= Solution
\b[Canonical <probabilities>.] A <canonical ensemble> describes a system of fixed <particle number> and fixed external parameters such as <volume (thermodynamics)>, weakly coupled to a much larger <thermal reservoir>. They exchange <energy> and reach <thermal equilibrium>, while the reservoir's <temperature> stays effectively constant. The interaction <energy> is negligible in the state counting. The combined isolated system is described by equal <probabilities> for its accessible <microstates>.
If its total <energy> is $E_{\mathrm{tot}}$, the <probability> of a particular system <microstate> $i$ is proportional to the reservoir multiplicity at $E_{\mathrm{tot}}-E_i$. Writing that multiplicity as $e^{S_R/k_B}$, expand its <entropy>:
$$
S_R(E_{\mathrm{tot}}-E_i)
=S_R(E_{\mathrm{tot}})-\left(\frac{\partial S_R}{\partial E_R}\right)E_i+\cdots
=S_R(E_{\mathrm{tot}})-\frac{E_i}{T}+\cdots.
$$
The large-reservoir limit suppresses the higher terms. Normalization therefore gives
$$
\boxed{P_i=\frac{e^{-\beta E_i}}{Z},\qquad
Z=\sum_j e^{-\beta E_j},\qquad\beta=\frac1{k_BT}.}
$$
The sum counts individual <microstates>; when grouping them by <energy>, their <state degeneracies> must be included. This $Z$ is the <canonical partition function>.
\b[Classical orientational <heat capacity>.] Hold the <magnetic field> fixed. Orientations of the <magnetic dipole moment> have the <solid angle> <measure> $d\Omega=\sin\theta\,d\theta\,d\varphi$. With $x=\beta mB$, the angular <partition function> is
$$
Z_{\mathrm{ang}}=\int e^{x\cos\theta}d\Omega
=2\pi\int_{-1}^{1}e^{xu}du
=4\pi\frac{\sinh x}{x}.
$$
Choosing a normalized angular <measure> would remove $4\pi$, with no effect on the <energy> <derivatives>. Differentiating at fixed $B$,
$$
U=-\frac{\partial\ln Z_{\mathrm{ang}}}{\partial\beta}
=-mB\left(\coth x-\frac1x\right)
=-mB\coth x+k_BT.
$$
The angular average $\langle\cos\theta\rangle=\coth x-1/x$ is the <Langevin function>. Since $dx/dT=-x/T$,
$$
\boxed{C_B=\left(\frac{\partial U}{\partial T}\right)_B
=k_B\left[1-\left(\frac{x}{\sinh x}\right)^2\right].}
$$
This is the <heat capacity of a classical magnetic dipole>, in $\mathrm{J\,K^{-1}}$ for one <magnetic moment>. The word specific here refers to that single moment; $N$ independent <magnetic moments> have $N$ times this <heat capacity>. At $B=0$ the <energy> is independent of orientation and $C_B=0$, obtained by the continuous $x\to0$ limit. At fixed nonzero $B$, the classical low-temperature limit instead gives $C_B\to k_B$.
\b[Equipartition and its proof.] The <equipartition theorem> states that an independent quadratic term $a q^2/2$ in a classical <Hamiltonian>, with $a>0$, contributes $k_BT/2$ to its mean <energy>. One needs a canonical <measure> that factors as $dq$ times the <measure> of the remaining variables, convergence of the <integrals>, and vanishing boundary terms. The coordinate must not be restricted by a thermally important hard boundary or a coupled <constraint>.
For $H=a q^2/2+H_0$, <integration by parts> gives
$$
0=\int_{-\infty}^{\infty}\frac\partial{\partial q}
\left(qe^{-\beta H}\right)dq
=\int_{-\infty}^{\infty}\left(1-\beta q\frac{\partial H}{\partial q}\right)e^{-\beta H}dq.
$$
Integrate over all other variables and divide by $Z$. It follows that
$$
\left\langle q\frac{\partial H}{\partial q}\right\rangle=k_BT,
\qquad
\boxed{\left\langle\frac{a q^2}{2}\right\rangle=\frac12k_BT.}
$$
The same reasoning applies to quadratic momenta among the <canonical variables>. It is the number of quadratic <energy> contributions that is counted, not an unrestricted count of coordinate labels.
\b[Two soft orientational directions.] When $x\gg1$, the moment is concentrated near alignment. Introduce local <Cartesian coordinates> in the <tangent plane> $q_1=\theta\cos\varphi$ and $q_2=\theta\sin\varphi$. To leading order,
$$
E=-mB+\frac{mB}{2}(q_1^2+q_2^2)+O(mB\theta^4),
\qquad d\Omega\simeq dq_1dq_2.
$$
The two quadratic terms each contribute $k_BT/2$, so
$$
\boxed{U=-mB+k_BT+o(T),\qquad C_B\longrightarrow k_B.}
$$
This agrees with the exact angular calculation. Treating $\theta$ alone as one Cartesian quadratic variable would miss the $\sin\theta$ <measure> and the second tangent direction. There are no extra rotational kinetic terms in the stipulated <Hamiltonian>; adding such terms would be a different model.
\b[Quantum <spin> one.] The <spin> projections are $m_s=-1,0,1$, with $S_z=\hbar m_s$. Put $\Delta=\hbar|\gamma|B>0$, where $\gamma$ is the <gyromagnetic ratio>. The three <energies> are $-\Delta,0,+\Delta$, irrespective of the sign of $\gamma$. Relative to the <ground state> they are $0,\Delta,2\Delta$. Thus, with $q=e^{-\beta\Delta}$,
$$
Z=e^{\beta\Delta}(1+q+q^2),\qquad
U=-\Delta+\Delta\frac{q+2q^2}{1+q+q^2}.
$$
Differentiation, or the <canonical energy fluctuation> identity $C_B=k_B\beta^2\operatorname{Var}(E)$, gives
$$
C_B=k_B(\beta\Delta)^2
\frac{q(1+4q+q^2)}{(1+q+q^2)^2}.
$$
In the requested limit,
$$
\boxed{C_B\sim k_B\left(\frac{\Delta}{k_BT}\right)^2e^{-\Delta/(k_BT)},
\qquad\Delta=\hbar|\gamma|B.}
$$
This <spin-1 paramagnet> has a nondegenerate <ground state> and a finite <spectral gap>. Excitations are exponentially rare, so its <heat capacity> tends to zero. Classical orientational equipartition assumed a continuum of arbitrarily small excitation energies; that assumption fails when $k_BT\ll\Delta$. This is a change in the accessible state spectrum, not a violation of the <equipartition theorem> within its classical domain.
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