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For freezing, the liquid concentration decreases away from the interface, so the liquidus $T_L=-mC$ increases there. Constitutional supercooling means $T_l<T_L$ somewhere ahead of the interface. Since these temperatures agree at the interface, its local onset isFor the similarity solution and salt rejection condition, this is exactlyIt is also the global criterion for this freezing solution when $D<\kappa$ and $T_{\infty }>T_i$: the ratio $T_L^{\prime }(x)/T_l^{\prime }(x)$ is a positive constant times $\exp [-x^2(1/D-1/\kappa )/(4t)]$. For $x\ge a\ge 0$ this ratio decreases with $x$. If the ratio is at most one at the interface, $T_l-T_L$ cannot become negative; if it is greater than one there, a supercooled interval appears immediately ahead. Equality marks onset, not a finite supercooled interval.

For fixed positive $\Delta \theta$ and $\lambda =O(1)>0$, the left side of the inequality is $O(ϵ)$ and the right side is positive of order one, so freezing is constitutionally supercooled for sufficiently small $ϵ$. To locate the boundary between the two regimes, resolve the smaller scale $\Delta \theta =O(ϵ)$, $\lambda =O(ϵ)$. The salt equation and liquidus then giveHere $ϵ\lambda =O({ϵ}^2)$, so the thermal equation improves to $T_i=(T_{\infty }+T_{-\infty })/2+O({ϵ}^2)$; this justifies the second expansion including latent heat. Also $T_{\infty }-T_i={\theta }_{+}+O(ϵ)$ and the left side of the exact criterion is $ϵ{\theta }_{+}/\sqrt{\pi }+O({ϵ}^2)$. Therefore the small-diffusivity constitutional-supercooling threshold isThis printed condition is an asymptotic onset condition, rather than an exact finite-$ϵ$ inequality. In its $O({ϵ}^2)$ transition window the complete algebraic system and exact gradient criterion must be used.

Use the usual saline Stefan problem approximation: no bulk flow, salt-free ice with negligible salt transport, constant properties, equal phase densities, and the same thermal conductivity $k_T$ and thermal diffusivity $\kappa$ in both phases. Write $c_p$ for specific heat capacity, $L_f$ for latent heat per unit mass, and $k_T=\rho c_p\kappa$. These thermal symmetries are needed for the arithmetic-mean interface temperature requested in the paper; unequal phase conductivities would give a weighted balance instead.

Set $\eta =x/(2\sqrt{\kappa t})$, $\xi =x/(2\sqrt{Dt})$, $ϵ=\sqrt{D/\kappa }$ and $a(t)=2\lambda \sqrt{Dt}$. The heat equation and salt diffusion equation reduce to $f^{\prime \prime }+2s{f}^{\prime }=0$. Their similarity solutions, in terms of the complementary error function, areThese have the required interface values and far-field limits. At each fixed $x\ne 0$ they recover the initial data as $t\downarrow 0$. The liquidus condition is $T_i=-m{C}_i$.

Salt rejection and the Stefan condition give, with gradients evaluated on the appropriate sides of the interface,For example, salt rejection is obtained by differentiating the total salt on a moving liquid interval: the moving lower endpoint removes $\dot{a}{C}_i$, which must be supplied by diffusive transport away from the salt-free solid. Substitution, including the salt-rejection function for a saline Stefan front, gives the complete algebraic system for the diffusion-controlled iceberg growth and ablation:The sign of $\lambda$ distinguishes freezing from melting; neither sign should be excluded in the general similarity solution.

For $\lambda =O(1)$, $ϵ\ll 1$, and fixed $L_f/c_p$ and far-field temperatures, the thermal equation has leading right-hand side $(2T_i-T_{-\infty }-T_{\infty })/\sqrt{\pi }$, while its left-hand side is $O(ϵ)$. HenceThis is a leading-order balance, not an exact cancellation of latent heat. The resulting $C_i=-T_i/m$ and the first algebraic equation determine the leading $\lambda$. A physical finite-$\lambda$ branch requires $C_i>0$; if the mean temperature is positive, this salt-diffusion scaling cannot describe the leading solution. Likewise, a latent-to-sensible heat ratio diverging as $1/ϵ$ changes the leading thermal balance.

