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In the three-dimensional Debye model, the lattice has $3N$ normal modes for $N$ atoms. Replacing their dispersion by a linear acoustic dispersion up to a cutoff gives the Debye heat capacityAt low temperature the upper limit can be replaced by infinity. Since the integral then equals $4{\pi }^4/15$,At high temperature each harmonic mode contributes $k_B$ by the equipartition theorem, giving the Dulong-Petit lawFor one mole replace $N{k}_B$ by $R$. The curve starts with zero slope, rises as $T^3$, and levels off toward the classical limit. This is the lattice contribution; a metal's electronic heat capacity is a separate low-temperature term.

The Debye wavevector follows by counting three polarizations inside the cutoff sphere:With the supplied speed of sound treated as the effective Debye speed, ${\omega }_D=v{k}_D$ andHere $k_D\simeq 1.52\times 10^{10}{\mathrm{m}}^{-1}$. Real crystals have different longitudinal and transverse sound speeds; the one-speed estimate uses the stated value, rather than inferring those unprovided branches.

For a degenerate Fermi gas, the Pauli exclusion principle restricts thermally changed occupations to an energy shell of width $k_{B}T$ around the Fermi energy. The number of active electrons is of order $g(E_F)k_{B}T\sim N{k}_{B}T/E_F$; each carries an excitation energy of order $k_{B}T$. Thus the thermal energy grows as $N(k_{B}T{)}^2/E_F$ and the electronic heat capacity is proportional to $T$.

More precisely, the fixed-particle-number Sommerfeld expansion givesHere $g$ is the total spin-summed density of states for the sample. For a three-dimensional parabolic band, $g(E_F)=3N/(2E_F)$, yieldingThe linear law concerns the electronic contribution; the low-temperature lattice heat capacity usually adds a $T^3$ term.

The Wiedemann-Franz law relates the electronic thermal conductivity of a normal degenerate metal to its electrical conductivity:An isotropic kinetic theory estimate gives $K_e=c_e{v}_F^2{\tau }_Q/3$. With electronic heat capacity density $c_e={\pi }^{2}n{k}_B^{2}T/(2E_F)$, $v_F^2=2E_F/m^{\ast }$ and $\sigma =n{e}^2{\tau }_{\sigma }/m^{\ast }$, the ratio is $L_0{\tau }_Q/{\tau }_{\sigma }$. Approximately elastic scattering with a smooth energy dependence gives equal heat and charge relaxation times. Inelastic electron-phonon scattering can make them different; the law also concerns the electronic contribution, not additional lattice heat conduction.^[ref]