Respuesta :
The rate of energy absorption from the outdoor air is 4400 KJ/h
Explanation:
Coefficient of performance (COP), is an expression of the efficiency of a heat pump. When calculating the COP for a heat pump, the heat output from the condenser (Q) is compared to the power supplied to the compressor (W).
The coefficient of performance, [tex]C O P=\frac{Q_{H}}{W_{i n}}[/tex]
We replace the value in the equation with given data,
[tex]C O P=\frac{(8000 K J / h)}{1 K W}=8000 K J / h[/tex]
By making a conversion, we get
[tex]C O P=\frac{(8000 \mathrm{KJ} / \mathrm{h})}{1 \mathrm{KW}} \times \frac{(1 \mathrm{KW})}{3600}=2.22[/tex]
The energy balance, [tex]Q_{L}=Q_{H}-W_{i n}[/tex]
Now, replace the values in the equation, we get
[tex]Q_{L}=(8000 \mathrm{KJ} / \mathrm{h})-(3600 \mathrm{KJ} / \mathrm{h})=4400 \mathrm{KJ} / \mathrm{h}[/tex]
