#### Answer

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