= Solution
The conventional <face-centered cubic lattice> cell contains four atoms, so
$$
n=\frac4{(0.36\times10^{-9})^3}=8.57\times10^{28}\ \mathrm{m^{-3}}.
$$
The preceding free-electron and Debye expressions give
$$
\boxed{T_F=8.23\times10^4\ \mathrm K,\qquad \theta_D=354\ \mathrm K},\qquad E_F\simeq7.09\ \mathrm{eV}.
$$
At low <temperature> the <electronic and lattice heat-capacity crossover in a monovalent metal> is found from
$$
\frac{C_e}{C_{\rm ph}}=\frac{5\theta_D^3}{24\pi^2T_FT^2}=\left(\frac{T_*}{T}\right)^2,
\qquad
\boxed{T_* =\left(\frac{5\theta_D^3}{24\pi^2T_F}\right)^{1/2}=3.38\ \mathrm K}.
$$
Thus the electronic term dominates for $0<T\lesssim3.4$ K within the normal-metal model. Both low-temperature approximations are well controlled at that crossover. At exactly zero <temperature> both <heat capacities> vanish; below sufficiently small finite-size <energy> scales a continuum model also needs revision.
Room <temperature> is not $T\ll\theta_D$: using the cubic formula at 300 K would incorrectly predict $C_{\rm ph}\simeq142nk_B$, far above the harmonic high-temperature limit $3nk_B$. Instead use the <Dulong-Petit law> as an estimate:
$$
\boxed{\frac{C_e}{C_{\rm ph}}\simeq\frac{(\pi^2/2)nk_B(300/T_F)}{3nk_B}
=6.0\times10^{-3}}.
$$
Keeping the finite cutoff in the <Debye heat capacity> gives
$$
C_{\rm ph}=9nk_B\left(\frac{T}{\theta_D}\right)^3
\int_0^{\theta_D/T}\frac{x^4e^x}{(e^x-1)^2}\,dx.
$$
At 300 K this is $2.8006nk_B$, whereas $C_e=0.017994nk_B$, giving $\boxed{C_e/C_{\rm ph}=6.42\times10^{-3}}$. Either appropriate room-temperature estimate shows that the lattice contribution dominates. The electronic approximation remains excellent because $300\ll T_F$.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2015-copper-heat-capacity.png]
{title=Electronic and full Debye lattice heat capacities of the stated copper model, with the low-temperature crossover and room temperature marked}
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