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

My answer:

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?

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    $\begingroup$ 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". $\endgroup$ – alephzero Jul 3 '19 at 10:36
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    $\begingroup$ This is why anything using that type of cycle has a "Coefficient of Performance", not an efficiency... $\endgroup$ – user207455 Jul 3 '19 at 11:08
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    $\begingroup$ See my updated answer which includes figures that may be of help. $\endgroup$ – Bob D Jul 3 '19 at 12:37
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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.

enter image description here

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