If we take heat capacity to be defined as "the ratio of the heat added to the temperature rise":

$$ C=\frac{\text{d} Q_{rev}}{\text{d}\theta}$$

then this leads me to ask: can this ever be negative? That is to say, are there any materials which cool as you add heat to them?

  • $\begingroup$ Are you set on materials, or would any system do? $\endgroup$
    – user10851
    Feb 1, 2016 at 1:17
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    $\begingroup$ check out negative temperatures and population inversion, e.g., en.wikipedia.org/wiki/Population_inversion $\endgroup$ Feb 1, 2016 at 1:18
  • $\begingroup$ @ChrisWhite materials would be most interesting to me but if you have a system then I'll take that too :) $\endgroup$ Feb 1, 2016 at 1:19
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    $\begingroup$ See, for instance, this SE question or the Wikipedia article. $\endgroup$
    – ACuriousMind
    Feb 1, 2016 at 1:44
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    $\begingroup$ a black hole temperature decreases as its mass increases : $T = {\hbar \, c^3 \over 8 \pi G M k_\text{B}}$ $\endgroup$
    – user46925
    Feb 1, 2016 at 2:14

4 Answers 4


There are certainly systems that have negative heat capacities, and in fact they come up all the time in astrophysics.

As a general rule, gravitationally bound systems have negative heat capacities. This is because in equilibrium (and remember we can't do classical thermodynamics without equilibrium anyway), some form of the virial theorem will apply. If the system has only kinetic energy $K$ and potential energy $U$, then the total energy is of course $E = K + U$, where $E < 0$ for bound systems. In virial equilibrium where the potential energy is purely gravitational, then we also have $K = -U/2$. As a result, $K = -E$, and so adding more energy results in a decrease in temperature.

Examples include stars and globular clusters. Imagine adding energy to such systems by heating up the particles in the star or giving the stars in a cluster more kinetic energy. The extra motion will work toward slightly unbinding the system, and everything will spread out. But since (negative) potential energy counts twice as much as kinetic energy in the energy budget, everything will be moving even slower in this new configuration once equilibrium is reattained.

At some level, this all comes down to what you're defining as temperature. Recall that temperature simply accounts for the flow of heat into whatever you've defined as your thermometer. If your thermometer couples to translational kinetic energy but not to gravitational potential energy, then you get the situation above.

I'll leave it to someone else to answer in terms of solid materials or inverted populations.

  • $\begingroup$ Could you give some references concerning this subject? $\endgroup$ Dec 14, 2017 at 12:43

We don't need to go to astrophysics for this. In the reversible expansion of a plain vanilla ideal gas, if one does not add sufficient heat, the temperature will drop (and, by this definition, the heat capacity will be negative). This can happen any time work is done such that there is not enough heat added to increase the internal energy. This is why $dQ/d\theta$ is such a poor way of defining heat capacity. When defined this way, heat capacity is not even a physical property of the material. In classical thermodynamics, heat capacity is more properly defined in terms of the partial derivatives of internal energy and enthalpy with respect to temperature.

  • $\begingroup$ So the be clear you're referring to a scenario where we are adding heat to a gas but it is expanding at a rate great enough to lower the temperature faster than the added heat will raise the temperature? $\endgroup$ Feb 1, 2016 at 2:36
  • $\begingroup$ No. It doesn't depend on rate. I said "reversible," so the expansion rate is very slow. In an adiabatic reversible expansion, the temperature of the gas drops (even though no heat is added or removed). If heat were to be added during the expansion, it might not be enough to entirely cancel out the temperature drop. $\endgroup$ Feb 1, 2016 at 2:52
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    $\begingroup$ "not add sufficient heat, the temperature will drop.." not exactly what the OP asked. Your system will cool regardless of external heat application. The question is: take a stable system and add heat. Can the temperature go down? $\endgroup$ Feb 1, 2016 at 14:05
  • $\begingroup$ Is this a more accurate interpretation of what the OP asked: Can the temperature of a pure substance or a mixture of constant composition decrease as its internal energy increases at constant volume? $\endgroup$ Feb 1, 2016 at 14:41

There are two different definitions of heat capacity, heat capacity at constant volume and heat capacity at constant pressure. The reversible expansion of an ideal gas cannot be done at constant volume. It cannot be done at constant pressure without adding heat.


