# Why does the electronic contribution to the specific heat of a conductor not depend on the heat supplied to it?

The usual explanation given for why electrons do not contribute much to the specific heat of a material (for the purposes of the question consider a conductor) is that the fermi energy of conductors is pretty large (for copper, it's 7 eV) compared to the thermal heat produced (0.026 eV) in the conductor at about 300 K, so that only electrons in the band within 0.026 eV of the fermi level can thermally participate.

I have an obvious question here - if we are actually supplying heat to such a conductor, what would happen if the energy we're supplying is large compared to the 7 eV of the Fermi level? When I think of gases, it makes sense that the specific heat should be independent of the heat supplied, but in the case of conductors (considering the electronic contribution) I think it's more like:

greater the heat supplied ---> more electrons participate thermally ---> more electrons absorb that heat and thus increase the heat capacity.

Of course, I am making a foolish mistake here, but I can't tell where exactly. Also, I don't know anything about solid state physics, since I'm only in high school.

• Well, if you are supplying 7eV per electron in the conduction band you will have a very energetic metal gas on your hands... Mar 22, 2021 at 15:27

This is a slightly more formal way of saying, I think, the same thing as you mean by “greater the heat supplied … increase the heat capacity”. Actually this is easiest seen at very low temperatures, where the heat capacity from phonons (vibrations of the atomic nuclei) is small. Under these circumstances, the heat capacity goes as $$C = \beta T + \gamma T^3$$, where the $$T$$ contribution comes from the electrons and the $$T^3$$ from the vibrations. Thus a plot of $$C/T$$ against $$T^2$$ gives a straight line, which can be seen in experimental data.