Latent Heat and Phase Coexistence

Put a pot of water on the stove and watch the thermometer. It climbs steadily to 100 °C — and then it stops. You keep pouring energy in, the burner roars, bubbles churn, but the temperature refuses to budge until the last of the water has boiled away. Where does all that heat go? It goes into a hidden reservoir called latent heat: the energy required to tear molecules out of the liquid and set them free as vapor, paid at a fixed price per molecule while the two phases coexist. This article is about that plateau and the physics behind it. At a first-order phase transition, two phases — liquid and gas, solid and liquid — sit in equilibrium with equal chemical potentials, $\mu_1 = \mu_2$ , but with different entropies and volumes. The entropy difference $\Delta s$ is the key: multiply it by the temperature and you get the latent heat $L = T\,\Delta s$ , the energy exchanged reversibly as the substance converts from one phase to the other at fixed $T$ and $p$ . This is not a heat capacity — it is a finite lump of energy delivered at a single temperature, an infinitely sharp spike in $C_p$ . We will derive $L = T\,\Delta s$ from the coexistence condition, show how the lever rule tracks the fraction of each phase along a coexistence tie-line, connect the latent heat to the slope of the coexistence curve through the Clausius–Clapeyron relation, and compute real numbers for water. Latent heat is the quantitative fingerprint that distinguishes a first-order transition from a continuous one — and it vanishes exactly at the critical point, where the two phases finally become one.

Key takeaways

Latent heat is the energy $L = T\,\Delta s = h_2 - h_1$ absorbed or released when a substance crosses a first-order transition at fixed $T$ and $p$ , while two phases coexist with equal chemical potentials $\mu_1 = \mu_2$ .
It arises from a discontinuity in the first derivatives of the Gibbs free energy: entropy $s$ and volume $v$ jump across the transition, while $\mu$ , $T$ , $p$ stay continuous.
During coexistence the temperature is pinned: added heat converts phase 1 into phase 2 rather than raising $T$ — the boiling plateau.
The lever rule $x_2 = (v - v_1)/(v_2 - v_1)$ sets the fraction of each phase from the overall volume along the coexistence tie-line.
Latent heat is not a heat capacity: it is a finite energy at one temperature, a Dirac- $\delta$ spike in $C_p$ whose integrated weight is exactly $L$ .
KEY IDEA: at coexistence the two phases' free energies are equal but their entropies differ; that entropy jump $\Delta s$ , times $T$ , is the latent heat — and it dies as $(T_c - T)^\beta \to 0$ approaching the critical point.

The core insight

Two phases coexist when their chemical potentials are equal, $\mu_1 = \mu_2$ , but the slopes of $\mu(T,p)$ differ: since $s = -(\partial\mu/\partial T)_p$ and $v = (\partial\mu/\partial p)_T$ , both the entropy and the volume jump. The entropy jump $\Delta s$ carries the latent heat $L = T\,\Delta s$ — the energy that reorganizes the microscopic structure instead of raising the temperature.

Where this lives today

Latent heat is one of the most economically consequential numbers in physics. The enormous $L_{\text{vap}}$ of water underpins the Rankine cycle: steam carries huge energy per kilogram into a turbine and releases it on condensing, the basis of most of the world's electricity. Run in reverse, the same physics drives refrigerators and heat pumps — a working fluid absorbs latent heat by evaporating in the cold reservoir and dumps it by condensing in the hot one, moving heat with modest mass flow. In the atmosphere, latent heat released when water vapor condenses into clouds powers convection, thunderstorms, and hurricanes; the Clausius–Clapeyron scaling of saturation vapor pressure predicts wetter, more intense precipitation in a warming climate. In the lab, differential scanning calorimetry measures the area under a heat-flow peak to read off $L$ — a sharp delta-like spike flags a genuine first-order transition, while a divergent-but-integrable cusp signals a continuous one. And engineered first-order transitions, from martensitic shape-memory alloys to the magnetocaloric materials chased for solid-state refrigeration, all trade on their latent heats and order-parameter jumps.

