Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 19 - Heat Engines and Refrigerators - Exercises and Problems - Page 551: 39

Answer

Cold-reservoir temperature= $-33^{\circ}C$ Hot-reservoir temperature= $47^{\circ}C$

Work Step by Step

For a carnot engine, efficiency is $\eta_{carnot}=1-\frac{T_{C}}{T_{H}}$ where $T_{C}$ is the temperature in Kelvin of cold reservoir and $T_{H}$ is the temperature of hot reservoir. Given: $\eta=0.25$, $T_{H}=T_{C}+80$. $\implies 0.25=1-\frac{T_{C}}{T_{C}+80}$ $\implies 0.25=\frac{80}{T_{C}+80}$ $\implies 0.25T_{C}+20=80$ $\implies 0.25T_{C}=80-20=60$ Or $T_{C}=\frac{60}{0.25}=240\,K=(240-273)^{\circ}C=-33^{\circ}C$ $T_{H}=T_{C}+80\,K=(240+80)\,K=320\,K$ $=(320-273)^{\circ}C=47^{\circ}C$
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