_C = \frac{335 J}{550 J} 620K $ $T_C = 378K$ (c) $\epsilon = 1 - |\frac{T_C}{T_H}|$ $\epsilon = 1 - |\frac{378K}{620K }|$ $\epsilon = 0.39 = 39 \%$
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