Answer:(b) 0.75 kW
Explanation:
Given
temperature of cold region [tex]T_L=3^{\circ}C[/tex]
Heat removal at the rate of [tex]Q_1=5400 kJ/h[/tex]
temperature of hot region [tex]T_H=30^{\circ}C[/tex]
COP of refrigerator =2
COP of a refrigerator is given by ratio of desired effect to the Power input
COP[tex]=\frac{Q_1}{Power}[/tex]
[tex]2=\frac{5400}{P}[/tex]
[tex]P=2700 kJ/h[/tex]
[tex]P=0.75 kW[/tex]