Answer: 1.79 kW
Explanation:
Given
The temperature of the refrigerator is [tex]T_L=2^{\circ}C\approx 275\ K[/tex]
The temperature of the kitchen is [tex]T_H=22^{\circ}C[/tex]
COP of the refrigerator is
[tex]COP=\dfrac{T_L}{T_H-T_L}=\dfrac{\text{Desired effect}}{\text{Power}}\\\\\Rightarrow COP=\dfrac{275}{22-2}=\dfrac{89}{P}\\\\\Rightarrow P=\dfrac{20}{275}\times 89=6.47\ MJ/h\ \text{or}\ 1.79\ kW[/tex]
