Answer
$8.8\%$
Work Step by Step
Use the definition of the efficiency of a heat engine.
$$e=\frac{W}{Q_H}=\frac{W}{(25.0kcal)(4186J/kcal)}= 0.0879=8.8\%$$
Physics: Principles with Applications (7th Edition)
by Giancoli, Douglas C.
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: 21
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