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 - General Problems - Page 441: 67

Answer

60 K.

Work Step by Step

Find the original high temperature from the information given. The original Carnot efficiency was 25 percent. $$0.25=1-\frac{T_L}{T_{H1}}=1-\frac{293K}{T_{H1}}$$ $$T_{H1}=390.67K$$ Find the new high temperature from the desired Carnot efficiency of 35 percent. $$0.35=1-\frac{T_L}{T_{H1}}=1-\frac{293K}{T_{H2}}$$ $$T_{H2}= 450.77 K$$ The difference is 60 kelvin.
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