Latent heat ($L=T\Delta S=\Delta H$)

Latent heat is the heat absorbed or released in a first-order phase transition through its entropy jump. For an equilibrium conversion from thermodynamic phase $\alpha$ to thermodynamic phase $\beta$ at fixed pressure, evaluate both thermodynamic phases at the same temperature and pressure on their phase coexistence curve:Use one normalization throughout: $L$, $S$ and $H$ may all be per mole, giving $L$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or all per unit mass, giving $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Here $S$ is entropy and $H$ is enthalpy on that basis. Equal Gibbs free energies of the coexisting thermodynamic phases give $\Delta H=T_{\mathrm{tr}}\Delta S$. Identifying this enthalpy difference with supplied heat assumes a closed system and an equilibrium conversion with only pressure-volume work. ^[ref]

For a pure substance with positive latent heat per unit mass $L$, let $x$ be the phase fraction of the higher-enthalpy thermodynamic phase, measured by mass. While both thermodynamic phases coexist at the fixed pressure,Adding heat changes $x$ rather than the temperature; this creates the horizontal portion of a heating curve. Once $x=1$, additional heat raises the temperature according to specific heat capacity at constant pressure. Reversing the conversion releases heat. The plateau represents changing enthalpy, rather than a finite single-thermodynamic phase heat capacity satisfying $Q=C_p\Delta T$. ^[ref]

For a constructed fixed-pressure model, take $m=1\mathrm{kg}$, initial temperature $280\mathrm{K}$, transition temperature $T_{\mathrm{tr}}=300\mathrm{K}$, solid specific heat capacity at constant pressure $c_s=2\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, latent heat $L=100\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, and liquid specific heat capacity at constant pressure $c_l=3\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. These are stipulated model parameters, not empirical properties of a named material. Heating the solid to $300\mathrm{K}$ takes $40\mathrm{kJ}$ of sensible heat; converting it fully to liquid takes another $100\mathrm{kJ}$ of heat; heating the liquid to $320\mathrm{K}$ takes $60\mathrm{kJ}$. The total is $200\mathrm{kJ}$. After a total heat input of $90\mathrm{kJ}$, $x=(90-40)/100=1/2$ and the temperature is still $300\mathrm{K}$. The corresponding per-unit-mass entropy jump is $L/T_{\mathrm{tr}}=\frac{1}{3}\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$.

A continuous phase transition has no latent heat. Its heat capacity can have a jump or a singularity without a finite interval in which added heat converts coexisting thermodynamic phases at a constant temperature. A heating plateau therefore requires the stated phase coexistence conditions; it is not a feature of every phase transition. ^[ref]

The energy per unit mass required to change a saturated liquid to vapour at a stated pressure and coexistence temperature. In a steam-condensation calorimetry experiment, sensible cooling of the condensate must be subtracted from the measured receiver heat gain.

The Stefan number compares sensible heat $c_p\Delta T$ with latent heat $L$ per unit mass. Some solidification literature uses its reciprocal as the Stefan parameter, so the defining formula must accompany the convention.

The latent-to-sensible heat ratio is the reciprocal of the Stefan number defined by $\mathrm{Ste}=c_p\Delta T/L$. Large $S$ means that phase change uses much more energy than changing temperature by $\Delta T$.

The Clapeyron equation connects the slope of a phase coexistence curve to the latent heat and the change in volume during a first-order phase transition. A steep boundary can indicate a large latent heat, a small change in volume, or both. This lets a measured boundary constrain caloric properties and predicts how changing pressure shifts a melting temperature.

For a pure substance in bulk phase coexistence, label the initial and final thermodynamic phases $\alpha$ and $\beta$. Both have the same absolute temperature $T$, pressure $p$, and chemical potential. With $s$, $h$, and $v$ denoting entropy, enthalpy, and volume on one common basis, define $\Delta s=s_{\beta }-s_{\alpha }$, $\Delta h=h_{\beta }-h_{\alpha }$, and $\Delta v=v_{\beta }-v_{\alpha }$. ThenThe derivative follows the phase coexistence curve; it is not a partial derivative within one thermodynamic phase. Use molar volume $v_m$ with molar latent heat $L_m$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or specific volume $v$ with latent heat $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Both give $dp/dT$ in $\mathrm{Pa}{\mathrm{K}}^{-1}$. For molar mass $M$, $L_m=ML$ and $\Delta v_m=M\Delta v$, so the predicted slope is independent of the chosen basis. This formulation assumes a simple compressible system, fixed composition, and negligible interface contributions; a curved interface can instead require the Young–Laplace equation.

