= Solution
For a complete single-particle state, the <Fermi-Dirac distribution> is
$$
\boxed{\langle n(\epsilon)\rangle=\frac1{\exp[(\epsilon-\mu_{\rm ch})/(k_BT)]+1}.}
$$
Here $\mu_{\rm ch}$ denotes the <chemical potential>, distinct from the magnetic-moment symbol in (b). At zero <temperature> it approaches the <Fermi energy> $E_F$ and the occupation has a sharp edge. At $T\ll T_F=E_F/k_B$, thermal changes occur within an <energy> interval of order $k_BT$ about that edge.
The <Fermi-edge heat-capacity counting> argument uses the fact that only order $g(E_F)k_BT$ <electrons> can be rearranged because states well below the edge are blocked by the <Pauli exclusion principle>. Each active excitation carries <energy> of order $k_BT$, so $U(T)-U(0)\sim g(E_F)(k_BT)^2$ and \b[$C_{V,e}\propto T$]. For a smooth <density of states>, the <Sommerfeld expansion> gives
$$
\boxed{C_{V,e}=\frac{\pi^2}{3}g(E_F)k_B^2T.}
$$
Here $g$ is the total single-particle <density of states>, including <spin>. In a three-dimensional free-electron gas, $g(E_F)=3N/(2E_F)$, giving $C_{V,e}=(\pi^2/2)Nk_B(T/T_F)$. This is the low-temperature normal-state electronic contribution, not the lattice <heat capacity> or a description across a superconducting transition.<Quantum Gases by David Tong>[Sections 3.6.2–3.6.4: Fermi edge, low-temperature heat capacity and Sommerfeld expansion]{ref}
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2016/paper-1-fermi-edge.png]
{title=Finite temperature rounds the Fermi edge over an energy scale of order kBT}
{source=paper-1-fermi-edge.py}
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