In differential form, the First Law of thermodynamics can be phrased as

$$ dU = T\,dS - P\,dV + \sum_{j = 1}^N \mu_j\, dn_j, \quad (1)$$

or equivalently as

$$ dU = \left( \frac{\partial U}{\partial S}\right)dS - \left( \frac{\partial U}{\partial V} \right) dV + \sum_{j = 1}^N \left(\frac{\partial U}{\partial n_j}\right)dn_j. \quad (2)$$

This latter expression shows how each of the intensive variables of the system (i.e. temperature, pressure and chemical potential) represent something akin to energy per unit extensive quantity. For example, one can think of pressure as the energy of the system per unit volume. Likewise, one can consider chemical potential to be the energy of the system per unit particle. Now, the analogy seems to break down when one tries to think of temperature in the same way. The temperature is… the system energy per unit of entropy? What the heck does that mean?

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    $\begingroup$ Note that the premise of your question is slightly wrong: pressure is not actually energy per unit volume, the chemical potential is only the (Gibbs free) energy per unit particle in the Gibbs ensemble (AFAIK this is the only time such a simple slogan holds true in thermodynamics). $\endgroup$
    – 5th decile
    Aug 20, 2023 at 15:26
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    $\begingroup$ There is a sign error in your second form $dU=...+\left(\frac{\partial U}{\partial V}\right)_{S,n}dV$ $\endgroup$
    – mike stone
    Aug 20, 2023 at 16:55
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    $\begingroup$ It might (or might not) be helpful to think of it as instead saying that inverse temperature is the system entropy per unit of energy. $\endgroup$
    – tparker
    Aug 21, 2023 at 15:40
  • $\begingroup$ In simpler times, temperature was defined from PV=RT in the limit as the pressure of a perfect gas goes to zero. Later it got defined in terms of statistical mechanics. I think both approaches mark the variable "temperature " as not a fundamental quantity but rather a basket holding more fundamental variables. $\endgroup$ Aug 23, 2023 at 17:13
  • $\begingroup$ @ZacharyCandelaria: have you heard of or encountered the equipartition theorem? $\endgroup$
    – TLDR
    Aug 24, 2023 at 0:53

3 Answers 3


The analogy does not break down at all, on the contrary it means exactly that: temperature is the system's internal energy per unit entropy increase while the other extensive parameters are kept constant. Entropy is the active agent of thermal interaction whose transport from one potential (temperature) to another is the reason for work performed, just as gravitational mass or electric charge are the agents of gravitational work or electric work, resp., mass or charge moves from one potential to another.

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    $\begingroup$ A small caveat: temperature is the derivative of the internal energy w.r.t. entropy not the ratio. It means that it can be interpreted as the increase of the energy per unit increase in the entropy not the total internal energy per unit of entropy. $\endgroup$ Aug 21, 2023 at 9:06
  • $\begingroup$ I appreciate this. However, I'm still having a hard time making sense of this. I can easily conceptualize a unit of volume or a unit mass of some substance, but what is a unit of entropy? $\endgroup$ Sep 2, 2023 at 20:34
  • $\begingroup$ A body has internal energy but can you conceptualize it? A "substance" or a thing has gravitational mass, has inertial mass (these two are the same), has electric charge, has magnetization and has entropy. An electron has electric charge, spin and magnetic dipole, but has no entropy, a large collection does. You can visualize a unit of mass by what it does, say you hang it on a rope and let it drop, then notice that by moving down it can charge a battery via a dynamo. You drop the temperature of an amount of entropy adiabatically and notice that by it you can also charge the same battery. $\endgroup$
    – hyportnex
    Sep 2, 2023 at 21:03

The answers saying that temperature is the energy per unit entropy are fine, but I think you must have already considered this and found it unsatisfying. Let me try to develop some intuition about entropy (which is a minefield, I know). Entropy is roughly proportional to the number of degrees of freedom or, more colloquially, with the number of moving parts.

For instance, for a molecule with N rotatable bonds and 3 accessible rotation states for each bond, the conformational entropy is $S = k \ln (3^N) = kN \ln(3)$. This is consistent with the idea that the entropy is proportional to the number of moving parts (in this case, rotatable bonds).

So, we can think of temperature as being roughly like the amount energy per degree of freedom or the amount of energy per moving part of the object.

Does this intuitive picture work? Let's say we have a small solid object, like a block of iron. Compared to a larger block of iron, it has few degrees of freedom. If the small block and large block have the same energy, the small block has a higher temperature than the large block. So, for simple situations like this, the picture gives the right intuition.

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    $\begingroup$ I like this intuition, but to clarify a bit --- $\ln \Omega$ is actually roughly equal to the number of degrees of freedom, whereas the number of microstates $\Omega$ goes up exponentially with the DOF. Think of two identical independent systems brought together --- the total DOF of the composite system doubles, while the number of microstates is squared. So thinking about temperature as energy per DOF is spot on, and also concurs with the intuition from kinetic gas theory. $\endgroup$
    – ratsalad
    Aug 21, 2023 at 20:52
  • $\begingroup$ @ratsalad Thanks, that was a mistake on my part. If you have N rotatable bonds each with 3 rotamers, you have $3^N$ microstates, so the entropy comes out proportional to the number of "moving parts". I'll adjust the explanation. $\endgroup$ Aug 21, 2023 at 21:33
  • $\begingroup$ I like this a lot. Thanks, @WaterMolecule! $\endgroup$ Sep 2, 2023 at 20:36

As hyportnex's nice answer says, there is nothing wrong with this analogy at all. The thermodynamic definition of temperature is, $$T = \frac{\partial U}{\partial S} $$ The fact that this is an incredibly weird and unintuitive statement is actually one of the reasons that thermodynamics is so powerful; it tells us surprising things about the behavior of the world in a truly fundamental way.

Just to build off of why there is some rationality behind the connection of entropy and temperature, one definition of the entropy in thermodynamics, attributed to Rudolf Clausius, is, $$ dS = \frac{\delta q_{rev}}{T}$$ where $\delta q_{rev}$ is the inexact differential for heat transfer in a reversible process. By doing some further work, you can arrive at the Clausius inequality for any process, $$ dS \geq \frac{\delta q}{T}$$ The reason that I bring this up is that the only things you need to argue that heat flows from hotter bodies to colder ones are the Clausius inequality and the second law of thermodynamics. This means that because of the spontaneity condition $dS \geq 0$ for an isolated system coming to equilibrium, we will always have changes in temperature that correspond to our physical intuition; heat will flow into the system if it is colder than the surroundings and out of it if it is warmer.


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