Answer
a. 17
b. $7.1\times 10^7J$
Work Step by Step
a. The COP for an ideal heat pump can be derived from equation 15–7, or by looking at problem 35.
$$COP=\frac{Q_H}{W}=\frac{Q_H}{Q_H-Q_L}=\frac{T_H}{T_H-T_L}=\frac{(24+273)K}{18K}=16.5\approx17$$
b. $COP=\frac{Q_H}{W}$
$$Q_H=\frac{W}{t}(t)(COP)=(1200watts)(3600s)(16.5)=7.1\times 10^7J$$
