Answer
The more expensive one will pay off in about 2 years and 5 months of use.
Work Step by Step
$COP_R=\dot{Q}_L/\dot{W}_i$
$\dot{W}_i=\dot{Q}_L/COP_R$
$C=P+p_e\times\dot{W}_i\times\Delta t$
Both: $p_e=\$0.10/kWh,\ \dot{Q}_L=40000\ kWh/year$
A: $P=\$5500,\ COP=2.3$
B: $P=\$7000,\ COP=3.6$
Time of same cost:
$\Delta t=(P_B-P_A)/(p_e\dot{Q}_L(\frac{1}{COP_A}-\frac{1}{COP_B}))$
$\Delta t =2.39\ years$
The more expensive one will pay off in about 2 years and 5 months of use.
