Temperature is the average of the kinetic energies of all molecules of a body. Then, why do we consider it a different fundamental physical quantity altogether [K], and not an alternate form of energy, with a dimensional formula derived from three initial fundamental quantities, length [L], mass [M] and time [T]?


Temperature is nothing else than energy per degree of freedom. It is purely for historical reasons that energy per degree of freedom is measured in Kelvin, and not in, say, micro-eV. It is just that these systems of units got fixed and became widely used before the statistical meaning of temperature became clear.

For the same reason, mass measured in kg and not in, say, Tera-eV.

If you would correct all of this, and apply more rational choices of units, you would end up with a natural system of units. This is what many physicists do in their professional lives. In such a system constants like the speed of light and Boltzmann's constant end up as being defined equal to unity. This makes it clear that these are not constants of nature, but man-made artifacts caused by the use of clumsy systems of units. In that respect Boltzmann's constant k is no different than the constant measuring the number of cubic inches in a gallon.

  • $\begingroup$ Do you happen to know of a page that translates legacy cruft like kelvin to the future standards you speak of? $\endgroup$ – Cees Timmerman Aug 23 '16 at 21:39

Particle physicists indeed do like to express temperature in the units of energy, usually in their most favorite unit of energy, a gigaelectronvolt.

Kelvins are used for historical reasons as well as for the sake of having reasonable numbers in everyday conditions. Before the temperature-energy relationship $$ E \sim kT$$ was realized in the late 19th century ($k=1.38\times 10^{-23} {\rm J/K}$), people used various temperature scales such as the Celsius scale (they still do). The Celsius scale divides the interval between the freezing and boiling points of water to 100 parts (percent of the length of the interval). The linearity is given by the volume of an ideal gas at fixed pressure, $V\sim T$.

Later, it was appreciated that the freezing point isn't a terrible fundamental special temperature worth the label "zero", so a shifted Celsius scale, the Kelvin scale, was defined. It's still convenient to use the Celsius degrees because it's sensible to talk about temperatures near 0 and not 300 kelvins.

Today, we know it's sensible to express the temperature via the equivalent energy per degree of freedom (times two), $E=kT$, but it's still reasonable to use kelvins and Celsius degrees because the room temperature is of order $10^{-20}$ joules.

  • $\begingroup$ So my room, being about 25.8 degree Celsius, is 258 zJ? $\endgroup$ – Cees Timmerman Aug 24 '16 at 13:10
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    $\begingroup$ You're on the right track but your number isn't right. The temperature as energy is calculated from the absolute temperature. 273.15+25.8 is 298.95 kelvins. Multiply it by 1.38 e-23 J/K and you get 4.12 e-21 joules or 4.12 zJ. How did you get the much higher figure? By the way, particle physicists etc. would prefer to express the temperature in electronvolts. Divide 4.12 e-21 by 1.602 e-19 and you get 0.0026 electronvolts for your room temperature. This gives you an idea how hard it is for the heat to thermally excite molecules etc. $\endgroup$ – Luboš Motl Aug 25 '16 at 9:12
  • $\begingroup$ I wondered why it was unreasonable, but now I know a joule is just too much. Thanks for the explanation. $\endgroup$ – Cees Timmerman Aug 26 '16 at 17:23
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    $\begingroup$ Joule is not too much for a human. But the temperature-as-energy means that the corresponding energy is given to every atom (more precisely every direction or "degree of freedom" in which it may move). And one joule per atom is huge, indeed. Even the LHC only pumps 6.5 GeV per proton. It's some $10^{10}$ eV or $10^{-9}$ joules, a nanojoule. The equivalent energy to the LHC energy-per-proton is some tens of billions of kelvins, you may calculate it precisely. $\endgroup$ – Luboš Motl Aug 27 '16 at 15:08
  • $\begingroup$ Yet the "negative kelvin / beyond infinite hot" experiment probably used less energy than the LHC. $\endgroup$ – Cees Timmerman Aug 30 '16 at 0:13

You actually explained why already by yourself and well done for noticing the problem. Dimensional analysis does show that temperature and energy are not the same.

Just to convince ourselves of this, lets look at the ideal gas equation for a single mole, (i.e. number moles n=1 below) of ideal gas:


With $nR=8.31\mathrm{m^2kg s^{-2}K^{-1}}$ looking like it has the units we use for energy over those for temperature (i.e. the units are $ \mathrm{\text{ } m^2kg s^{-2}K^{-1}}=JK^{-1}$).


What happens to the units now? The units in this expression do not amount tot units of energy. This means $\text{Energy}\neq\text{Temperature}$ and is what we expected.

As others have said, in various ways; temperature is a function of state. The sum of variables the function of state depends on tells us the number of degrees of freedom it has. These are represented by the number of invariables in the function of the state. i.e. general form:


Where subscript n, is a positive integer which gives the degree of freedom. The Empirical temperature is a single valued result found at a particular time (i.e. so we have a result that is independent of time) when we examine our function of state. Look at the function of state variables volume, V and pressure, p which are represented as follows:



Proof for existence of temperature (derivation is from C. J. Adkins equilibrium thermodynamics and paraphrased a bit here to illustrate the concept) added for anyone who wants to convince themselves of what I just said.

The condition for thermal equilibrium is described by the zeroth law of thermodynamics that is:

If two bodies A and B are in equilibrium with a third, C then they are also in equilibrium with each other.

Consider a simple example where we model a thermodynamic process involving fluid of fixed mass (i.e. independent of mass).

