Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 18 - The Laws of Thermodynamics - Problems and Conceptual Exercises - Page 646: 45

Answer

$0.28$

Work Step by Step

As we know that Total heat input= Work done+Heat exhausted Total heat input= 340+870=1210 J Now we can find the efficiency as follows: $\eta=\frac{Work \space done}{Heat \space input}$ We plug in the known values to obtain: $\eta=\frac{340}{1210}=0.28$
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