I'm learning about coefficient of performance (COP) and am trying to understand its implications for heating/cooling.

By intuition I assumed heating or cooling a room by a set amount requires the same amount of energy regardless of the absolute temperature. For example, does the amount of energy used to heat/cool a room 1°C depend on its absolute temp (e.g. 20°C vs 22°C)?

The maximum theoretical heating COP is

$COP_{heating} = \frac{T_{hot}}{T_{hot} - T_{cool}}$

where the temps are in Kelvin.

So for the COP for raising 20°C to 21°C is

$294.15 = \frac{(21 + 273.15)}{ (21 + 273.15) - (20 + 273.15) }$

and the COP for raising 22°C to 23°C is

$296.15 = \frac{(23 + 273.15)}{ (23 + 273.15) - (22 + 273.15) }$

So it appears my assumption was wrong. What is an intuitive explanation for why this the case? If this is wrong, why so?

  • $\begingroup$ Just look at the link you provided. That ratio of temperatures is really a ratio of heats. The thermodynamical laws used to calculate these heats require the use of absolute temperatures. $\endgroup$
    – Gert
    Jul 15, 2016 at 19:35
  • $\begingroup$ @Gert I'm still trying to understand the law. I understand that the ratio of heats causes the COP to increase when the absolute temperatures are larger, but I do not understand why the law uses the ratio. $\endgroup$ Jul 15, 2016 at 19:45
  • $\begingroup$ You need to study what makes a heat engine tick: en.wikipedia.org/wiki/Heat_engine $\endgroup$
    – Gert
    Jul 15, 2016 at 21:14

3 Answers 3


Let's try in logical, non-technical terms. When you are attempting to raise the temperature of a room one degree you are raising the energy level of the room. If the starting state is lower, it takes less total effort to do this that if the starting state is already high. While this difference might be small when attempting to raise the temperature one degree and comparing starting points of 20C and 21C, it is very noticeable when comparing starting points of 20C to 5000C.

Think of it in terms of a car. It takes much less effort to speed a car up from 20 to 21 MPH than it takes to speed it from 100 to 101 MPH even though it is the same 1 MPH increase. You are not just adding 1 MPH to the speed or 1 degree to the temperature, you are making a increase to the total system energy state which becomes an exponential function, not linear.


The Carnot Cycle is particularly important here. Carnot's Theorem states:"Any cycle [engine] operating between temperatures Th and Tc will never exceed the efficiency of a Carnot cycle. The efficiency of the CC = W/Qh = 1 - Tc/Th = (Th-Tc)/Th The derivation for the efficiency can be found in Wikipedia https://en.wikipedia.org/wiki/Carnot_cycle, and the 4 legs of the cycle are also there. Additional information would be found under various Thermodynamic subjects, Energy, Entropy, Temperature, Isothermal, reversible adibiatic (= isentropic), etc. You'll note the relationship between thermodynamic efficiency and the COP. ... As far as dealing with your intuition, I'd suggest taking a look at the major thermodynamic cycles Carnot, Rankine, Stirling, Diesel, Otto and noting that they (esp. the Carnot cycle) involve Entropy and Entropy is based from absolute zero. You aren't wrong that the energy contained in a volume (of an ideal gas) depends only on temperature (at equilibrium, of course), but dE = dQ-dW = TdS-dW which means you have to neglect the work (dW) to arrive at a relationship where a change in Energy is proportional to a change in temperature. And neglecting the work done is exactly what you don't want to do. HTH


Your title is different from the question you ask in the text.

You are correct that the amount of energy needed to raise the temperature of a fixed mass of ideal gas by $1^\circ C$ is independent of the absolute temperature.

But this is not the same as calculating the coefficient of performance of a heat engine operating between those temperatures.

You need to be clear about which question you are asking.


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