= Solution
{c}
For the degenerate <electron gas>, let $D_g=Am/(\pi\hbar^2)$ include both <spin> states. The <Sommerfeld expansion> gives
$$
U_g(T)-U_g(0)=\frac{\pi^2}{6}D_g(k_BT)^2,\qquad
C_v=\frac{\pi^2}{3}D_gk_B^2T.
$$
For the gapless symmetric <intrinsic semiconductor>, part (d) gave the same coefficient with $D_s=Am^*/(\pi\hbar^2)$. Thus for equal sample areas and equal <spin> and <valley degeneracies>,
$$
\boxed{\frac{C_v^{\rm tot}}{C_v}=\frac{m^*}{m};\qquad
\frac{C_v^{\rm tot}}{C_v}=1\text{ if }m^*=m.}
$$
The equality often intended in this comparison requires the same <density of states>. The source sets $m_e^*=m_h^*$ but does not explicitly equate them to the <electron gas> <mass> $m$. Without that additional assumption the ratio is $m^*/m$, or more generally $D_s/D_g$ if sample areas or <degeneracy factors> differ. For example, $m^*=2m$ with equal areas satisfies the stated equal-<effective mass> condition in the <semiconductor> and gives ratio two. The filled <Fermi sea> does not add a factor $N_e/N_c$: only the thermal shell near its <Fermi energy> contributes to its <heat capacity>.
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