Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 15 - The Laws of Thermodynamics - Problems - Page 439: 24

Answer

1200 MJ

Work Step by Step

Use the definition of the efficiency of a heat engine. $$e=\frac{W}{Q_H}=\frac{W}{W+Q_L}$$ $$Q_L=W(\frac{1}{e}-1)$$ In one second, W is the work output, 580 MJ. Calculate the heat discharged per second. $$Q_L=(580MJ)(\frac{1}{0.32}-1) =1232.5 MJ\approx 1200MJ $$
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