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:

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

 

[Moetasim]

?

I would like to clarify a few things before providing a response. First, it would be helpful to know the units of the values given, as they are not specified in the question. Additionally, the second equation given, e = QH/W, is not a correct representation of efficiency. The correct equation for efficiency is e = (QH - QL)/QH, where QH is the heat added to the system and QL is the heat removed from the system.

Given the information provided, it seems that the COP of 5.0 and the heat removal of 25kJ are referring to the same process, in which case QH and QL would be equal. This would result in an efficiency of 0%, which is not possible for a reversible heat engine. Therefore, further clarification or additional information would be needed to accurately calculate the efficiency of this system.