# Can a heat pump have efficiency greater than 100%?

A heat pump has an efficiency greater than 100%. Does this violate the laws of thermodynamics?

I'm assuming that "heat pump" means heat engine.

$$e>1$$

$$\displaystyle\implies\frac{W}{Q_H}>1$$

$$\implies W>Q_H$$

$$\implies W-Q_H>0$$

$$\implies (Q_H+Q_C)-Q_H>0$$

$$\implies Q_C>0$$

This violates the laws of thermodynamics.

Is this correct?

• You said you are "assuming" what a heat pump is, and apparently your assumption is wrong. See en.wikipedia.org/wiki/Heat_pump. Heat pumps do have a COP (coefficient of performance) greater than 1 - typically 3 or 4, which is why they are useful! This does not violate the laws of thermodynamics. Note that "the COP" is not the same as "the thermodynamic efficiency". Jul 3, 2019 at 10:36
• This is why anything using that type of cycle has a "Coefficient of Performance", not an efficiency...
– user207455
Jul 3, 2019 at 11:08
• See my updated answer which includes figures that may be of help. Jul 3, 2019 at 12:37

A heat pump is not a heat engine. It is a heat engine in reverse. It's Coefficient of Performance (the term used instead of efficiency) is normally greater than 1 and does not violate the laws of thermodynamics. Your first equation is for a heat engine. For the heat pump it is $$\frac{Q_H}{W}$$.

The figures below show a heat engine and heat pump operating between two temperatures.

A heat engine takes heat $$Q_H$$ from a higher temperature environment $$T_H$$, produces output work $$W_{OUT}$$, and rejects heat $$Q_L$$ to a lower temperature environment, where

$$W_{OUT}=Q_{H}-Q_{L}$$

The efficiency of a heat engine is the work output divided by the gross heat input, or

$$e=\frac{W_{OUT}}{Q_H}$$

It's efficiency is always less than 1. An efficiency of more than 1 violates the first law of thermodynamics (conservation of energy). The second law limits the maximum efficiency to $$1-\frac{T_L}{T_H}$$

A heat pump takes heat $$Q_L$$ from a lower temperature environment, uses work input $$W_{IN}$$ to move heat $$Q_H$$ to a higher temperature environment $$T_H$$. The desired output of the heat pump is the heat $$Q_H$$ and is

$$Q_{H}= Q_{L}+W_{IN}$$

We don't use the term efficiency for a heat pump, but rather the Coefficient of Performance $$COP$$. The COP is the desired heat transferred to the higher temperature environment divided by the work in required, or

$$COP=\frac{Q_H}{W_{IN}}$$

The COP is normally greater than 1 and does not violate the laws of thermodynamics.

Hope this helps.