Define ${\theta }_{-}=-m{C}_0-T_{-\infty }$ and ${\theta }_{+}=T_{\infty }+m{C}_0$. Put $T_0=-m{C}_0$ and $\Delta \theta ={\theta }_{-}-{\theta }_{+}$. Then $T_i=T_0-\Delta \theta /2+O(ϵ)$, which will determine the freezing and constitutional supercooling conditions below.

Introduce specific heat capacity $c_p$, latent heat per unit mass $L_f$, and $\Delta T=T_m-T_s>0$. Write the prescribed convective heat flux as $F=K(\overline{T}-T_m)$, whereThere is a dimensional convention to specify: with $Q$ measured in $\mathrm{W}\mathrm{k}{\mathrm{g}}^{-1}$ and dimensionless $\lambda$, this printed expression makes $K$ a heat transfer coefficient only if $\mu$ is kinematic viscosity. If $\mu$ denotes dynamic viscosity, the numerator needs ${\rho }_0^{2}g\alpha Q$ instead. The question says only “viscosity”; use the consistent kinematic interpretation, or equivalently the corrected dynamic formula. The following balances apply to either convention once $K$ has units $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.

For a thin lava lake crust with a linear temperature profile, Fourier's law gives the outward conductive flux $k\Delta T/a$. The Stefan condition balances it against upward thermal convection and freezing, while the well-mixed liquid loses the convective flux and gains radiogenic heating. To leading order in $a/H$ the radiogenically heated lava-lake crust model isThe fixed surface temperature already incorporates radiative cooling, so no extra radiative loss is added to the interior balance. The linear conductive profile is interpreted as a quasi-steady approximation, so sensible-heat storage within the crust is not evolved. Thinness alone does not establish this approximation: rapid thermal adjustment in the crust is also needed, for example a small Stefan number during the initial conduction-driven growth. Replacing the liquid depth by $H-a$ and retaining the moving liquid-volume enthalpy would introduce thin-crust corrections: the requested constant-temperature branch refers to the leading model above. A transient thermal profile in a thick or newly nucleating crust would require the full Stefan problem, rather than this linear-profile closure.

Using the thermal diffusion time $H^2/\kappa$, defineThe three independent dimensionless parameters are the convective-conductive coefficient $B$, the heating parameter $R$, and the Stefan number $S$. For positive $B,R,S$ the evolution equations becomewhere prime denotes $d/d\tau$. Temperature has the exact solutionConsequently the unique constant-temperature initial condition isAt this temperature, the convective flux equals the column's radiogenic heat production. Starting with no crust, the constant-temperature lava-crust growth law is obtained by separating $h^{\prime }=S(1/h-R)$:The crust grows towards $h_{\ast }=1/R$, or $a_{\ast }=k\Delta T/({\rho }_{0}QH)$, where conduction carries all the radiogenic input. The logarithm makes reaching this equilibrium take infinite time. For an initial thickness $h_0>0$, subtract $[-R{h}_0-\log (1-R{h}_0)]/(S{R}^2)$ on the same side of the equilibrium; the corresponding antiderivative uses $\log |1-Rh|$ on either branch. The thin-crust description through equilibrium requires $R\gg 1$.

For general initial temperature, the full early and intermediate history is specified by the exact temperature above and one scalar ordinary differential equation. A particularly convenient integral formulation for an initially bare surface isThis removes the infinite initial derivative of $h$. The right-hand side points into $y>0$ at zero, and is decreasing in $y$ when ${\theta }_0\ge 0$; two nonnegative solutions therefore cannot separate, which gives uniqueness. Its solution, together with the temperature formula, determines the entire model evolution and can be computed without a phase-boundary search. The early expansion follows by substitution:The dimensional leading crust thickness is $a\sim \sqrt{2k\Delta Tt/({\rho }_0{L}_f)}$. Thus hotter liquid reduces the first correction to conductive crust growth, whereas colder liquid permits more rapid thickening.

