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: 20.4

Answer

(a) $Q_H = 6.43 \times 10^5 J$ (b) $Q_C = 4.63 \times 10^5 J$

Work Step by Step

(a) For one second, $ W = 180 \times 10^3 J$ $Q_H = \frac{W}{\epsilon}$ $Q_H = \frac{180 \times 10^3 J}{0.280 }$ $Q_H = 6.43 \times 10^5 J$ (b) $𝑄_𝐶 = 𝑄_𝐻 - W$ $𝑄_𝐶 = 6.43 \times 10^5 J - 180 \times 10^3 J $ $Q_C = 4.63 \times 10^5 J$
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