Pomeranchuk entropy-balance model ($T(x)=T_0-(1-x_0)(T_0-T_i)/(1-x)$)

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 entropiesThe 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 giveswhen $b>aT$ and $x<1$. Define $T_0=b/a$; solving between two states givesIn particular, stipulate $x_0=0$, $T_i=T_0/4$ and $x=1/8$. ThenThe last expression is the local reversible heat requirement per mole frozen at the final state, divided by $b{T}_0$; 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.

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 $dT/dx=-T(S_s-S_l)/[(1-x)C_l+x{C}_s]$. 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.