If ${\theta }_0>{\theta }_{\ast }$, the temperature decreases and the instantaneous stationary lava-crust thickness $h_e(\tau )=1/[B\theta (\tau )]$ increases towards $h_{\ast }$. Starting from zero, the crust always remains below this rising curve: at a first contact $h^{\prime }=0$ but $h_e^{\prime }>0$, so it cannot cross from below. Hence hot-start crust grows monotonically. It also stays thinner than the constant-temperature solution: the early expansion puts it below, and at a proposed first crossing their derivative difference is $-SB({\theta }_0-{\theta }_{\ast })e^{-B\tau }<0$, preventing crossing from below. If $0\le {\theta }_0<{\theta }_{\ast }$, the temperature increases and the instantaneous stationary lava-crust thickness decreases. The crust stays thicker than the constant-temperature solution by scalar comparison, but it may cross the decreasing $h_e$, reach a maximum and subsequently melt back towards $h_{\ast }$. Such overshoot is possible, not inevitable; the two relevant relaxation rates are $B$ and $S{R}^2$. The numerical plot uses $B=4$, $R=20$, $S=0.1$, so crust adjustment is faster than temperature relaxation and the cool-start example does overshoot:

The curves use ${\theta }_0/{\theta }_{\ast }=0.5,1,1.5$ and initially $h=0$. They solve the complete reduced time-dependent equations, rather than replacing the moving stationary thickness by the actual crust. At late times, set $h=h_{\ast }+v$, $\delta ={\theta }_0-{\theta }_{\ast }$; linearization gives $v^{\prime }=-S{R}^{2}v-SB\delta e^{-B\tau }$. For $S{R}^2\ne B$,At equal rates the forced term is $-SB\delta \tau e^{-B\tau }$. In particular, slow warming with $B<S{R}^2$ approaches equilibrium from a thicker crust, explaining the plotted remelting. These long-time expressions describe the linearized response; nonlinear corrections can have their own faster exponential rates. All predictions are restricted to $a/H\ll 1$, nonnegative liquid superheat, and the assumed vigorous-convection flux law. If crust growth violates those conditions, the reduced model must be replaced rather than extrapolated to total freezing.

Use a quasistatic surface energy balance with negligible heat storage in the surface or ice, a linear ice temperature profile, and a fixed freezing temperature at the ice-ocean interface. These are the assumptions behind the single-category sea-ice thermodynamic model used below. Write $\theta =t/\tau$, $F_O={\lambda }_O(T_O-T_m)>0$ andHere ${\lambda }_A$ and ${\lambda }_O$ are heat transfer coefficients, $k$ is thermal conductivity, $\kappa =k/(\rho c_p)$ is thermal diffusivity, and $\mathcal{S}$ is the inverse Stefan number. Here $L$ is latent heat per unit mass. The thermal $k$ denotes conductivity; the previous questions use $k$ for permeability.

While there is open water, the upward ocean heat supply balances the loss to air: ${\lambda }_O(T_O-T_s)={\lambda }_A(T_s-T_A)$. ThusFreezing begins when this surface reaches $T_m$ on the cooling part of the annual cycle, which requires $r<1$ and givesFor $r\ge 1$ the surface never cools strictly below freezing and no positive ice thickness develops in this model. The limit $r\to 0$ gives $t_1\to \pi \tau /2$.

With ice present and a cold surface, thermal conduction and Newton's law of cooling give $k(T_m-T_s)/h={\lambda }_A(T_s-T_A)$. The surface temperature isDuring surface melting the temperature is capped at $T_s=T_m$; the ice is not permitted to follow the above formula to a temperature higher than its melting point. The upward conductive heat flux during the cold phase isThe Stefan condition at the ocean interface gives $\rho L\dot{h}=F_C-F_O$. Under the permitted approximation of negligible ocean heat during growth,Starting from $h(t_1)=0$, integration yields the implicit growth lawThe leading maximum occurs when the air returns to freezing from below:The equality for $h_m$ is within the ocean-heat-neglected growth model, with the freezing time set by the open-water balance. If $aH\gg 1$, conductive resistance through the ice dominates the atmospheric boundary resistance at maximum thickness. ConsequentlyThis reproduces the displayed large-conductance limit. A sufficient condition for the quasistatic temperature assumption is $h_m^2/\kappa \ll \tau$; large $\mathcal{S}$ ensures this in the thick-ice scaling.

