= Solution
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 capacity>
$$
C_m(T)=9Nk_B\left(\frac{T}{\theta_D}\right)^3\int_0^{\theta_D/T}\frac{x^4e^x}{(e^x-1)^2}\,dx.
$$
At low <temperature> the upper limit can be replaced by infinity. Since the integral then equals $4\pi^4/15$,
$$
\boxed{C_m(T)\sim\frac{12\pi^4}{5}Nk_B\left(\frac{T}{\theta_D}\right)^3\quad(T\ll\theta_D).}
$$
At high <temperature> each harmonic mode contributes $k_B$ by the <equipartition theorem>, giving the <Dulong-Petit law>
$$
\boxed{C_m(T)\longrightarrow3Nk_B\quad(T\gg\theta_D).}
$$
For one mole replace $Nk_B$ 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:
$$
3\frac{\mathcal V}{(2\pi)^3}\frac{4\pi k_D^3}{3}=3N,\qquad k_D=(6\pi^2n)^{1/3}.
$$
With the supplied <speed of sound> treated as the effective Debye speed, $\omega_D=vk_D$ and
$$
\boxed{\theta_D=\frac{\hbar v}{k_B}(6\pi^2n)^{1/3}\simeq\frac{\hbar(3600)}{k_B}(6\pi^2\times5.9\times10^{28})^{1/3}\simeq4.17\times10^2\,\mathrm K.}
$$
Here $k_D\simeq1.52\times10^{10}\,\mathrm{m^{-1}}$. Real crystals have different longitudinal and transverse sound speeds; the one-speed estimate uses the stated value, rather than inferring those unprovided branches.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2016-heat-capacity-and-conductivity.png]
{title=Debye lattice heat capacity and the squared magnitude of the Drude conductivity}
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