The physical picture

Imagine the chemical potential of each phase as a surface floating above the temperature–pressure plane. The stable phase is whichever surface lies lower. Where the two surfaces cross, the phases have equal $\mu$ and coexist — and because they cross at an angle, their slopes differ. The slope in the $T$ -direction is $-s$ and in the $p$ -direction is $+v$ , so at the crossing both the entropy and the volume change discontinuously. That entropy discontinuity, multiplied by $T$ , is the latent heat.

Figure

Two chemical-potential sheets crossing: the stable phase is the lower one; at the crossing μ₁=μ₂ but the slopes −s differ, so entropy jumps

The everyday example is boiling. As you heat water at atmospheric pressure it warms until 100 °C, then stalls. Every joule you add now goes into converting liquid into vapor — breaking hydrogen bonds and pushing molecules apart against the surrounding pressure — not into faster thermal motion. The thermometer reads a flat 100 °C until the last drop is gone. That plateau is latent heat being absorbed at constant $T$ . On a pressure–volume isotherm below the critical temperature, the same physics shows up as a flat segment. Crossing it left to right, you traverse a mixture of liquid and gas at a single fixed coexistence pressure; the lever rule sets how much of each you have, and the width of the segment is exactly the volume jump $\Delta v$ . This is the punchline plot: on a temperature–entropy diagram, the transition is a horizontal line — the temperature is fixed while the entropy jumps by $\Delta s$ , and the area $T\,\Delta s$ under that line is the latent heat.

Figure

Punchline: the T–s coexistence plateau. Temperature is pinned while entropy jumps by Δs; the shaded area T·Δs is the latent heat delivered at fixed T

Heating along a coexistence curve is like walking a ridge between two valleys. The closer you get to the critical summit, the shallower the valleys and the smaller the entropy step you must climb to switch sides — until at the summit the valleys merge, $\Delta s \to 0$ , and no latent heat is needed at all. Latent heat is the order-parameter-discontinuity hallmark of a first-order transition; where it vanishes, the transition has become continuous.

The core equations

The starting point is the coexistence condition: two phases in mutual equilibrium at the same $T$ and $p$ can still exchange particles unless their chemical potentials match. Only when the per-particle Gibbs free energies are equal does net conversion stop, defining the coexistence curve in the phase diagram.

\mu_1(T,p) = \mu_2(T,p)

where: $\mu_i$ is the Gibbs free energy per particle of phase $i$ , and $T,p$ are the shared temperature and pressure. Physically, this is the equilibrium that pins the boiling temperature at a given pressure: at 1 atm, water and steam have equal $\mu$ only at 373.15 K. The latent heat follows from writing $\mu = h - Ts$ for each phase. Equal $\mu$ and shared $T$ turn the enthalpy difference directly into an entropy difference times temperature.

L = T\,\Delta s = T\,(s_2 - s_1) = h_2 - h_1

where: $L$ is the heat absorbed to convert one unit of phase 1 into phase 2, $\Delta s = s_2 - s_1$ is the entropy jump per particle, and $h_i$ is the enthalpy per particle. At constant pressure the reversible heat equals the enthalpy change, $\delta Q = T\,\mathrm{d}S = \mathrm{d}H$ , so $L$ is literally the enthalpy step across the transition. The latent heat is an enthalpy change, not an internal-energy change. Converting liquid to vapor also does $p\,\Delta v$ work of expansion, so the internal energy changes by less.

\Delta u = L - p\,\Delta v

where: $\Delta u$ is the internal-energy jump and $p\,\Delta v$ is the work done pushing back the atmosphere as the volume expands. For vaporization this term is large (the gas volume dwarfs the liquid); for melting it is negligible. The fraction of each phase in the coexistence region is fixed by the lever rule. If the overall volume per particle is $v$ , somewhere between the two pure-phase values $v_1$ and $v_2$ , then the split follows a simple balance.

x_2 = \frac{v - v_1}{v_2 - v_1}

where: $x_2$ is the fraction in phase 2 and $x_1 = 1 - x_2$ in phase 1. The name comes from mechanics: the system "balances" on the tie-line like a lever, with the distances to each end setting the phase fractions. It is the same rule that reads off the liquid/vapor ratio anywhere on the flat isotherm segment. Finally, latent heat and coexistence-curve geometry are tied together by the Clausius–Clapeyron relation, derived in full in the sibling article `aq-sm-clausius-clapeyron-relation`.