The approximation concerning ocean heat needs care. Its ratio to atmospheric heat removal during the cold phase isIt is uniformly small away from the beginnings and ends of winter providedFor the thick-ice limit this requires both $r\ll 1$ and $raH\ll 1$. This also makes the integrated ocean heat small relative to the latent heat of the maximum ice column. It is not possible to neglect $F_O$ pointwise throughout the closed growth interval: at the actual freezing onset it balances the atmospheric removal, and at the true maximum the growth rate is zero because the two fluxes again balance. The ocean-heat correction to seasonal ice growth therefore gives narrow endpoint corrections. In the full model the maximum is slightly earlier than $3\pi \tau /2$, withThus $t_2=3\pi \tau /2-O(\tau r[1+a{h}_m])$ in the stated approximation, rather than an exact maximum time for finite ocean heat.

For $3\pi \tau /2\le t\le 2\pi \tau$, while ice survives, both the warm air and ocean melt it. The surface is at $T_m$, and the total Stefan condition isIntegrating from the leading maximum givesuntil its first zero. The physical thickness is then zero for the rest of this warm interval and the open-water temperature applies. Therefore at the end of the first year,Positive ice at that time is equivalent, in the specified approximation, toThis establishes the algebraic criterion printed in the paper.

There is, however, an actual qualification to the paper's word “perennial.” This inequality tests peak-temperature sea-ice survival, since $t=2\pi \tau$ is the next atmospheric temperature maximum. Surface melting continues until at least $t=5\pi \tau /2$, so positive thickness at the maximum does not guarantee survival throughout the summer. In the same approximation, survival through that full warm half-cycle would require the stronger inequalityIn fact, starting ice-free under the specified zero-mean sinusoidal forcing, even the zero-ocean-heat upper bound on winter growth cannot meet it. In that limit $t_1=\pi \tau /2$ andsince $h_m^2/2>0$. The expression on the right is already the atmospheric melt over the entire subsequent warm half-cycle. Positive ocean heat can only reduce the winter ice and increase summer melt. Thus the printed condition is a first-year peak-temperature survival test, not a sufficient criterion for literal year-round ice in this model. For example, with $aH=1/2$ and very small positive $r$, it gives $h_m/H\simeq 2(\sqrt{2}-1)\simeq 0.828$, while first-year melt is approximately $0.5H$ but full warm-season melt is approximately $H$. Ice is present at the year's endpoint and disappears later that same summer. No reinterpretation of the printed inequality as guaranteed perennial cover is justified without an additional seasonal assumption.

Use the common-density, equal-volume-heat-capacity model implicit in the stated speed law. Let the liquid fraction be $\varphi$ ahead of the melting front and $\varphi +\Delta \varphi$ behind it. Melting adds exactly the liquid needed to fill the newly created pores, so total mass conservation makes Darcy flux continuous. Consequently,Here $u$ denotes Darcy velocity and $v$ the pore-liquid velocity in the stationary rock frame. Treating the pore-space increase as storage without its simultaneous melting source would incorrectly change the Darcy flux.

Relative to the cold unmelted material, the bulk enthalpy increase behind the front is $\rho c_p\Delta T+\rho L\Delta \varphi$: all phases gain sensible heat and the melted ice consumes latent heat. The advective heat-flux difference is $\rho c_{p}U\Delta T$. Applying the Rankine-Hugoniot condition to this energy balance therefore givesso the advection-driven melting front in a porous matrix moves atThe advected latent contribution of the liquid is the same on both sides and cancels; the denominator measures sensible heating plus phase-change energy per unit bulk volume.

