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: 23

Answer

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Work Step by Step

$e=\frac{T_H-T_L}{T_H}$ $e_i=\frac{T_H+10^oC-T_L}{T_H+10^oC}=1-\frac{T_L}{T_H+10^oC}$ $e_d=\frac{T_H-T_L}{T_H}=\frac{T_H-(T_L-10^oC)}{T_H}=1-\frac{T_L-10^oC}{T_H}$ $\frac{T_L-10^oC}{T_H}>\frac{T_L}{T_H+10^oC}$ Decreasing the lower temperature has a large impact on the efficiency.
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