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?

// Ok, I get the feeling, that it may be the wrong place to aks? Maybe the engineering forum would be better suited? I just figured physics fits quite well, because I asked about the fundamentals, not a specific implementation. Is there mabye a way to move this question? Sorry, I'm new here!


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 '17 at 7:46
  • $\begingroup$ @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 '17 at 12:43
  • $\begingroup$ Maybe you should rewrite the formulas using T_l and T_h for T-low and T-high $\endgroup$ – JanKanis Sep 2 '19 at 8:59
  • $\begingroup$ @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 '19 at 9:59

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