Electron heat capacity vs phonon heat capacity For metals, the heat capacity has two contributions. The phonon contribution is proportional to $T^3$, while the electron contribution is proportional to $T$, as $T$ goes to zero. Therefore, for sufficiently low $T$, the electron contribution will dominate, while for sufficiently high $T$, the phonon contribution will dominate.
Does anyone know any specific values of such transition or crossover temperature? Is it typically on the order of $K$ or $10K$?
 A: Heat capacity data for elements is well researched. Picking Cu as a good metal where one would expect the electron contribution to be large, Google
(Googled 'Cu heat capacity vs temperature') leads quickly to nist.gov with a copy of (from Web Of Science search):
HEAT-CAPACITY OF REFERENCE MATERIALS - CU AND W
By: WHITE, GK; COLLOCOTT, SJ
JOURNAL OF PHYSICAL AND CHEMICAL REFERENCE DATA   Volume: ‏ 13   Issue: ‏ 4   Pages: ‏ 1251-1257
As noted in the introduction,

For the reference solids discussed here, the vibrational energy is the major contribution above liquid-helium temperatures.

One does not normally consider W to be a great metal, electron conduction wise, so I think the principle applies broadly, so under 10K, closer to 4K.
A: I believe you can make an approximate calculation of this temperature for a free electron case.  From Kittel (or many other sources), the approximation for heat capacity of phonons as low temperature is:  $$C_{ph}=234N_{ph}k_B(\frac{T^3}{\theta^3})$$  Where $\theta$ is the Debye Temperature and $k_B$ is Boltzmann's constant, and $N_{ph}$ is the number or primitive cells in the specimen.
For a metal, in the free electron case: $$C_{el} = 0.5\pi^2N_{el}k_B(\frac{T}{T_F})$$ Where $T_F$ is the metals Fermi Temperature and $N_{el}$ in this case is the total number of orbitals in the free electron sphere (including spin).
So for the crossover point you equate these two expressions and solve for T and you get: $$T=\sqrt{\frac{\pi^2\theta^3N_{el}}{468N_{ph}T_F}}$$  If you have one electron per unit cell, then I believe that $N_{ph}$ and $N_{el}$ should cancel out.  If you have  2 electrons per unit cell, then this ratio is 2 and etc.
Sodium, for which a free electron approximation should be good, has a Debye Temperature of 158 K, a Fermi Temperature of 37500 K and 1 electron per unit cell.  Thus T calculates out to be approximately 1.5 K - if I've done the math on my calculator correctly!
Note: The actual formula for the electron heat capacity is: $$C_{el}=\frac{1}{3}\pi^2D(\epsilon_F)k_B^2T$$  Thus if you know the density of states at the Fermi Level for the metal of interest, $D(\epsilon_F)$ you can make a better approximation.