Short answer is "no". The theory shows that heat capacities are positive. The negative heat capacities mentioned in the literature are based in misunderstandings of this theory.

For instance, the astrophysicists' argument uses the virial theorem to transform the sum of kinetic and potential energy $E=K+\Phi$ into $E= -K$ and then uses $K= \frac{3}{2} Nk_BT$ to get

$$C_V \stackrel{wrong}{=} \frac{dE}{dT} = -\frac{3}{2} Nk_B$$

which is a negative quantity, but is not the heat capacity of the system. The mistake is that the heat capacity $C_V$ is defined by a partial derivative at constant volume

$$C_V = \left( \frac{\partial E}{\partial T}\right)_V$$

The kinetic energy is a function of temperature, whereas the potential energy is a function of volume $E(T,V) = K(T) + \Phi(V)$, which means

$$C_V = \left( \frac{\partial E}{\partial T}\right)_V = \frac{3}{2} Nk_B$$

and we recover a positive heat capacity in agreement with both Schrödinger statistical mechanics theorem and with classical thermodynamic stability theory.


It seems my reply is being down-voted with invalid arguments such as "there is no confining volume", $(3/2)Nk_B$ is only for ideal gases, etcetera).

First, I am not considering the more general case, I am simply refutting the usual astrophysicists' argument and explaining (i) why they are not computing heat capacities, (ii) why finding regions where $(dE/dT) < 0$ does not identify systems with negative heat capacity, and (iii) why the statement "when heat is absorbed by a star, or star cluster, it will expand and cool down" is ignoring that volume is not held constant and therefore that $dQ \neq dE$.

My above answer used an abstract model, but we can be more specific. Consider the system discussed in [1]. The energy is $E = aT–b/r$, with $a = (3/2)Nk_B$ and $b$ an unspecified constant. Differentiating this energy and using the first numbered equation in [1], we obtain $dE = adT + PdV$ and a positive heat capacity for this gravitational system

$$C_V = \left( \frac{\partial E}{\partial T}\right)_V = a >0$$

D. Lynden-Bell and R. M. Lynden-Bell obtain the wrong result

$$C_V \stackrel{wrong}{=} \frac{dE}{dT} = -a$$

because they are not computing a heat capacity. Additional mistakes and physical inconsistencies in the arguments of LBLB are discussed in [2].

[1] Negative heat capacities do occur. Comment on “Critical analysis of negative heat capacities in nanoclusters” by Michaelian K. and Santamaría-Holek I. 2008: EPL 82(4), 43001. Lynden-Bell D.; Lynden- Bell, R. M.

[2] Reply to the Comment by D. Lynden-Bell and R. M. Lynden-Bell 2008: EPL 82(4), 43002. Michaelian K.; Santamaría-Holek I.

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    $\begingroup$ This counterargument against negative heat capacity in gravitational systems is wrong: first of all, usually there is no confining volume in gravitational systems. Even more important, $E$ is the average energy and usually the average value of $\Phi$ is a function of $T$ as well as of $V$. Otherwise, all systems would have the heat capacity of the ideal gas. $\endgroup$ Apr 29, 2019 at 11:00
  • $\begingroup$ @GiorgioP Above remarks are useless. (i) Lyndell-Bell considers systems with spherical volume. More general geometries can be considered. Even if we admit there is no "confining volume" for some systems, this would mean that $C_V$ is not defined for those systems, not it is negative. (ii) I have not considered the more general possible system, that is why I take kinetic energy as $(3/2)Nk_BT$ and potential energy as $r^{-n}$ as Lyndell-Bell does. $\endgroup$
    – juanrga
    May 7, 2019 at 17:29
  • $\begingroup$ (iii) I could consider a more general $\Phi(T,V)$; but still the partial derivative would be different than the total derivative than Lynden-Bell takes. I.e. the astrophysicists' argument continues being wrong. (iv) The heat capacity I have used as illustration is not exclusive to ideal gases. For instance, the internal energy of van der Waals gas is $E=(3/2)Nk_BT - a(N^2/V)$, with the potential energy not depending on temperature. Taking the partial derivative one can easily see that $C_V=(3/2) Nk_B$ is also valid for real gases of the Van der Waals kind. $\endgroup$
    – juanrga
    May 7, 2019 at 17:38

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