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The Antoine equation is an empirical vapor pressure fit whose coefficients depend on the pressure and temperature units and on a stated validity interval. Its slope is $d\ln p_{\mathrm{sat}}/dT=\ln (10)B/(T+C{)}^2$. In the ideal-vapor Clausius-Clapeyron approximation, this implies latent heat $L=R_v\ln (10)B[T/(T+C){]}^2$. An accurate pressure fit need not give an equally accurate derivative; extrapolation to the liquid-gas critical point cannot replace the full Clausius-Clapeyron relation.

Why can hot liquid wait for a charged particle rather than immediately turn into vapor? A small cavity costs surface energy. Consider an original macroscopic capillarity model at fixed temperature with surface tension $\sigma >0$, imposed liquid pressure $p_l$, and driving difference $\Delta p=p_{\mathrm{sat}}(T)-p_l>0$, where $p_{\mathrm{sat}}$ is the planar-interface vapor pressure. Neglect dissolved gas, interface thickness, heat transfer and curvature dependence of $\sigma$. The reversible formation-work approximation isThe first term charges for surface area; the second rewards the new vapor volume. This is a macroscopic free energy model, not the energy that an actual detector must receive as localized heat. It shares the capillarity balance used in the cloud-chamber expansion model, but the new phase is now a bubble.

Differentiating givesAt the critical nucleus, $d^2\Delta G/d{r}^2=-8\pi \sigma <0$. The barrier is a maximum: cavities smaller than $r_{\ast }$ tend to shrink, while larger ones can grow in this model. Doubling $\Delta p$ halves $r_{\ast }$ and quarters $\Delta G_{\ast }$ if $\sigma$ stays fixed. This explains the strong sensitivity to operating conditions without claiming a universal detector threshold.

Choose $\sigma =0.010\mathrm{J}{\mathrm{m}}^{-2}$ and $\Delta p=2.0\times 10^5\mathrm{Pa}$ solely for a teaching example. ThenFor the same chosen $\sigma$ and twice the driving pressure, $r_{\ast }=50\mathrm{nm}$ and $\Delta G_{\ast }\simeq 0.654\mathrm{keV}$. These are calculated formation-work scales, not measured bubble thresholds for diethyl ether or any other substance.

A simplified way to discuss localization is to stipulate deposited energy $E_{\mathrm{loc}}$ within a segment of length $\ell$ and a trial conversion fraction $0<\eta \le 1$. A toy criterion $\eta E_{\mathrm{loc}}\gtrsim \Delta G_{\ast }$ gives an energy per unit length scale $\Delta G_{\ast }/(\eta \ell )$. It is only a bookkeeping estimate: a real bubble also needs latent heat, must exchange heat, and is affected by the spatial pattern and time scale of energy deposition. Consequently neither a single stopping power value nor this barrier alone predicts detection efficiency or the number of bubbles. Two tracks with the same mean stopping power can distribute their energy differently on the scale of a critical nucleus.

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.

Consider an illustrative water–air mixture, initially saturated at $T_0=293.15\mathrm{K}$. Let $\lambda =V/V_0$ be the expansion ratio and approximate the carrier gas by a uniform ideal gas with constant heat capacity ratio $\gamma =1.40$. Before significant condensation, assume fixed numbers of carrier and vapor molecules, negligible incoming heat, and the reversible adiabatic expansion relation. The first law of thermodynamics, $n{C}_{V}dT=-pdV$, with the ideal gas law, givesThe vapor follows the same temperature, so its partial pressure scales as $T/V$. A rapid real piston motion need not be reversible; the formula is an idealized bulk estimate after neglecting flow and losses.

Use a physically motivated Integrated Clausius-Clapeyron relation for the equilibrium water vapor pressure, with stipulated rounded constants $p_0=2339\mathrm{Pa}$ at $T_0$, molar latent heat $L_m=45000\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$:These constants define the example, without fitting any historical measurements. Throughout the plotted interval $1\le \lambda \le 1.17$, the temperature stays above the freezing point of water. InitiallyThus the decrease in vapor pressure initially beats dilution. At $\lambda =1.15$, the model gives $T=277.21\mathrm{K}$, $p_v=1923.33\mathrm{Pa}$, $p_{\mathrm{sat}}=809.07\mathrm{Pa}$ and $S=2.3772$.