At fixed volume, entropy from heat capacity obeys $dS=C(T)dT/T$. The third law of thermodynamics fixes $S(0)=0$ when the ground state is nondegenerate, so below the transitionAbove the transition,For a transition with no latent heat, entropy is continuous, and comparison with $S=\beta \log T+\gamma$ gives the directly testable relationThus calorimetric measurements of $C(T)$ and $T_c$ verify $\gamma$. More generally, if a measurable latent heat $L$ is present, $S(T_c^{+})-S(T_c^{-})=L/T_c$, and the formula becomes

For a simple system of fixed composition, the first law of thermodynamics andgiveThusEquality of mixed derivatives gives the Gibbs free-energy Maxwell relation

The heat capacity at constant volume and heat capacity at constant pressure areRegarding $S$ as a function of $T,V$ and differentiating along a constant-pressure curve givesThe Maxwell relation from the Helmholtz free energy isTherefore the difference between constant-pressure and constant-volume heat capacities is

Along the liquid-gas phase coexistence curve, the molar Gibbs free energies agree. Differentiating $G_{\mathrm{l}}=G_{\mathrm{g}}$ along the curve givesHence the Clausius-Clapeyron relation iswhere $L=T(S_{\mathrm{g}}-S_{\mathrm{l}})$ is the molar latent heat.

If $V_{\mathrm{g}}\gg V_{\mathrm{l}}$ and the gas is ideal, then $V_{\mathrm{g}}=RT/P$, soFor constant $L$, integration yieldswhere $P_0$ is constant over the range in which the approximations apply.

A first-order phase transition occurs where two phases have equal Gibbs free energy but a first derivative of that free energy is discontinuous. Typical jumps occur inA nonzero entropy jump gives a nonzero latent heat $T\Delta S$.

The unperturbed temperature is linear in each material. Continuity of the conductive heat flux at $x=h$ givesOn definingthis becomes $T_m-T_h=ϵ(T_h-T_a)$. Therefore

Let ${\rho }_i$ be the ice mass density and $L_f$ its latent heat of fusion per unit mass. The ice is isothermal in this model, so the Stefan condition equates latent-heat production to the heat flux conducted through the water:Consequently the unperturbed lateral solidification speed isThe material parameters are the thermal conductivities $k_w,k_a$, ice density ${\rho }_i$, and specific latent heat $L_f$; the geometric thermal length is $\delta$.

Conservation of energy and Fourier's law giveAt steady state $k{T}_z=F_c$ is constant. Since $k=a/T$,The two temperature boundary conditions therefore giveAt the basal phase boundary, the Stefan condition isThe first term removes latent heat released by freezing, while the oceanic flux $F_w$ supplies heat to the interface. A steady shell has $\dot{h}=0$, hence $F_c=F_w$ andHere $T_s$ must simultaneously satisfy the surface balance from part a with $F_c=F_w$.

At a stated external pressure, the boiling point of a pure liquid is the temperature at which its saturation vapor pressure equals that pressure. A vapour bubble inside the liquid must also withstand the hydrostatic head and any capillary pressure; the simple free-surface boiling-point rule neglects those small local corrections.

Increasing external pressure raises the usual boiling point, while lowering it lowers the boiling point. The Clapeyron equation explains the direction:for ordinary vaporization with positive latent heat $L$ and a larger vapour specific volume. Consequently water can boil below its usual open-vessel temperature at altitude or under partial vacuum, and above it in a sufficiently pressurized vessel.

Composition also matters. A nonvolatile dissolved solute generally lowers the solvent's vapour pressure and produces boiling-point elevation at fixed external pressure. A volatile second component can instead raise or lower the boiling temperature according to the mixture's phase equilibrium; an azeotrope is a useful warning against applying the nonvolatile-solute rule to every mixture.

Finally, distinguish the equilibrium boiling point from the observed onset of bubbles. A smooth clean vessel or a shortage of nucleation sites can permit superheating; agitation or suitable existing sites can make boiling begin nearer equilibrium. Heating power mainly controls the rate of sustained boiling under otherwise fixed conditions, rather than setting an arbitrary new equilibrium boiling point. A thermometer must be placed to measure the intended liquid or saturated vapour region, rather than a locally overheated wall.

