Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 21 - Heat Engines and Refrigerators - Exercises and Problems - Page 594: 3

Answer

a)$\eta = 27.27\%$ b) $W = 15 kJ$

Work Step by Step

We have $Q_1 = 55kJ, Q_2 = 40kJ $ a) Thermal efficiency is given as $ \eta = \frac{Q_1 - Q_2 }{Q_1}$ $ \eta = \frac{55kJ - 40kJ}{55kJ}$ $\eta = 27.27\%$ b) We have $ W=Q_1 -Q_2 = 55kJ -40kJ$ $W = 15 kJ$
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