Why does $S>1$ not instantly make every molecule a visible drop? In the capillarity approximation of classical nucleation theory, a neutral spherical drop of radius $r$ has a positive surface energy cost and a negative bulk Gibbs free energy contribution. For molecular liquid volume $v_l$, surface tension $\sigma$ and Boltzmann constant $k_B$,Differentiating $\Delta G$ shows that the critical nucleus at $r_{\ast }$ is a maximum: smaller neutral drops tend to shrink, whereas larger ones can grow. With stipulated $\sigma =0.075\mathrm{J}{\mathrm{m}}^{-2}$, $v_l=2.99\times 10^{-29}{\mathrm{m}}^3$ and $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, the $\lambda =1.15$ example gives $r_{\ast }=1.353\mathrm{nm}$ and $\Delta G_{\ast }/(k_{B}T)=150.3$. A substantial neutral-drop barrier can coexist with supersaturated vapor.

This is deliberately a neutral-drop comparison. An ion changes the cluster free energy through electrical interactions and molecular hydration; the formula does not calculate the ion-assisted nucleation threshold. At nanometre radii, using bulk surface tension is itself approximate. There is no predicted droplet count, growth time or detection efficiency here. Once many drops grow, latent heat release, vapor depletion and heat transfer from the chamber walls invalidate the dry adiabatic expansion trajectory. The example therefore explains the direction of the effect and the need for nucleation, without extrapolating its curves to Wilson's cloud limits.

Compositional convection is buoyancy-driven motion caused by composition-dependent density variations. In a planetary core, freezing can expel light elements into the surrounding liquid, making that liquid buoyant and supplying mechanical power for a planetary dynamo. It can operate alongside thermally driven convection and release of latent heat.

Convective adjustment replaces an excessively steep vertical temperature gradient by an allowed profile while respecting the selected energy budget. It is a closure for unresolved convection, not a direct simulation of individual rising parcels. A prescribed critical atmospheric lapse rate is a modeling assumption; it need not equal the dry adiabatic lapse rate.

For an original dry two-slab example, take fixed equal heat capacities and slab centers separated by $\Delta z=2\mathrm{km}$. Initially $T_1=300\mathrm{K}$ below and $T_2=280\mathrm{K}$ above. Stipulate a permitted atmospheric lapse rate ${\Gamma }_c=6\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$. The initial difference $20\mathrm{K}$ exceeds the allowed $12\mathrm{K}$. Conservation of the sum of the two sensible energies requires $T_1^{\prime }+T_2^{\prime }=580\mathrm{K}$, while the adjusted gradient requires $T_1^{\prime }-T_2^{\prime }=12\mathrm{K}$. Hence $T_1^{\prime }=296\mathrm{K}$ and $T_2^{\prime }=284\mathrm{K}$. Heat removed below equals heat added above. Simply holding the lower slab fixed and raising the upper slab to $288\mathrm{K}$ would introduce external energy.

For unequal constant heat capacities $C_1,C_2$, the same two constraints instead giveApply this adjustment only when the original gradient exceeds the threshold. A moist adjustment also has to account for latent heat and changing composition; conservation of this dry sensible-energy expression alone is then insufficient.

Cryogenics studies the production and use of very low temperatures. Its experimental challenge is a balance of rates: a refrigerator must remove energy from a cold region faster than unwanted heat transfer supplies it. A low temperature reading alone does not establish a useful cooling capacity. A maintained cold liquid bath is especially valuable because latent heat allows it to absorb heat while its phase fraction changes at nearly fixed temperature and pressure.

Helium liquefaction illustrates why staged cooling and recovery of a returning stream's cooling capacity matter. A counterflow heat exchanger transfers heat from the incoming stream to the colder returning stream without mixing them. This reduces the incoming specific enthalpy before expansion. The heat exchanger recovers cooling already produced; the surrounding refrigerator still requires input work and rejection of heat to a warmer environment.

Curvature depresses the equilibrium melting temperature of solid convex into liquid. Here $\mathcal{K}$ is the sum of principal curvatures, equal to twice a common convention for mean curvature, and $T_m$ is absolute temperature. The coefficient uses solid-liquid surface energy, mass density and latent heat. This stabilizes fine corrugations of a solidification front.

Fixed water vapor mole fraction gives $d\ln e_v=d\ln p$. Combining hydrostatic balance with the Clausius-Clapeyron relation yields $d\ln T_D/dz=-(g{R}_v/L_v{R}_d)(T_D/T)$. If $T_D/T\simeq 1$, then $T_D(z)\simeq T_D(0)e^{-z/H}$. All ratios use absolute temperature; constant latent heat, fixed composition and approximate parcel–environment temperature equality are required.