\frac{\mathrm{d}p}{\mathrm{d}T} = \frac{s_2 - s_1}{v_2 - v_1} = \frac{L}{T\,\Delta v}

where: $\mathrm{d}p/\mathrm{d}T$ is the slope of the coexistence line and $\Delta v = v_2 - v_1$ the volume jump. The latent heat is the numerator: a steep or shallow phase boundary directly reports $L$ .

Derivation ALatent heat as the entropy jump at coexistence

We want to show, cleanly and from first principles, that the heat absorbed while a substance converts entirely from phase 1 to phase 2 at fixed temperature and pressure is exactly $L = T\,\Delta s = h_2 - h_1$ , and that this same quantity equals the enthalpy jump. The whole result rests on two facts: the phases coexist with equal chemical potentials, and at constant pressure the reversible heat is the enthalpy change. We assemble them here so that every later formula — the lever rule, the Clausius–Clapeyron slope, the delta-spike in $C_p$ — inherits a rigorously grounded latent heat.

Step 1. Write the chemical potential of each phase as $\mu_i = h_i - T s_i$ , the per-particle Gibbs free energy (recall `aq-sm-free-energies-enthalpy-gibbs`: $G = H - TS$ , and $\mu = G/N$ for a single component). Why this form: $\mu$ is the natural potential in the variables $T$ and $p$ that are held fixed during coexistence.

Step 2. Impose coexistence, $\mu_1 = \mu_2$ , at shared $T$ : $h_1 - T s_1 = h_2 - T s_2$ . Rearranging gives $h_2 - h_1 = T(s_2 - s_1)$ . Why this matters: it converts the enthalpy jump — an energy — into the entropy jump times temperature, with no approximation.

Step 3. Identify this with the heat exchanged. At constant pressure the first law gives $\delta Q_{\text{rev}} = T\,\mathrm{d}S = \mathrm{d}H$ . Converting one unit of phase 1 fully to phase 2 therefore absorbs $L = \Delta H = h_2 - h_1$ . Why constant pressure is essential: only then does the reversible heat equal $\mathrm{d}H$ rather than $\mathrm{d}U$ ; the $p\,\Delta v$ work is folded into the enthalpy.

Step 4. Combine Steps 2 and 3: $L = h_2 - h_1 = T(s_2 - s_1) = T\,\Delta s$ . The latent heat is the entropy discontinuity scaled by $T$ . Why this is the whole story: the sign of $\Delta s$ (positive going to the higher-entropy phase, e.g. liquid→gas) fixes the sign of $L$ , and its magnitude is set entirely by how disordered the new phase is relative to the old.

Step 5. Separate internal energy: since $h = u + pv$ , the internal-energy jump is $\Delta u = \Delta h - p\,\Delta v = L - p\,\Delta v$ . Why this distinction is worth flagging: students often equate $L$ with $\Delta u$ ; the expansion work $p\,\Delta v$ is substantial for vaporization and must not be dropped. Verified symbolically in verify_latent-heat-phase-coexistence.py (Step 1: $L = h_2-h_1 = T(s_2-s_1)$ from $\mu_1=\mu_2$ ; Step 5: $\Delta u = L - p\,\Delta v$ ).

Go deeper · Derivation B · 1 minThe lever rule from volume conservation

Derivation BThe lever rule from volume conservation

The goal here is to show precisely how much of each phase is present at a point inside the two-phase coexistence region, given only the overall molar volume. This is the quantitative content of the flat isotherm segment: knowing $v_1$ , $v_2$ , and the mixture's average $v$ is enough to read off the phase fractions, with no additional thermodynamic input. The rule is exact and follows from a single conservation statement, which is why it works identically for the T–s plateau and the p–v tie-line.

Step 1. Let a total amount of substance split into fraction $x_1$ in phase 1 and $x_2$ in phase 2, with $x_1 + x_2 = 1$ . Each phase occupies its own per-particle volume, $v_1$ and $v_2$ . Why fractions: along the tie-line only the proportion of each phase changes, not the pure-phase properties.

Step 2. Conserve total volume: the overall per-particle volume is the weighted average $v = x_1 v_1 + x_2 v_2$ . Why an average: volume is extensive, so the total is the sum of each phase's contribution, divided by the total particle number.

