# 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". – alephzero Jul 3 '19 at 10:36
• This is why anything using that type of cycle has a "Coefficient of Performance", not an efficiency... – user207455 Jul 3 '19 at 11:08
• See my updated answer which includes figures that may be of help. – Bob D Jul 3 '19 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. 