Answer:
[tex]COP_{HP} = 2.278[/tex], [tex]\dot Q_{L} = 1.278\,kW[/tex]
Explanation:
The heat supply is:
[tex]Q_{H} = (8200\,\frac{kJ}{h})\cdot (\frac{1\,h}{3600\,s} )[/tex]
[tex]Q_{H} = 2.278\,kW[/tex]
The Coefficient of Performance of a Heat Pump is:
[tex]COP_{HP} = \frac{\dot Q_{H}}{\dot W}[/tex]
[tex]COP_{HP} = \frac{2.278\,kW}{1\,kW}[/tex]
[tex]COP_{HP} = 2.278[/tex]
The rate of heat absorption from the outdoor air is:
[tex]\dot Q_{L} = \dot Q_{H} - \dot W[/tex]
[tex]\dot Q_{L} = 2.278\,kW-1\,kW[/tex]
[tex]\dot Q_{L} = 1.278\,kW[/tex]