{title2=$x=\mu B/(k_BT)$}

Use the <independent spin-one-half two-level system> with fixed $N$ distinguishable sites and <Hamiltonian> levels $E_\pm=\pm\mu B$, where $\mu>0$ is the magnitude of the field-axis <magnetic moment> and $B>0$ is a uniform externally controlled <magnetic field>. Assume positive <temperature>, no interactions between sites, and no appreciable lattice or load <entropy>. The two levels differ by $2\mu B$. With <inverse temperature> $\beta$, the <canonical ensemble> gives the one-site and total <partition functions>
$$
\beta=\frac1{k_BT},\qquad x=\beta\mu B,\qquad
z=e^x+e^{-x}=2\cosh x,\qquad Z=z^N,
\qquad p_-=\frac{e^x}{2\cosh x},\quad p_+=\frac{e^{-x}}{2\cosh x}.
$$
Here the minus subscript denotes the lower <energy> $-\mu B$. Factorization of the <partition function> requires additive <energy> and independent sites; there is no particle-exchange factor $N!$ for these fixed distinguishable sites.

Differentiate the <partition function>, holding $B$ fixed when differentiating with respect to $\beta$. The <internal energy>, <Helmholtz free energy>, total field-axis <magnetic moment> $m$, and <magnetic entropy> are
$$
U_{\rm mag}=-\partial_\beta\ln Z=-N\mu B\tanh x,\qquad
F=-Nk_BT\ln(2\cosh x),
$$
$$
m=-\left(\frac{\partial F}{\partial B}\right)_T=N\mu\tanh x,\qquad
\frac{S_{\rm mag}}{Nk_B}=\frac{\ln Z+\beta U_{\rm mag}}N
=\ln(2\cosh x)-x\tanh x=:s(x).
$$
The extensive moment $m$ is distinct from <magnetization>, the moment per unit <volume (thermodynamics)>. Direct substitution of $p_\pm$ into the <Gibbs entropy> gives the same $s(x)$ independently of the free-energy <derivative>.

At fixed $B$, the magnetic <heat capacity> is
$$
C_{B,{\rm mag}}=\left(\frac{\partial U_{\rm mag}}{\partial T}\right)_B
=Nk_Bx^2\operatorname{sech}^2x\geq0,\qquad
\frac{ds}{dx}=-x\operatorname{sech}^2x<0\quad(x>0).
$$
Strict decrease makes $s$ invertible on $x>0$. Constant <magnetic entropy> therefore fixes $x$, not $B$ or $T$ separately:
$$
\frac{B}{T}=\text{constant},\qquad
\frac{T_f}{T_i}=\frac{B_f}{B_i}\quad(B_i,B_f,T_i,T_f>0).
$$
The <partition function> may be used to label these equilibrium states even though the isolated cooling leg has no external <heat bath>. Actual thermalization of the lattice and load additionally requires a sufficiently fast <spin-lattice relaxation>; perfectly uncoupled spins by themselves cannot refrigerate an attached object.

The limiting cases are instructive. For $x\ll1$, $s(x)=\ln2-x^2/2+O(x^4)$, $m\simeq N\mu^2B/(k_BT)$, and $C_{B,{\rm mag}}\simeq Nk_Bx^2$. The moment response reproduces <Curie's law>. For $x\gg1$, $s(x)\simeq(1+2x)e^{-2x}$ and the upper-state occupation is exponentially small. Taking $B\to0$ at fixed $T>0$ gives $s\to\ln2$; taking $T\to0$ at fixed $B>0$ gives $s\to0$. The origin has a path-dependent limit. Extrapolating a line of constant $x$ to $B=T=0$ does not construct a usable zero-<temperature> state or prove that a real material retains independent degenerate spins there.


 Two-level paramagnetic cooling model (source code)