= Solution
The <Pauli exclusion principle> confines thermally excited conduction <electrons> to an <energy> interval of order $k_BT$ around the <Fermi surface>. Their excess <energy> is therefore proportional to $T^2$, giving an <electronic heat capacity> $\gamma T$. More precisely, $\gamma=(\pi^2/3)k_B^2g_{\rm mol}(\epsilon_F)$. In a three-dimensional <Debye model>, <acoustic phonons> have <density of states> proportional to $\omega^2$; their thermal <energy> scales as $T^4$, giving
$$
\boxed{C=\gamma T+\beta T^3,\qquad
\beta=\frac{12\pi^4R}{5\theta_D^3}}
$$
per mole of atoms. This assumes a normal <metal> without a lower-temperature superconducting or magnetic anomaly.
For $\theta_D\ll T\ll T_F$, the lattice approaches the Dulong-Petit value, so \b[$C_V\simeq3R+\gamma T$, usually dominated by $3R$]. Each classical vibrational mode contributes $k_B$ through kinetic and potential equipartition. This is a high-temperature solid-state regime, not a claim that the solid persists to arbitrarily high <temperature>; anharmonicity and $C_P-C_V$ can matter.
The <body-centred cubic> cell has two atoms, so $n_{\rm at}=2/a^3$. Counting three acoustic branches gives $k_D^3=6\pi^2n_{\rm at}$. By the Debye-temperature definition,
$$
\boxed{\omega_D=\frac{k_B\theta_D}{\hbar}=1.18\times10^{13}\ \mathrm{s^{-1}},
\qquad
v_D=\frac{k_B\theta_D}{\hbar(12\pi^2/a^3)^{1/3}}
\simeq1.25\times10^3\ \mathrm{m\,s^{-1}}}.
$$
This is an effective <Debye sound speed>; separate longitudinal and transverse speeds are not determined. For isotropic branches, $3/v_D^3=1/v_L^3+2/v_T^3$.
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