A heat pump is used to keep a house warm at 22∘C.

Question

A heat pump is used to keep a house warm at 22∘C∘C.

How much work is required of the pump to deliver 3400 JJ of heat into the house if the outdoor temperature is 0∘C∘C. Assume a COP of 3.0

Express your answer to two significant figures and include the appropriate units.

Answer 1

They key to this question is that the coefficient of performance (COP) is defined as the ratio of the heat delivered to a hot reservoir (Q_h) divided by the amount of work.

COP = Q_h / W

The rest is algebra:

W = Q_h / COP

I'll let you do the math yourself. If you want to check your answer, mine was between 1,100 Joules and 1,200 Joules.

There seems to be some extra information in the question. You could use the temperatures provided to calculate the coefficient of performance. In ideal circumstances

COP = T_H / (T_H - T_C)

You need to convert the temperature to Kelvins, which I'll do for you: T_H = 295 K, T_C = 273 K (note that the difference is the same, so you can use 22 K in the denominator to save time).

So the COP when calculated this way will be between 13 and 14, which will obviously change the answer to the question a lot.

I'm not sure why the question supplies contradictory information. Perhaps they want you to calculate it both ways and compare, since the COP of 3.0 that they assume is much less efficient than the ideal.