For positive latent heat in the stated conversion direction, the sign of the slope is the sign of $\Delta v$. Converting liquid to gas normally increases volume and gives a positive slope. The negative melting-point slope of water illustrates the opposite sign. At a thermodynamic critical point, the coexisting gas and liquid merge: both $L$ and $\Delta v$ vanish, so their ratio needs a limiting analysis rather than substitution of zero into the Clapeyron equation.^[ref]

A short derivation uses the molar Gibbs free energy $g_m$, equal to the chemical potential for a pure substance. Equality persists along the boundary, so $d{g}_{m,\alpha }=d{g}_{m,\beta }$ with $d{g}_m=-s_{m}dT+v_{m}dp$. Therefore $-\Delta s_{m}dT+\Delta v_{m}dp=0$. Equality of $g_m=h_m-T{s}_m$ also gives $\Delta h_m=T\Delta s_m$. Eliminating $\Delta s_m$ yields the Clapeyron equation.

The equilibrium partial pressure of a vapour in contact with its condensed phase at a specified temperature. With constant specific latent heat and negligible condensed-phase volume, $d\ln e_s/dT=L/(R_v{T}^2)$. The condensed phase and any metastability convention must be specified.

For a dilute ideal gas vapor whose specific volume greatly exceeds that of its condensate, the Clausius-Clapeyron relation integrates to the displayed formula. If $L(T)=L_{\ast }(T_c-T)/(T_c-T_{\ast })$, the result is $\ln (p/p_{\ast })=L_{\ast }[T_c(1/T_{\ast }-1/T)-\ln (T/T_{\ast })]/[R(T_c-T_{\ast })]$. Near the liquid-gas critical point, both the ideal gas approximation and the neglect of condensed-phase volume fail; zero latent heat alone does not determine the limiting coexistence slope.

Floating ice with submerged volume fraction $f$ has density $f{\rho }_l$ by Archimedes' principle. Its volume change on melting is therefore $v_l-v_s=(1-1/f)/{\rho }_l<0$. An enthalpy cycle gives $L_f=L_s-L_v$ for quantities consistently expressed per mass at the same temperature. Treating these latent heats and specific volumes as constant and integrating the Clausius-Clapeyron relation gives the displayed pressure estimate. Using a vaporization latent heat extrapolated from a different temperature carries that approximation into the inferred fusion heat.

Molar crystal volume follows from $V_m=N_A{V}_{\mathrm{cell}}/Z$ with $Z$ atoms per cell for an elemental solid. Combining two such values with their molar entropy difference gives $\mathrm{d}P/\mathrm{d}T=\Delta S_m/\Delta V_m$. Ambient values supply only an approximation to a high-pressure coexistence slope, not a phase-boundary intercept.

For liquid–gas phase coexistence, make two distinct approximations: the gas obeys the ideal gas law, $v_{m,g}=RT/p$, and its molar volume greatly exceeds the liquid molar volume, $v_{m,g}\gg v_{m,l}$. Here $R$ is the molar gas constant and $p$ is the saturation pressure of the pure substance. Substitution into the Clapeyron equation giveswhere $p_{\ast }>0$ is any fixed reference pressure, making the logarithm dimensionless. Constant latent heat is a further approximation, needed for the elementary Integrated Clausius-Clapeyron relation below.^[ref]

The neglected liquid molar volume has a transparent error estimate. If the gas is ideal but $ϵ=v_{m,l}/v_{m,g}$ is retained, thenThus ignoring the liquid molar volume underestimates this slope by a relative amount $ϵ$. This estimate does not bound errors from treating a real gas as ideal or from freezing $L_m(T)$. Near a thermodynamic critical point, the gas and liquid molar volumes approach each other and these dilute-vapor assumptions fail.

