# Heat engine

## Homework Statement

A machine with COP = 5.0 removes 25kJ of heat from a cold reservoar. If the machine is reversible and driven the other way as a heat engine, what is the efficiency?

## The Attempt at a Solution

COP = QL/W --> W = 5kJ

e = QH/W

Is QH the same as QL som that e = 20%

No, I don't believe that $Q_H$ and $Q_L$ are the same. Using conservation of energy, you can find out how Qh, QL, and W are related. What do you get?
$$e=\frac{W}{Q_H}$$