Evaporation cools the remaining material when the escaping vapour takes energy from it faster than the surroundings supply heat. For an observable demonstration, compare two similar thermometer bulbs at the same initial temperature, one dry and the other wetted with a volatile liquid. Expose both to the same controlled airflow. The wetted bulb falls in temperature; the effect depends on vapour removal and environmental conditions. Another demonstration is continued heating during boiling at constant pressure: vapour is produced while the boiling liquid's temperature remains nearly constant. The supplied energy has not vanished; it contributes to the latent heat and expansion associated with the phase transition.

One calorimetry method estimates the specific latent heat of vaporization by condensing a known mass $m_s$ of dry saturated steam at $T_s$ into a known mass $m_w$ of cooler water. Let $C_{\mathrm{cal}}$ be the vessel's heat capacity, and $T_i,T_f$ the initial and final water temperatures. Assuming complete condensation without escaped steam or thermal losses,soThe last term removes the sensible cooling of the condensed steam. Weighing the receiving vessel before and after gives $m_s$ only if splashing, evaporation and liquid carried in with the steam are controlled. A steam trap and an insulated transfer path address two distinct errors: droplets falsely counted as condensing steam, and cooling before the calorimeter.

The historical local application was Richardson's ether spray. A bellows atomizer dispersed diethyl ether into fine droplets, promoting rapid vaporization and withdrawal of heat from the surface; the cooling could produce local loss of sensation. Benjamin Ward Richardson's 1866 account describes this evaporative method and its apparatus. Its historical physiological language is not adopted as a modern description of nerve signalling, and the account does not establish that the method is appropriate for every operation.^[ref]

The radius is half the diameter, $r^{\ast }=1.5\times 10^{-9}\mathrm{m}$, and the latent heat density is $1.25\times 10^9\mathrm{J}{\mathrm{m}}^{-3}$. HenceUsing the diameter directly in the radius formula would underestimate the depression by a factor of two. The result is the stipulated continuum-model estimate; at only a few nanometres atomistic structure and support effects can matter substantially.

Let $g_l-g_s$ be the liquid-minus-solid free energy density at the same temperature. Near the bulk coexistence temperature $T_m$, linearization givesFor a sphere, an interface displacement changes area and volume with $dA/dV=2/r$. Equality of the bulk driving force and the capillary cost at a curved solid–liquid interface therefore gives $g_l-g_s=2\gamma /r^{\ast }$. Equivalently, differentiating $4\pi r^2\gamma -(4\pi /3)r^3(g_l-g_s)$ gives the same stationary radius. Combining the expressions yieldsThis form of the Gibbs-Thomson effect assumes isotropic interfacial energy, a spherical interface, bulk-like latent heat per volume and a near-coexistence free energy expansion. More detailed free-particle models must distinguish solid–vapor, liquid–vapor and solid–liquid energies.

In terms of diameter $d$, the model isThus melting temperature rises toward the bulk value as diameter increases, with a depression proportional to $1/d$. The analytic sketch below includes the calculated point. It deliberately does not extrapolate the continuum expression down to a zero or negative melting temperature.

Facet-dependent energies, non-spherical shapes, size-dependent interfacial energies and latent heat, structural changes or magic cluster sizes can modify the trend. Premelting can broaden the transition. A substrate or encapsulating matrix changes interface energetics and can stabilize or destabilize the solid. Ligands, impurities, strain, particle size distributions, coalescence during heating and nonequilibrium temperature/beam effects can shift the measured transition. These possibilities motivate separate thermal, structural and size controls rather than a fit to diameter alone.

Exchange constraint. Exchange of two identical particles cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the exchange operator obeys $P_{12}^2=1$, so its allowed eigenvalues are $\pm 1$. bosons use the symmetric overall wavefunction, $P_{12}\Psi =+\Psi$; fermions use the antisymmetric overall wavefunction, $P_{12}\Psi =-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the Spin-statistics theorem.

