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Using the concept of equipartition of energy is it possible to calculate the specific heat of aluminium at room temperature ?

For gas molecules it would have been easy as we could use $C_v=\dfrac{R}{\gamma-1}$ or $C_p=\dfrac{\gamma R}{\gamma-1}$. However, as we know aluminium is not a gas at room temperature. So is it possible to calculate it's specific heat at room temperature?

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At room temperature you have energy (per atom) of 3kT/2 from kinetic energy of translational degrees of freedom and 3kT/2 from potential energy of vibrations, so you have 3kT per atom. The electron system at room temperature is degenerate and does not provide significant contribution to heat capacity. So you have the Dulong -Petit law. Please see http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/Dulong.html#c1 and http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html (last column).

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  • $\begingroup$ What about rotational degrees of freedom ? And how do you know number of vibrational degrees of freedom is 3 ? $\endgroup$ – user139621 Mar 5 '17 at 13:28
  • $\begingroup$ Also, isn't $\frac{k}{2T}$ energy per degree of freedom only valid for gas molecules ? $\endgroup$ – user139621 Mar 5 '17 at 13:29
  • $\begingroup$ I read the links but it gives very vague explanations.. $\endgroup$ – user139621 Mar 5 '17 at 13:29
  • $\begingroup$ @Mystic: Rotational degrees of freedom of one-atom molecules are degenerate ("frozen") at room temperature -this is a quantum effect. $\endgroup$ – akhmeteli Mar 5 '17 at 13:34
  • $\begingroup$ Do you know any source/link which states the number of different types degrees of freedom for solid atoms ? By the way, thanks a lot =) $\endgroup$ – user139621 Mar 5 '17 at 13:39

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