Calculating Efficiency of a Heat Engine with COP = 5.0 and 25kJ Heat Removal

[kasse]

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 = Q_L/W --> W = 5kJ

e = Q_H/W

Is Q_H the same as Q_L som that e = 20%

 

[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:

<br /> e=\frac{W}{Q_H}<br />