Answer
43 cents per hour.
Work Step by Step
$$COP_{ideal}=\frac{T_L}{T_H-T_L}=\frac{273K+22K}{13K}=22.69$$
The air-conditioning unit is 18 percent as efficient as that.
The heat removed is the latent heat of fusion for 5 tons of water.
$$\frac{W}{t}=\frac{Q_L/t}{COP}=\frac{5(909kg/day)(3.33\times10^5J/kg)}{0.18 COP_{ideal}}=3.705\times 10^8J/d$$
Find the hourly cost for the air conditioning.
$$(3.705\times 10^8J/d) (\frac{1d}{24h})(\frac{kWh}{3.6\times10^6 J})(\$0.10/kWh)=\$0.43/h$$
