Question

A heat pump has a coefficient of performance equal to 4.45 and requires a power of 1.60 kW to operate.

(a) How much energy does the heat pump add to a home in one hour?
(b) If the heat pump is reversed so that it acts as an air conditioner in the summer, what would be its coefficient of performance?

Answer 1

Step-by-step explanation:

COP = Q/W

where Q is the heat into building and W is the work input = 1.60 kW

COPh = 4.45

Q = COPh x W

1) Q = 4.45 x 1.6 = 7.12 kW = 7120 W

In one hour there is 3600 sec

Therefore energy in one hour = 7120 x 3600 = 25632000 J

= 25.632 MJ

2) COPa = COPh - 1

= 4.45 - 1 = 3.45

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