Step 3. Substitute $x_1 = 1 - x_2$ : $v = (1 - x_2) v_1 + x_2 v_2 = v_1 + x_2 (v_2 - v_1)$ . Why substitute: eliminating $x_1$ leaves a single linear equation in $x_2$ .

Step 4. Solve for the phase fraction: $x_2 = (v - v_1)/(v_2 - v_1)$ , and correspondingly $x_1 = (v_2 - v)/(v_2 - v_1)$ . Why this is the "lever": the fraction in each phase is proportional to the distance from the opposite endpoint, exactly as a balance beam apportions weight by lever arm. Verified symbolically in verify_latent-heat-phase-coexistence.py (lever rule $x_2 = (v-v_1)/(v_2-v_1)$ from $v = x_1 v_1 + x_2 v_2$ ).

Limiting cases

Case	Result	Physical meaning
Near the critical point	$L = T_c\,\Delta s \to 0$	Both $\Delta s$ and $\Delta v$ vanish as $(T_c-T)^\beta$ ; the first-order line ends and the transition becomes continuous
Dilute-vapor (integrated CC)	$p \propto e^{-L/RT}$	With $\Delta v \approx v_{\text{gas}} = RT/p$ and $L \approx$ const, the vapor-pressure curve rises Arrhenius-fashion
Melting (two condensed phases)	$\mathrm{d}p/\mathrm{d}T = L_{\text{fus}}/(T\,\Delta v)$ , large	$\Delta v$ is tiny, so melting lines are nearly vertical; for water $\Delta v < 0$ , giving a negative slope
Absolute zero	$L = T\,\Delta s \to 0$	The Third Law forces $\Delta s \to 0$ as $T \to 0$ ; any transition surviving to $T = 0$ has vanishing latent heat

Common trapLatent heat is not a heat capacity

It is tempting to fold latent heat into $\int C_p\,\mathrm{d}T$ , but they are physically distinct. A heat capacity measures how much heat raises the temperature; latent heat is exchanged at fixed temperature. Formally, the transition contributes a Dirac- $\delta$ spike to $C_p$ : $C_p(T) = c_p^{\text{bg}}(T) + L\,\delta(T - T_0)$ , whose integrated weight is exactly $L$ — an effectively infinite $C_p$ over a zero-width interval. A second trap: do not equate $L$ with the internal-energy change. At coexistence $L = \Delta h$ (enthalpy), and $\Delta u = L - p\,\Delta v$ ; the expansion-work term is large for vaporization and small (but nonzero) for melting.

Worked examples

Example 1 — Latent heat of vaporization of water, from entropy. We check the internal consistency of the textbook value $L_{\text{vap}} = 40.66\ \text{kJ/mol}$ for water boiling at 1 atm, using $L = T\,\Delta s$ , and compare to Trouton's rule.

Step 1. Take $L_{\text{vap}} = 40\,660\ \text{J/mol}$ at $T_b = 373.15\ \text{K}$ . The entropy of vaporization is $\Delta s_{\text{vap}} = L_{\text{vap}}/T_b = 40\,660 / 373.15$ .

Step 2. Compute: $\Delta s_{\text{vap}} = \mathbf{108.96\ \text{J/mol}\,\cdot\,\text{K}}$ .

Step 3. Compare to Trouton's rule, the empirical observation that many liquids have $\Delta s_{\text{vap}} \approx 88\ \text{J/mol}\,\cdot\,\text{K}$ . Water sits well above this: $108.96 > 88$ . Why: hydrogen bonding gives liquid water anomalously low entropy, so the entropy jump on vaporization is larger than for a normal liquid. The elevation over Trouton's rule is a direct thermodynamic fingerprint of hydrogen bonding.

Step 4. Cross-check per kilogram: with molar mass $M = 18.015\ \text{g/mol}$ , $L_{\text{vap}} = 40\,660 / 0.018015 = \mathbf{2257\ \text{kJ/kg}}$ — the familiar steam value that makes water such an effective energy carrier in power cycles.

Example 2 — Latent heat of melting ice, and the lever rule at half-melt. We compute the entropy of fusion of ice and then use the lever rule to find the volume of a half-melted ice–water mixture.

