The "efficiency" for heat pumps and chillers is called coefficient of performance (COP), because there is no conversion of electrical energy to heat energy but only a transportation. This COP can be way higher than 1, a heat pump can supply more heat energy than you put electrical energy into it. I once read that typical systems achieve COPs up to 3 or 4.

But how is the COP limited? I know, it depends on the temperature difference, the cooling fluid and some more factors, but is there a theoretical maximum? If I had a super duper cooling fluid with all the properties I need, could I build a chiller (or heat pump, I guess it works both ways pretty similar, because it only depends on which site of the system you use) with a COP of 10? Or 50? Or a few hundred? Or is there some fundamental law of physicis (probably some thermodynamic stuff) that limits the COP to a maximum value? If so, how high would that approximatly be?

  • $\begingroup$ I think this is an interesting study. This is practically available that almost all heat pumps push 2 to 4.5 or in some cases even 6 times the heat energy the pump consume to do this. In lay man language a heat pump can push nE kw heat from a low temperature to high temperature side just by requiring E ke of energy as input. Now if some one reverse this and use a heat engine to produce energy then this nE ke heat available on hot side will do the work through a heat engine then the output given by the heat engine will be efficiency(say f) x nE ke. It means, if principle of conversation of ener $\endgroup$ Sep 5, 2021 at 4:24

1 Answer 1


In a heat pump, $Q_1$ amount of heat is removed from the system maintained at temperature $T_1$, and $Q_2>Q_1$ amount of heat is dumped into the ambient at temperature $T_2>T_1$. If work input is $W$, then energy conservation requires $Q_2-Q_1=W$.

The entropy change of the universe is given by \begin{align} \Delta S=-\frac{Q_1}{T_1}+\frac{Q_2}{T_2} \end{align} Second law requires $\Delta S\geq 0$ which implies \begin{align} -\frac{Q_1}{T_1}+\frac{Q_2}{T_2} & \geq 0 \\ \frac{Q_2}{Q_1} & \geq\frac{T_2}{T_1} \\ \frac{Q_2}{Q_1}-1 & \geq \frac{T_2}{T_1}-1 \\ \frac{W}{Q_1} & \geq \frac{\Delta T}{T_1}\quad (\Delta T\equiv T_2-T_1)\\ \therefore\quad \textrm{C.O.P.}& \leq\frac{T_1}{\Delta T}\quad (\textrm{C.O.P.}\equiv Q_1/W) \end{align}

Therefore for specified temperature of system and ambient above relation gives theoretical bounds on C.O.P. See that nothing forbids C.O.P. from becoming less than one.

  • $\begingroup$ Ah, thanks. That means, to bring an example, if I would want to heat my home to a temperature of 21°C at an ambient temperature of 10°C, my COP would be limited to COP < (21K + 273K) / (21 - 10) K = 26.7. Thats not so bad. Thanks again for that interesting insight =) $\endgroup$
    – jusaca
    Aug 9, 2017 at 7:46
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    $\begingroup$ @jusaca In your example, COP would be defined as $Q_2/W$, while the definition I have used in my answer applies to refrigerator viz. $Q_1/W$. In your case the bounds would be COP $\leq T_2/\Delta T$, as you have done correctly in your example. $\endgroup$
    – Deep
    Aug 9, 2017 at 12:43
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    $\begingroup$ Maybe you should rewrite the formulas using T_l and T_h for T-low and T-high $\endgroup$
    – JanKanis
    Sep 2, 2019 at 8:59
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    $\begingroup$ @jusaca The short version is ideal COP = T-high / delta-T. But for your high temperature you need to take the temperature of the heating element or water temperature, depending on what kind of heating system you use. This will be higher than the 21°C you want to keep your home at. $\endgroup$
    – JanKanis
    Sep 2, 2019 at 9:59
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    $\begingroup$ Or for the layman, to put it simply: TheoreticalMaxCOP = (desiredIndoorTempC + 273) ÷ (desiredIndoorTempC - outsideTempC) $\endgroup$
    – Dan W
    Mar 24, 2020 at 21:35

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