= Solution
Take $\mu>0$ to be the magnitude of either <spin> projection's <magnetic moment>, so the independent magnetic levels are $E_\pm=\mp\mu B$. Their <partition function> and magnetic <energy> are
$$
Z_1=2\cosh x,\qquad Z=(2\cosh x)^N,\qquad U_{\rm mag}=-\partial_\beta\ln Z=-N\mu B\tanh x,
$$
where $\beta=1/(k_BT)$ and $x=\beta\mu B$. The <canonical entropy> is consequently
$$
\boxed{S=k_B\ln Z+\frac{U_{\rm mag}}T=Nk_B[\ln(2\cosh x)-x\tanh x].}
$$
This <independent-spin magnetic entropy> counts the two orientation states of each independent moment; translational <entropy> and interactions are excluded.
Along the <magnetic-entropy demagnetization adiabat>, constant magnetic <entropy> fixes $x$, hence \b[$B/T$ is constant]. Lowering the field therefore gives the ideal <adiabatic demagnetization> relation
$$
\boxed{T_f=T_i\frac{B_f}{B_i}.}
$$
A useful cycle first magnetizes isothermally while releasing heat to a reservoir, then thermally isolates the sample and lowers the field. Cooling the entire sample requires its other <heat capacities> and spin-lattice equilibration to be accounted for. Magnetic interactions, residual fields and heat leaks prevent extrapolating the independent-moment formula to an attainable zero <temperature>.<The Fundamentals of Statistical Mechanics by David Tong>[Canonical ensemble: two-state partition function, entropy and thermodynamic derivatives]{ref}
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2016/paper-1-paramagnetic-cooling.png]
{title=Magnetic entropy decreases with B over T, while an ideal demagnetization adiabat has constant B over T}
{source=paper-1-paramagnetic-cooling.py}
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