Step 1. Ice melts at $T_m = 273.15\ \text{K}$ with $L_{\text{fus}} = 6.01\ \text{kJ/mol} = 6010\ \text{J/mol}$ . The entropy of fusion is $\Delta s_{\text{fus}} = 6010 / 273.15 = \mathbf{22.0\ \text{J/mol}\,\cdot\,\text{K}}$ — much smaller than vaporization, because solid and liquid are far more alike than liquid and gas.

Step 2. Per kilogram: $L_{\text{fus}} = 6010 / 0.018015 = \mathbf{334\ \text{kJ/kg}}$ , the standard value.

Step 3. Now the lever rule. Take molar volumes $v_{\text{ice}} = 19.66\ \text{cm}^3/\text{mol}$ and $v_{\text{water}} = 18.02\ \text{cm}^3/\text{mol}$ at 0 °C (ice is less dense, so $v_{\text{ice}} > v_{\text{water}}$ ). For a mixture exactly half water by mole, $x_{\text{water}} = 0.5$ .

Step 4. The overall molar volume is $v = x_{\text{water}} v_{\text{water}} + (1 - x_{\text{water}}) v_{\text{ice}} = 0.5(18.02) + 0.5(19.66) = \mathbf{18.84\ \text{cm}^3/\text{mol}}$ . Inverting the lever rule confirms $x_{\text{water}} = (v_{\text{ice}} - v)/(v_{\text{ice}} - v_{\text{water}}) = (19.66 - 18.84)/(19.66 - 18.02) = 0.82/1.64 = \mathbf{0.5}$ , as it must. The mixture floats — because the average volume exceeds that of pure water, a half-melted ice cube is still buoyant.

Origins · 1 minlatent heat and how this was discovered

latent heat

concept · Joseph Black 1761, Clausius formalization 1850 · The concept crystallized when Joseph Black, measuring how ice absorbs heat while staying at 0 °C, recognized a "latent" heat hidden from the thermometer — the quantitative seed of the modern coexistence picture.

The idea of latent heat predates thermodynamics itself. In the 1760s Joseph Black, weighing how much heat ice and boiling water could absorb without changing temperature, distinguished this hidden ("latent") heat from the "sensible" heat that a thermometer registers — a distinction that made the steam engine intelligible and, a century later, was recast by Clausius and Gibbs as the entropy jump $\Delta s$ at a first-order transition. Today "latent heat" names both the everyday quantity engineers tabulate and the sharp order-parameter-discontinuity signature $L = T\,\Delta s$ that defines first-order phase transitions in statistical mechanics.

History

1761

— Joseph Black distinguishes latent from sensible heat by measuring the melting of ice and boiling of water

1834

— Émile Clapeyron casts phase equilibrium into the coexistence-slope relation later named after him and Clausius

1850

— Rudolf Clausius gives the thermodynamic form using entropy, tying the coexistence slope to the latent heat

1873

— van der Waals's equation of state provides a microscopic model with a liquid–gas coexistence region

1875

— James Clerk Maxwell formulates the equal-area construction fixing the coexistence pressure on a van der Waals loop

Going further

For deeper treatments: Pathria & Beale, Statistical Mechanics, 3rd ed., Ch. 12 (phase transitions, exact results); Kardar, Statistical Physics of Particles, Ch. 5; Reif, Fundamentals of Statistical and Thermal Physics, Ch. 8; Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 11; Landau & Lifshitz, Statistical Physics Part 1, §81–83. Provenance: the equations are standard — coexistence $\mu_1 = \mu_2$ , $L = T\,\Delta s$ , Clausius–Clapeyron $\mathrm{d}p/\mathrm{d}T = L/(T\,\Delta v)$ , the Maxwell equal-area construction, and the lever rule (Pathria & Beale Ch. 12; Reif Ch. 8; Kardar Ch. 5; Landau–Lifshitz §81–83). Numbers verified independently: $L_{\text{vap}}(\text{water}) = 40.66\ \text{kJ/mol} \Rightarrow \Delta s = 108.96\ \text{J/mol}\,\cdot\,\text{K}$ and $2257\ \text{kJ/kg}$ ; $L_{\text{fus}}(\text{ice}) = 6.01\ \text{kJ/mol} \Rightarrow \Delta s = 22.0\ \text{J/mol}\,\cdot\,\text{K}$ and $334\ \text{kJ/kg}$ (see verify_latent-heat-phase-coexistence.py). Prerequisites: First-Order Phase Transitions · Order Parameter · Clausius–Clapeyron Relation. Leads to: Continuous Phase Transitions · Landau Theory of Phase Transitions · Critical Exponents.

