## Essential University Physics: Volume 1 (3rd Edition)

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: $20 \ kW\times \frac{gal}{30\ kWh}\times \frac{3.6 \ dollars}{gal}=2.4 \ dollars \ per\ hour$