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: 7

Answer

a) $ W = 12.5 kJ$ b) $Q_2 = 50kJ$

Work Step by Step

We have $ P =\frac{W}{t}$ For $2400 rpm$ $P= 500kW$ a) For one cycle $ P = \frac{500kW}{2400}$ $P = 5/24kW $ Thus $W = P*t = 5/24 kW *1m$ $ W= 5/24kW *60s$ $ W = 12.5 kJ$ b)Also $ \eta =\frac{ W}{Q_1}$ $20\% = \frac{12.5kJ}{Q_1}$ $Q_1 = 62.5kJ$ and $W= Q_1-Q_2$ $Q_2 = 62.5kJ - 12.5kJ$ $Q_2 = 50kJ$
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