= Pomeranchuk entropy-balance model
{c}
{title2=$T(x)=T_0-(1-x_0)(T_0-T_i)/(1-x)$}
An original deliberately local model makes the cooling quantitative. Fix a total amount $n$ of material, let $x$ be the mole fraction of <solid>, and stipulate molar <entropies>
$$
S_l(T)=aT,\qquad S_s(T)=b,\qquad a,b>0,
\qquad \frac{S_{\rm total}}n=(1-x)aT+xb.
$$
The constants $a$ and $b$ have units of molar <entropy> per <temperature> and molar <entropy>, respectively. Ignore their <pressure> dependence and the cell's own <heat capacity>. This is a local <thermodynamics> model, not an extrapolation to zero <temperature>. A constant positive $b$ cannot remain valid once the <solid>'s <spins> order.
For a <reversible thermodynamic process> with no <heat> exchange, total <entropy> is constant. Differentiating gives
$$
(1-x)a\,dT+(b-aT)\,dx=0,
\qquad \frac{dT}{dx}=-\frac{b-aT}{a(1-x)}<0
$$
when $b>aT$ and $x<1$. Define $T_0=b/a$; solving between two states gives
$$
T(x)=\frac{b}{a}-\frac{(1-x_0)(b-aT_i)}{a(1-x)}.
$$
In particular, stipulate $x_0=0$, $T_i=T_0/4$ and $x=1/8$. Then
$$
\frac{T}{T_0}=\frac17,\qquad
\frac{S_{\rm total}}{nb}=\frac14,\qquad
\frac{Q_{l\to s}}{bT_0}=\frac{T}{T_0}\left(1-\frac{T}{T_0}\right)=\frac6{49}.
$$
The last expression is the local reversible <heat> requirement per mole frozen at the final state, divided by $bT_0$; it is not the <heat> exchanged during the isolated compression. That compression exchanges no <heat>. This distinction separates the latent-<heat> sign from the actual boundary <heat> budget.
\Image[/helium-three-pomeranchuk-entropy-balance.png]
{title=Why <freezing> can cool: an idealized <Pomeranchuk cooling> <entropy> balance}
{description=Original model in arbitrary units, not data. Left: stipulated $S_l=aT$ and $S_s=b$. Right: the isolated reversible path from $x=0$, $T/T_0=1/4$ to $x=1/8$, $T/T_0=1/7$, with fixed $S_{\rm total}/(nb)=1/4$. Only the illustrated local regime is modeled.}
The utility of the model is its explicit condition: $S_s>S_l$ makes $dT/dx<0$, while positive total <heat capacity> supplies a stable thermal response. More generally, for molar <entropies> depending only on <temperature>, let $C_l,C_s$ be the molar <heat capacities>. Then $dT/dx=-T(S_s-S_l)/[(1-x)C_l+xC_s]$. Incoming <heat> or <entropy production> adds terms to the balance and can overwhelm the ideal cooling. Neither the numerical ratios above nor the straight <entropy> lines are measurements of <helium-3>.
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