In a constant-thickness laminar boundary layer at an ablating glacier, approximate fluxes by $\kappa (T_{\infty }-T_i)/\delta$ and $D(C_{\infty }-C_i)/\delta$. Neglect solid diffusion and sensible heating relative to latent heat. With ablation speed $U>0$, solute balance gives $U(C_i-C_s)=D(C_{\infty }-C_i)/\delta$, so $C_i=C_s+(C_{\infty }-C_s)/(1+P)$, $P=U\delta /D$. The liquidus $T_i=T_m-\Gamma (C_i-C_s)$ and Stefan condition $UL/c_p=\kappa (T_{\infty }-T_i)/\delta$ then give the displayed law, where $A_T=LD/(c_p\kappa )$. Its positive root exists when the ocean is above its own freezing point. In the small-$P$ regime, interface salinity approaches ocean salinity; in the large-$P$ regime it approaches the ice composition.

Enthalpy is the state function obtained by adding pressure times volume to internal energy:Here $U$ denotes internal energy, also written $E$ in some references. Like internal energy, enthalpy is measured in joules and is an extensive variable. Its definition packages the internal energy change and expansion work into one useful quantity for fixed-pressure calculations. It does not define a body's stored heat. ^[ref]

For a fixed amount and composition of a simple compressible system, the relation between neighboring equilibrium states isHere $T$ is absolute temperature and $S$ is entropy. Thus entropy and pressure are the natural variables of this thermodynamic potential, with partial derivatives $T=(\partial H/\partial S{)}_p$ and $V=(\partial H/\partial p{)}_S$. Fixed composition matters: changing composition adds further terms. The formula follows in one line from the equilibrium first law of thermodynamics identity:The cancellation of $pdV$ is the reason enthalpy replaces volume with pressure as a natural variable. This identity concerns equilibrium states; it does not identify $TdS$ with actual irreversible heat transfer. ^[ref]

For a closed system undergoing a quasistatic process at constant pressure, require fixed amount and composition, only pressure-volume work, no friction, and unchanged bulk kinetic energy and potential energy. Then the actual first law of thermodynamics balance givesThe symbol $Q_p$ means transferred heat for this fixed-pressure process. If the piston example in heat additionally meets these conditions, its $20\mathrm{kJ}$ of incoming heat equals $\Delta H$, while $\Delta U=15\mathrm{kJ}$ and $p\Delta V=W=5\mathrm{kJ}$. The example shows exactly what enthalpy adds to internal energy: the expansion work required at the imposed pressure.

Within a single thermodynamic phase, heat capacity at constant pressure giveswith the amount and composition fixed. Constant $C_p$ gives $\Delta H=C_p\Delta T$; dividing by mass gives specific heat capacity at constant pressure. At equilibrium melting, enthalpy can instead change through latent heat while temperature stays fixed. The sensible heat example therefore adds two single-thermodynamic phase enthalpy changes and one latent heat contribution.

The usefulness of enthalpy lies in matching the imposed constraint: a fixed-pressure experiment naturally reports $\Delta H$, while fixed-volume heating with no other work reports $\Delta U$. Neither equality is universal. Additional work or changing pressure alters the relation between supplied heat and $\Delta H$; irreversible entropy production also prevents identifying actual $\delta Q$ with $TdS$. The separate Joule-Thomson coefficient branch uses enthalpy to describe throttling under its own assumptions.

The molar enthalpy difference between vapor and liquid at specified conditions, evaluated on coexistence when it is a latent heat. Treating it as constant and treating the vapor as ideal gives $\ln (P_2/P_1)=(\Delta H_{\mathrm{vap}}/R)(1/T_1-1/T_2)$.

Along a reversible path at fixed volume, $dS=C_V(T)dT/T$. Hence entropy differences can be measured macroscopically fromwith an additional $L/T_c$ across a first-order transition of latent heat $L$ at $T_c$.

Evaporation transfers molecules from a liquid surface into the surrounding vapour. It can occur below the boiling temperature, and competes with condensation from the vapour. Its net rate depends on the surrounding vapor pressure and transport conditions. Removing latent heat through evaporation can cool the remaining liquid.

Freezing is a liquid-to-solid phase transition. A moving phase boundary releases latent heat; its rate is constrained by heat transport and the Stefan condition. In a solution the liquidus depends on composition, so being below the pure solvent's melting temperature does not imply that the solution must freeze.

