Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 19 - Exercises and Problems - Page 350: 39


a) 5.7 b) 3.5 kW c) Pump: 54 cents per hour; Oil furnace: 2.4 dollars per hour

Work Step by Step

a) Converting to Kelvin and using equation 19.4b, we find: $COP=\frac{343K}{343K-283K}=5.7$ b) We know that the power consumed is equal to the power produced divided by the COP: $P_c=\frac{20}{5.7}=3.499 \ kW$ c) We compare the costs: Pump: $3.499 \times .155 = \fbox{.54 dollars per hour}$. Furnace: $20 \ kW\times \frac{gal}{30\ kWh}\times \frac{3.6 \ dollars}{gal}=2.4 \ dollars \ per\ hour$
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