# Heat engine

1. May 24, 2008

### kasse

1. The problem statement, all variables and given/known data

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?

3. The attempt at a solution

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

e = QH/W

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

2. May 25, 2008

### alphysicist

Hi kasse,

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?

Also, your efficiency formula needs to be:

$$e=\frac{W}{Q_H}$$