Heat is energy transferred across a thermodynamic system's boundary because of a temperature difference. It is measured in joules, like work and internal energy. A hot cup transfers heat to cooler surroundings; its temperature describes its state, while the transferred heat describes a process. The cup stores internal energy, rather than a substance called heat. ^[ref]

For a closed system with unchanged bulk kinetic energy and potential energy, choose $Q>0$ for heat entering and $W>0$ for work done by the system. The first law of thermodynamics givesHere $U$ is the stored internal energy. For example, a sealed piston cylinder receiving $20\mathrm{kJ}$ of heat and delivering $5\mathrm{kJ}$ of work gains $15\mathrm{kJ}$ of internal energy. A closed system retains its matter while exchanging energy; it need not be an isolated system. ^[ref]

Unlike internal energy, heat is not a state function: different paths between the same equilibrium states can transfer different $Q$, with compensating differences in work. The symbol $\delta Q$ records this dependence on the path; there is no stored quantity whose difference is always $Q$. The two paths in first law of thermodynamics illustrate this directly. Supplied heat can raise temperature as sensible heat, change a phase fraction through latent heat, or help supply expansion work. Measuring temperature alone therefore does not determine transferred heat.

For a reversible thermodynamic process, $\delta Q_{\mathrm{rev}}=TdS$, where $T$ is absolute temperature and $S$ is entropy. This identification requires reversibility: an irreversible adiabatic process has $Q=0$ and can nevertheless increase entropy. Likewise $Q=\Delta H$, with enthalpy $H$, requires the particular fixed-pressure conditions in enthalpy. These distinctions make heat useful as a precise boundary-transfer quantity in heat engines, refrigerators and heating calculations.

Historically, James Prescott Joule's paddle experiments showed that stirring water with input work could raise its temperature, helping establish conservation of energy. The lesson is still useful: a temperature rise can result from work even without incoming heat. ^[ref]

A heating curve records temperature against accumulated supplied heat under a specified constraint. Within one thermodynamic phase, its slope is $dT/dQ=1/C_X$, where $C_X$ is the constrained heat capacity. A latent heat plateau instead records changing phase fraction at the phase coexistence temperature. The sensible heat example combines two sloping intervals and a melting plateau in one heating budget.

As $t\to 0$, the conditional mean rotational energies approach $U_{\mathrm{p}}/N=0$ and $U_{\mathrm{o}}/N=2k_{\mathrm{B}}{\theta }_r$, using the para $J=0$ level as the common zero. A frozen mixture with ortho fraction $x$ therefore retains $U_{\mathrm{fr}}/N\to 2x{k}_{\mathrm{B}}{\theta }_r$. If a bath keeps the sample at low temperature while a catalyst permits conversion to almost pure parahydrogen, the rotational heat delivered to the bath tends to $Q_{\mathrm{out}}/N=2x{k}_{\mathrm{B}}{\theta }_r$, or $3k_{\mathrm{B}}{\theta }_r/2$ for normal hydrogen.

One may shift the ortho ladder to its own lowest level for computing its conditional heat capacity, because a constant energy offset disappears on differentiation at fixed composition. That offset must be restored before comparing species or calculating conversion heat. Otherwise the calculation loses precisely the stored rotational energy that conversion releases. The expression above accounts for the internal rotor contribution at fixed bath temperature; it does not model liquid intermolecular energies, latent heat, or an adiabatic temperature rise.

A catalyst changes the conversion rate and allows heat removal during cooling; it does not change the free-gas equilibrium weights at a specified temperature. This connects the microscopic exchange rule to refrigeration: delayed conversion supplies a continuing internal heat source. Rotational Raman spectroscopy offers a complementary population probe because its $\Delta J=\pm 2$ transitions stay within the even or odd ladder; line intensities require the appropriate species populations and transition response, not a bare $3:1$ intensity rule.

Transfer of thermal energy between regions. In a constant-property thermal Boussinesq approximation, the advective heat flux per unit area is ${\rho }_0{c}_{p}T\mathbf{u}$ and transport across a section is its normal component integrated over that section. With zero net volume transport, adding a constant reference temperature does not change that integral. Otherwise the reference and the return-flow budget must be specified. Atmospheric energy transport also includes latent heat and gravitational potential energy.

At fixed relative humidity and fixed excited-oxygen concentration, a water-reaction hydroxyl radical source scales as $e_s(T)/T$ when concentrations are number densities. Thus $d\ln P_{\mathrm{OH}}/dT=\Delta H_{\mathrm{vap}}/(R{T}^2)-1/T$ under constant latent heat and rate coefficient. Using only $e_s$ instead gives a vapor pressure or fixed-pressure mole-fraction approximation. Neither derivative follows without a humidity condition.