Why is the entropy of vaporization so much larger than the entropy of fusion?

Because a gas is enormously more disordered than a liquid, whereas a liquid and a solid are structurally similar. For water, $\Delta s_{\text{vap}} = 108.96\ \text{J/mol}\,\cdot\,\text{K}$ versus $\Delta s_{\text{fus}} = 22.0\ \text{J/mol}\,\cdot\,\text{K}$ — a factor of five. Melting only loosens the crystal lattice into a still-dense, still-correlated liquid; vaporization frees molecules to fly apart into a volume roughly a thousand times larger. The latent heat, $L = T\,\Delta s$ , inherits this hierarchy: vaporizing water costs about seven times more energy per mole than melting ice.

What exactly is "latent" about latent heat?

The word (from Latin latēre, to lie hidden) captures that this heat does no visible thermal work — it does not raise the temperature. You pour energy in and the thermometer sits still; the energy is "hidden" in the reorganization of the phase. It reappears — becomes "sensible" again — only when the reverse transition occurs, as when condensing steam scalds far worse than hot water: the latent heat stored in vaporization is dumped back onto your skin.

Is the latent heat exactly

\Delta h

\Delta u

At constant pressure — the usual condition for melting and boiling in the open — the reversible heat equals the enthalpy change, so $L = \Delta h = h_2 - h_1$ . The internal-energy change is smaller by the expansion work: $\Delta u = L - p\,\Delta v$ . For vaporization the gas volume is so much larger than the liquid that $p\,\Delta v \approx RT$ per mole (about 3.1 kJ/mol at 373 K), a noticeable fraction of the 40.7 kJ/mol latent heat. For melting, $\Delta v$ is tiny, so $\Delta u \approx L$ to excellent accuracy.

Why does a first-order transition make an infinitely sharp spike in

C_p

Because the entropy has a genuine step at the transition: as phase 2 fully forms, $S$ jumps by $\Delta S$ at a single temperature $T_0$ . Since $C_p = T\,(\partial S/\partial T)_p$ , and the derivative of a step function is a Dirac delta, $C_p$ acquires a term $L\,\delta(T - T_0)$ . Its integral is exactly $L = T_0\,\Delta S$ , so the "infinite" heat capacity has finite total weight. In practice, imperfections and finite heating rates smear the spike into a tall narrow peak, and calorimeters measure its area to extract $L$ .

How is latent heat related to the slope of the phase boundary?

Through the Clausius–Clapeyron relation, $\mathrm{d}p/\mathrm{d}T = L/(T\,\Delta v)$ , derived in full in the sibling article. The latent heat is the numerator: measuring the slope of a boiling or melting curve, together with the volume change, hands you $L$ . This is why a plot of $\ln p$ against $1/T$ for a vapor-pressure curve is a straight line whose slope gives the latent heat of vaporization directly.

Why is a half-melted ice cube still buoyant?

The lever rule with the actual volumes tells the story. Ice is less dense than liquid water ( $v_{\text{ice}} = 19.66 > v_{\text{water}} = 18.02\ \text{cm}^3/\text{mol}$ ), so a 50/50 mixture has average molar volume $18.84\ \text{cm}^3/\text{mol}$ , greater than pure water's $18.02$ . Its mean density is therefore below that of the surrounding water, and it floats. Only when the last of the ice melts does the volume drop to the pure-liquid value. This same density anomaly is why the water–ice coexistence line has a negative slope (see the Clausius–Clapeyron article).

Does every transition have a latent heat?

No — only first-order ones. A continuous (second-order) transition has $\Delta s = 0$ and $\Delta v = 0$ at $T_c$ , so $L = T\,\Delta s = 0$ : no latent heat, no coexistence plateau. Instead its heat capacity diverges (an integrable cusp), rather than showing a delta spike. This is the sharpest operational distinction between the two kinds of transition: measure $C_p$ across it — a delta spike means first order with latent heat, a divergent cusp means continuous.