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 550: 28

Answer

a) $6.325$ b) $ 31.62J/s$ c) $231.62J/s$

Work Step by Step

We have $ T_H = 20^oC =293K$ $T_C = -20^oC = 253K$ $P_C = Q_C / t = 200J/s$ a) We know that $ K= \frac{T_C}{T_H-T_C}$ $K= \frac{253K}{40K}$ $K= 6.325$ b) We also know $ K= \frac {Q_C}{W}$ Dividing numerator and denominator on RHS by $t$ we get $K = \frac{P_C}{P_W} $ where $P_W$ is rate at which work is done on refrigerator $6.325 = (200J/s)/(P_W)$ $P_W = 31.62J/s$ c) Also we have $Q_H= Q_C +W$ Dividing throughout by$t$ we get $P_H = P_C +P_W$ where $P_H$ is rate at which heat is exhausted to hot side $P_H = 200J/s + 31.62J/s$ $P_H = 231.62J/s$
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