If the molar latent heat $L_m$ is approximately constant on the chosen temperature interval, the ideal-vapor Clausius-Clapeyron approximation integrates toFor $L_m>0$, higher absolute temperature gives higher saturation pressure. A plot of $\ln (p/p_1)$ against inverse temperature $1/T$ is a straight line of slope $-L_m/R$. One reference pair $(T_1,p_1)$ fixes its intercept. On a per-unit-mass basis replace $L_m/R$ by $L/R_s$, with $R_s=R/M$ and $M$ the molar mass; using $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$ with the molar $R$ would give incompatible units.

If only the constant-latent heat approximation is relaxed, the same ideal-vapor model instead givesConsequently an apparent straight-line fit across a broad temperature range need not represent a single material constant. The simple exponential also approaches a finite value as $T\to \infty$; this is a property of the stipulated formula, not a prediction that liquid–gas phase coexistence extends beyond a thermodynamic critical point.

Consider an unnamed model substance with molar latent heat $L_m=40.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ and saturation pressure $p_1=10.0\mathrm{kPa}$ at $T_1=300\mathrm{K}$. Stipulate the ideal-vapor Clausius-Clapeyron approximation throughout $280\le T\le 340\mathrm{K}$, constant $L_m$, and $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These parameters define an illustrative model, not a measured dataset or a fit to water. At $T_2=330\mathrm{K}$,At the reference state the gas molar volume is $R{T}_1/p_1=0.24942{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, and the local slope is $dp/dT\simeq 0.5346\mathrm{kPa}{\mathrm{K}}^{-1}$. The tangent estimate $p_2\approx 10.0+30(0.5346)=26.04\mathrm{kPa}$ is much lower than the integrated result: a local derivative should not be extrapolated over this whole interval. By contrast, over a small temperature increment, the fractional sensitivity is $d\ln (p/p_1)/dT=L_m/(R{T}_1^2)\simeq 0.05346{\mathrm{K}}^{-1}$.

Within the ideal-vapor Clausius-Clapeyron approximation, a local slope of the natural logarithm of saturation pressure against inverse temperature estimates the molar latent heat:In the stipulated vapor-pressure calculation, the slope is $-4811.16\mathrm{K}$, which recovers $40.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. For variable $L_m(T)$, a two-point estimate gives an average weighted by $T^{-2}$, rather than the latent heat at an arbitrary midpoint. Differentiating noisy pressure measurements amplifies their errors. Outside the ideal-vapor regime, use independently determined molar volumes in $L_m=T\Delta v_{m}dp/dT$; a straight-looking graph alone cannot establish that the gas is ideal.

A concrete experiment turns the phase coexistence curve into pairs of temperature and saturation pressure. A 1969 paper reports water measurements made in 1942 and 1947–1949 using a boiler and precision mercury manometer. Platinum resistance thermometers read the boiler temperature through a resistance bridge. The pressure-transfer line passed through an ice-water condenser and a dry-ice trap; helium filled connecting sections. Corrected mercury column height gave the pressure at the thermometer coils, including hydrostatic pressure contributions from water vapor, helium, mercury, and mercury vapor.^[ref]

Controls included normal/reversed-lead readings, readings at two thermometer electric currents extrapolated to zero current to remove self-heating, repeated bridge calibration, and daily triple point thermometer calibration checks. Cellar temperature measurements corrected fluid-column and gauge-block thermal expansion. One measurement series was excluded because a trapped gas bubble was suspected of corrupting the mercury head. The paper adjusted measurements to round temperatures using previously tabulated slopes; those adjusted values therefore are not independent measurements of those slopes.^[ref]

The thermodynamic inference requires liquid and gas in phase coexistence, a well-defined common temperature, and a pressure corrected to that same location. Leaks, contamination, or a temperature gradient can spoil that interpretation. A separate calorimetry measurement of latent heat, together with the two specific volumes, allows a test of the full Clapeyron equation. Converting a measured phase coexistence curve slope directly into latent heat instead assumes that equation; it is not an independent test.

When the solid and liquid have the same pressure, equality of chemical potentials implies $-(s_l-s_s)dT+(1/{\rho }_l-1/{\rho }_s)dp=0$. With latent heat of fusion $L=T_m(s_l-s_s)$, this gives the displayed slope. If solid is less dense than liquid, raising the common pressure depresses the melting temperature. $T_m$ is absolute temperature. This coexistence-curve derivative differs from a thermomolecular pressure difference between equal-density phases.

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.