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An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]

For a local illustrative ice–water melting calculation, stipulate $T_m=273.15\mathrm{K}$, latent heat of fusion $L=334\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, solid mass density ${\rho }_s=917\mathrm{kg}{\mathrm{m}}^{-3}$, and liquid mass density ${\rho }_l=1000\mathrm{kg}{\mathrm{m}}^{-3}$. These are explicitly stipulated rounded inputs, not measurements made here. The specific volume change on melting isThe common-pressure melting-point slope therefore givesA small $1\mathrm{MPa}$ increase in the common pressure thus lowers the melting temperature by about $0.074\mathrm{K}$ in this local model. Positive latent heat does not force a positive phase coexistence curve slope: ice contracts on conversion to liquid water. Increasing pressure favors the smaller specific volume. This calculation compares equilibrium states of the same two thermodynamic phases; extending its tangent to large pressures would require updated mass densities, latent heat of fusion, and the relevant ice phase diagram.

Write $ϵ=1-{\rho }_s/\rho >0$ and $\Delta v=1/{\rho }_s-1/\rho =ϵ/{\rho }_s$. When the ice sphere grows by volume $4\pi a^2\dot{a}dt$, it consumes liquid volume $({\rho }_s/\rho )4\pi a^2\dot{a}dt$. The difference is expelled from the fixed cavity. Thus mass conservation fixes the outward Darcy flux in the surrounding porous medium:The Darcy law $u_r=-(\Pi /\mu ){\partial }_{r}p$ and $p(\infty )=p_m$ implyMatching at the cavity wall, whose internal pressure is uniform, givesThe solid and liquid pressures are equal when the Gibbs--Thomson relation is neglected; nevertheless the coexistence temperature changes with their common pressure because their specific volumes differ. Along coexistence, equality of chemical potentials givesHence the common-pressure melting-point slope is $d{T}_m/dp=-T_m\Delta v/L$. To first order in the pressure change,The $T_m$ multiplying this coefficient is absolute temperature, approximately $273.15\mathrm{K}$ for water; a temperature difference may equally be measured in kelvin or degrees Celsius. Substituting the numerical value $0$ from the Celsius scale into a thermodynamic denominator would be incorrect.

In the quasi-stationary thermal conduction approximation, neglect advective heat transport and use the stated effective conductivity $k$ outside the sphere. The radial Laplace equation gives the bounded solid temperature and the liquid/porous-region temperature:This treats the exterior as a uniform conducting medium, as required by the printed reduction; distinct cavity and porous-medium conductivities would give an additional thermal resistance. There is no solid-side heat flux in the bounded quasi-stationary solution. The Stefan condition therefore becomesEliminate $T_i$ and $p$ to obtainFor $x=a/R$, $t=t_0\tau$ and $t_0=R^2{\rho }_{s}L/(k\Delta T)$, this isThe pressure follows either from the Darcy flux or the Stefan condition:In particular $T_i-T_{\infty }=\Delta TK/(x+K)$: rising pressure progressively reduces the available undercooling. This is spherical freezing inside a porous cavity, whose fixed hydraulic outlet radius $R$ distinguishes it from freezing directly through a porous matrix.

For the idealized zero-radius start, integration givesA finite seed replaces the right side by its difference from the initial value. The reduction ends when the sphere fills the cavity.

For $K\gg 1$, hydraulic drainage is easy and the pressure feedback is small throughout the cavity. The leading expressions areThis is ordinary conduction-controlled spherical solidification, with a completion time asymptotic to $t_0/2$.

For $K\ll 1$, distinguish the initial crossover from the pressure-dominated regime. At $x\ll K$, equivalently $\tau \ll K^2$, the same square-root growth applies and $p-p_m\simeq p^{\ast }x/K$. At $K\ll x\le 1$, equivalently $K^2\ll \tau \lesssim 1/K$,The interface approaches $T_{\infty }$, the pressure approaches the finite coexistence value $p_m+p^{\ast }$, and growth becomes hydraulically limited. The leading completion time is $t_0/(3K)$. The cube-root approximation is not valid at the instant of nucleation, where the initial square-root regime must be retained.

First estimate the excess pressure as weight divided by contact area:For the equilibrium estimate, use the Clapeyron equation for melting. The specific volume change on melting is $\Delta v=1/{\rho }_w-1/{\rho }_i$, and the latent heat of fusion $L$ gives $\Delta s=L/T_m$. Thus the common-pressure melting-point slope isTake $T_m\simeq 273.15\mathrm{K}$ and the usual rounded water density ${\rho }_w\simeq 1000\mathrm{kg}{\mathrm{m}}^{-3}$, in addition to the supplied ${\rho }_i=920\mathrm{kg}{\mathrm{m}}^{-3}$ and $L=3.3\times 10^5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. ThensoThe sign is negative because ice contracts on melting. This is a linear equilibrium estimate treating the load as a common local pressure on coexisting ice and water. The actual contact stress is nonuniform, and frictional heating, deformation and drainage also affect skating; the estimate is not a complete account of the lubricating water film.

A spherical solid of radius $a$ growing inside a fixed cavity of radius $R$ expels volume at rate $4\pi (1-{\rho }_s/{\rho }_l)a^2\dot{a}$. The Darcy flux outside the cavity gives $p-p_m=\mu (1-{\rho }_s/{\rho }_l)a^2\dot{a}/(\Pi R)$. The common-pressure melting-point slope and quasi-steady spherical thermal conduction give ${\rho }_{s}La\dot{a}=k(T_i-T_{\infty })$. Define $x=a/R$, $\tau =tk(T_m-T_{\infty })/(R^2{\rho }_{s}L)$ and $K=L^2\Pi /[k{T}_m\mu (1/{\rho }_s-1/{\rho }_l{)}^2]$. Elimination yields $(x+x^2/K)x_{\tau }=1$ and the displayed zero-seed integral. The pressure rise is $p^{\ast }x/(x+K)$, with $p^{\ast }=L(T_m-T_{\infty })/[T_m(1/{\rho }_s-1/{\rho }_l)]$. Easy drainage gives square-root growth; low permeability gives an initial square-root regime followed by cube-root growth and pressure saturation. The model stops at $x=1$ and uses a uniform exterior conductivity.