The condition for thermodynamic equilibrium for function, F of two arbitrary states,with body A at state 1 and and body C at state 3. (You may have plotted a p/V indicator diagram so it will help you to think of this now)


Next the condition for equilibrium for body B at state 2 and C at 3.


Solving both for $p_3$



and equating the two:


Solve for $p_1$. The reasoning for this change of variables did actually confuse me for a little while, but it is again just a matter of being comfortable manipulating expressions.







Which gives the conditions for thermal equilibrium the existence of temperature as a function of p and V.

Hence the result expressed earlier.


This should convince you that temperature and energy are not the same thing!

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    $\begingroup$ Downvoter care to explain? $\endgroup$ – Magpie May 5 '13 at 6:41
  • $\begingroup$ Concerning the first question, I think the first dimensional argument why temperature and energy are not the same is a good one, and indeed it could be that this is all what the OP wanted since the question is titled "Dimensional Analysis". $\endgroup$ – Dilaton May 8 '13 at 11:05
  • $\begingroup$ the rest confuses me a little bit. For example as Johannes said in his answer, temperature is the kinetic energy of a degree of freedom, so I think it should not depend on the number of degrees of freedom as I see it. In a system with different kinds of degrees of freedom in extreme nonequilibrium each degree of freedom can have its own temperature. In theormodynamics and statistical mechanics, the temperature is the intensive quantitiy (it does not depend on the size of the system) which corresponds to the energy of the system, so they are related. $\endgroup$ – Dilaton May 8 '13 at 11:10
  • $\begingroup$ When deriving the equilibrium state by maximazing the entropy, the temperature is the Lagrange multiplier corresponding to the energy. $\endgroup$ – Dilaton May 8 '13 at 11:10
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    $\begingroup$ I saw ;-) I do also see what you mean now about the degrees of freedom bit now. I think I misused the word depend there. I'll change it. $\endgroup$ – Magpie May 9 '13 at 20:57

Heat is the total energy of the molecules in a body caused due to there random kinetic movement. Thus, heat depends on the size of the body, as more the size of the body, more the molecules in the body, and more the molecules moving, more their combined kinetic energy, more the heat. Thus, a hot tub and a pan which have water with the same temperatures, may have varying amounts of heat, because the hot tub has more water and thus, more vibrating molecules.

Temperature, on the other hand, is the quantity which actually tells the hotness or coldness of a body. It DEPENDS on the AVERAGE kinetic energy of the molecules constituting the body (which is misinterpreted by some who think that temperature IS the average kinetic energy of the molecules constituting the body, who argue that it should be measured using the unit of energy.) Two bodies with the same amount of heat, may have varying temperatures, because the heat may either be caused due to there size or their temperature. For example, a large hottub and a pan which have the same amounts of heat, may vary in temperatures because the heat in the hottub is caused due to the large number of molecules while the heat in the pan is caused by the high temperature of water. Thus, heat depends on the size of the body and its temperature.

Here is a real life situation which might give you the difference between heat and temperature:

As discussed earlier, heat depends on size of the body and the bodies temperature and the temperature of the body. But in that definition, shouldn't a large iceberg have more heat then a hot pan (assuming you have not heated the water to like, a million degrees kelvin/ celsius/ fahrenheit)? Suprisingly, yes! But then why does the iceberg feel so cold and the pan so hot to touch? This is because we do not feel the heat of a body as much as its temperature. Temperature is the quantity which actually tells us the thermal state (hotness or coldness) of a body. It is true that increase in temperature results in increase in heat, but temperature is definitely not a measurement of heat!

  • $\begingroup$ See Magpie's answer. What you're saying is very unclear; are you saying the same thing as Magpie? . $\endgroup$ – Abhimanyu Pallavi Sudhir Sep 1 '13 at 6:52
  • $\begingroup$ Ok, with the anonymous paragraphing, I think it's clear enouggh now. $\endgroup$ – Abhimanyu Pallavi Sudhir Sep 1 '13 at 7:54
  • $\begingroup$ This is the only answer that explains it clearly to layman me. $\endgroup$ – Cees Timmerman Aug 24 '16 at 13:21

From the phenomenological point of view of thermodynamics, the unit of temperature is somewhat arbitrary and has been fixed by historical 'accident' - as long as $$ Q = T\cdot\Delta S $$ ends up being an energy, everything works out.

From the point of view of statistical mechanics, it makes sense to make the entropy $S$ unitless, same as the probabilities in terms of which it is defined.

This corresponds to setting Boltzmann's constant $k_B=1$ and makes temperature an energy. This as well as further unifications (like setting speed of light $c=1$ to unify the dimensions of space and time as well as mass and energy) are features of 'natural' systems of units.


Kinetic temperature is the average of the kinetic energies. The thermodynamic concept of temperature $T$ is more general.

Temperature measures the partial ratio of energy to entropy changes $T\equiv (\partial E / \partial S)$. Temperature cannot be considered an equivalent of energy, because the concept of entropy is needed for the definition as well. Notice from the definition that energy is an extensive quantity whereas temperature is an intensive quantity. It is possible to have a composite system with a given energy, but without a thermodynamic temperature (e.g. a composite system made of two thermally isolated solids: one hot and the other cold)

Using the fundamental equation of thermodynamics we find that $T=T(E,V,N,\dots)$, which implies that the concept of temperature surpasses that of energy. For instance, temperature can change whereas energy remain constant.


protected by Qmechanic Sep 1 '13 at 6:59

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