Factored fermion states. If $\Psi ={\psi }_{\mathrm{space}}{\chi }_{\mathrm{spin}}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$; the latter factors are the three triplet states. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.

hydrogen spin isomers. The two protons are identical spin-one-half fermions. parahydrogen has total nuclear spin $I_{\mathrm{n}}=0$, the antisymmetric singlet; orthohydrogen has $I_{\mathrm{n}}=1$, the symmetric triplet. Here $I_{\mathrm{n}}$ is a spin quantum number, distinct from the moment of inertia used below.

Rotational restriction. In the electronic and vibrational ground state, these two factors are symmetric under nuclear exchange. Exchanging the protons reverses the internuclear axis, $(\theta ,\varphi )\mapsto (\pi -\theta ,\varphi +\pi )$. The spherical harmonic rotational state contributes the exchange sign $(-1{)}^{\ell }$. Overall antisymmetry therefore requiresHence parahydrogen has even $\ell$; orthohydrogen has odd $\ell$. The $\ell$ here labels molecular rotation, rather than the angular momentum obtained by also coupling the nuclear spins.

Conversion heat. The ortho-para conversion heat follows from the linear rigid rotor approximation, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ with $I=\mu r^2\simeq m_p{r}^2/2$. Thus conversion of one molecule from $\ell =1$ to $\ell =0$ releasesNeglecting the small electron contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ molecules. Only the initial orthohydrogen fraction releases this energy, givingThe estimate uses $r=74$ pm, $m_p\simeq 1.67262\times 10^{-27}$ kg and $\hslash \simeq 1.05457\times 10^{-34}$ J s. Handbook-level rounding and the rigid rotor approximation justify about three significant figures, not a high-precision thermochemical prediction.^[ref]

Storage consequence. The internal conversion is a delayed heat source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external heat transfer but does not remove this source. Converting to predominantly parahydrogen during liquefaction, while removing the released heat, reduces the subsequent storage heat load. No evaporation fraction can be calculated without a latent heat and a model for how the heat is removed.

At coexistence, the two phases have equal chemical potential. For a single component, $d\mu =-sdT+vdp$, with entropy and volume measured on the same molar or mass basis. Equating the phase derivatives along coexistence gives $\Delta vdp=\Delta sdT$. Reversible latent heat satisfies $L=T\Delta s$, henceHere $p$ and $T$ are coexistence pressure and absolute temperature, $L$ is latent heat for transforming phase 1 to phase 2, and $\Delta v=v_2-v_1$ uses the same amount of material as $L$.

For vaporization, neglect liquid molar volume compared with ideal-vapour volume $RT/p$. The ideal-vapour Clausius-Clapeyron approximation becomesUsing the normal boiling reference $T_b=373.15\mathrm{K}$, $p_b=1.01325\times 10^5\mathrm{Pa}$ and the supplied constant $L_e$ givesThis deliberately constant-latent-heat extrapolation does not reproduce the supplied triple-point pressure exactly. It predicts the liquid-vapour branch, metastable against ice below the triple-point temperature; it is not a precision sublimation-pressure prediction.

The phase diagram has solid, liquid and vapour regions meeting at the triple point. The sublimation and vaporization boundaries have positive slopes, whereas the ordinary ice–water boundary slopes negatively because ice has larger specific volume. The liquid–vapour line terminates at a critical point; the original schematic below emphasizes topology and marks the supplied triple point.

For the melting-pressure estimate, Archimedes' principle gives ${\rho }_i=0.90{\rho }_w=900\mathrm{kg}{\mathrm{m}}^{-3}$, soConvert the molar evaporation heat using water molar mass $M\simeq 0.0180\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, rounded from the tabulated value $18.0153\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref] An enthalpy cycle through vapour then givesThis uses the question's temperature-independent evaporation heat all the way to the triple point; no replacement experimental fusion heat is inserted. The melting pressure inferred from flotation and latent heats, on holding $L_f$ and $\Delta v$ constant, isusing $T_t=273.16\mathrm{K}$ and $T=270.15\mathrm{K}$. Linearizing the logarithm about the triple point gives $53.4\mathrm{MPa}$ instead of $53.7\mathrm{MPa}$. Both are estimates within the stated constant-property approximation; the negative melting slope makes a positive pressure necessary to lower the melting temperature.