# Why doesn't a Coefficient Of Performance > 1 violate the second law of thermodynamics?

Apparently, if I understand correctly, one can move heat from a cold to a hot reservoir using less energy than the heat your are moving. This can be measured by the Coefficient Of Performance (COP). The COP of heating is given by $$\text{COP}_H=\frac{Q_H}{W}$$ where $$Q_H$$ is the heat added to the hot reservoir and $$W$$ is the work required to do this. Because of the first law of thermodynamics we have $$Q_H=Q_C+W$$ so we get $$\text{COP}_H=\frac{Q_H}{Q_H-Q_c}.$$ For an idealised cooler we can use a Carnot process which means $$\frac{Q_H}{T_H}=\frac{Q_C}{T_C}$$. So this means $$\text{COP}_H=\frac{T_H}{T_H-T_C}.$$ This equation seems to imply that if our reservoirs start at the same temperature we can move arbitrarily high amounts of heat with a finite amount of work. So if $$T_H=T_C$$ we could supply $$W=1\,\text{j}$$ and use that to move $$Q_H=10^{10}\,\text{j}$$ to give an arbitrarily high number. This would increase the temperature of the hot reservoir so if the final temperature $$T_H$$ is large enough we could extract work using another Carnot cycle which would be higher than the amount of work we supplied initially.

So where does my reasoning go wrong? There are other question similar to this one but they don't answer my question.

• I have posted an answer. What do you think of it. It would be nice, if not expected, to get some feedback. Mar 5, 2021 at 23:01
• It looks you don't care. I guess that's OK.. Mar 8, 2021 at 3:43

So where does my reasoning go wrong?

If I understand you correctly, your reasoning is wrong because you are not taking into account that when the temperature of the high temperature environment ($$T_H$$) increases, the COP of the heat pump decreases and it takes more work input to move the same amount of heat to the high temperature environment. So you can't compare the Carnot work output using that heat to the work input of the heat pump based on the original value of $$T_H$$. The work output of a Carnot engine always equals the work input of a Carnot heat pump when operating between the same temperatures.

The Carnot heat pump COP, as you know, is

$$COP_{hp}=\frac{T_H}{T_{H}-T_C}$$

The efficiency of the Carnot heat engine is

$$\eta_{he}=1-\frac{T_C}{T_H}=\frac{T_{H}-T_C}{T_H}$$

Therefore

$$COP_{hp}=\frac{1}{\eta_{he}}$$

So the higher the heat engine efficiency the lower the heat pump COP, and vice-versa.

Also, since

$$\eta_{he}=\frac{W_{out}}{Q_{H}}$$

and

$$COP_{hp}=\frac{Q_H}{W_{in}}$$

we get

$$W_{out}=W_{in}$$

Hope this helps.

• Hey sorry for the delayed reaction. This answer explains my question but I still think the physics behind it is weird so I guess I was waiting for more answers so I could pick the best one and while waiting I kind of forgot about this post. Mar 8, 2021 at 11:04
• @AccidentalTaylorExpansion. No Problem. What about the physics is it you find weird? Is it the amount of heat involved when the temperatures are close? Perhaps I can elaborate. Mar 8, 2021 at 13:17