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Does any known substance have a heat capacity at constant volume ($C_V$) per mole of atoms greater than $3k_B$ ~ 24.94 J/(mol K)?

In order to count, the substance must actually be made of atoms, that is, ordinary nuclei and electrons.

If so, what are the extra degrees of freedom responsible for this unusually large heat capacity?

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    $\begingroup$ ""the substance must actually be made of atoms,"" :=) not easy to avoid that! By restriction to "monoatomic" crystals or metals You do not get the interesting exceptions. For those primitive solids there are no cases higher than 3kB $\endgroup$
    – Georg
    Commented May 13, 2011 at 9:36
  • $\begingroup$ I'd imagine that electronic degrees of freedom would be extra on top? $\endgroup$
    – genneth
    Commented May 13, 2011 at 10:15
  • $\begingroup$ @genneth Think where (temperature) "electronic" degrees are relevant $\endgroup$
    – Georg
    Commented May 13, 2011 at 11:21
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    $\begingroup$ @Georg I don't think we're understanding each other. By saying "the substance must actually be made of atoms, that is, ordinary nuclei and electrons", I didn't mean to exclude any ordinary matter. I only meant to exclude things like neutron star material, quark-gluon plasma, etc., because those are not made of just nuclei and electrons. Anything made of nuclei and electrons is fine. $\endgroup$
    – Keenan Pepper
    Commented May 13, 2011 at 23:01
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    $\begingroup$ @Georg: I know; but that doesn't preclude interesting things when the electrons are strongly interacting, when the Fermi energy is massively reduced. Also, even if there were no itinerant electrons, local spins can definitely contribute a bit --- after all, we're looking for any violation, not just "big" ones? $\endgroup$
    – genneth
    Commented May 14, 2011 at 12:23

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As genneth said in a comment, any metal around (or maybe somewhat above) room temperature should have a higher heat capacity than $3k_B$ per atom.

Each vibrational degree of freedom (a.k.a. phonon mode) has a heat capacity of $k_B$ as long as the temperature $T$ and vibrational frequency $\nu$ satisfy $k_BT\gg h\nu$ (if this is not satisfied, the heat capacity is less than $k_B$). There are three phonon modes per atom, so phonons give you $3k_B$, as long as the temperature is high enough. For example, in gold, all the phonon frequencies are less than 5 THz; 5 THZ corresponds to 240K; so at room temperature the phonon heat capacity is almost $3k_B$ (but a bit less).

(I chose gold as an example because its atoms are heavy so they vibrate slowly. Metals with lighter atoms have higher vibration frequencies so a higher required temperature to get the full $3k_B$.)

On top of the phonons, a metal also has heat capacity from kinetic energy of the free electrons. So altogether it can be more than $3k_B$.

For example, I looked up gold's heat capacity (.128 or .129 J/gK) and atomic mass (196.97) and got $3.03k_B$ to $3.06k_B$ per atom.

(I'm a bit surprised it's not higher, since each atom should contribute at least one free electron, and a free electron would be expected to have $1.5k_BT$ of translational kinetic energy. I guess it's too simplistic to treat the electrons like non-interacting free particles. For example, maybe there is a ceiling on electron kinetic energy because of the band structure, or because of velocity-dependent phonon scattering? I'm not sure.)

Other possible degrees of freedom that provide extra heat capacity in some solids include plasmons, magnons, excitons, polaronic excitations, and many others. :-)

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    $\begingroup$ The electronic part of the heat capacity is so low because at normal temperatures (~300 K), the electron gas is highly degenerated and most of the electrons are in the ground state. For the electrons to become important, we would have to heat the metal to the Fermi temperature, which is for various metal several tens of thousands of Kelvin and thus far above its melting/boiling point and thus irrelevant. Interestingly, the simplification of treating electrons as non-interacting particles is a very good one, because the electric charges are shielded by the nuclei. $\endgroup$
    – Kasper
    Commented Jun 26, 2011 at 21:03

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