Determine the cop of a heat pump that supplies energyto a house at a rate of 8000 kj/h for each kw of electric power it draws. also, determine the rate of energy absorption from the outdoor air.

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,

We replace the value in the equation with given data,

By making a conversion, we get

The energy balance,  

Now, replace the values in the equation, we get

vapour or gas assuming atmospheriv pressure

so easy. just google the electromagnetic waves and it’ll pop up.