A window air conditioner unit is placed on a laboratory bench and tested in cooling mode using 750 W of electric power with a COP of 1.75. What is the cooling power capacity and what is the net effect on the laboratory?

Q=1312.5 W

Explanation:

Given that

Cooling load or power input = 750 W

COP=1.75

We know that COP can be given as

COP is the ratio of cooling effect to the input power.

Lets take cooling effect is Q. So now by using COP formula

COP=Q/750

1.75=Q/750

Q=1312.5 W

So the cooling effect produce by air conditioning will be 1312.5 W.

The net effect on laboratory, the laboratory temperature will reduce.