Answer:
[tex]COP_{R} = 1.406[/tex], [tex]\dot Q_{H} = 2.406 W[/tex]
Explanation:
The COP of a refrigerator is modelled after this expression:
[tex]COP_{R} = \frac{\dot Q_{L}}{\dot W} \\COP_{R} = \frac{(5060 \frac{kJ}{h})(\frac{1 h}{3600 s})}{1 kW}\\COP_{R} = 1.406[/tex]
The rate of heat rejection to the outside air is:
[tex]\dot Q_{H} = \dot Q_{L} + \dot W\\\dot Q_{H} = (5060 \frac{kJ}{h} )\cdot (\frac{1}{3600 s} ) + 1 kW\\\dot Q_{H} = 2.406 kW[/tex]
