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## Homework Statement

Consider a heat pump used to heat a house to 70°F when the outside temp is 5°F. The total heat loss from the house is 10000 BTU/hr. Efficiency of the heat pump is 45% of the theoretical maximum. What will be the power used by the pump?

## Homework Equations

e (or cop) = Q_H/W

ideally, e = 1/(1-T_C/T_H)

## The Attempt at a Solution

I am given to understand that a heat pump is a kind of refrigerator, which is a heat engine run backwards...what I don't understand is what to do with the heat loss I am given. I have absolutely no idea where the heat lost from the house would fit into the equations. Which leaves me stumped on where to even begin to start finding the work. Which is what I'm assuming is meant by "power"; however, I am not sure about that, either.