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The horizontal area is $A=10^6{\mathrm{m}}^2$ and the total mass from part (a) is $m=2.55\times 10^{11}\mathrm{kg}$. The iceberg melt heat budget includes warming to $0^{\circ }\mathrm{C}$ followed by fusion. With temperature $T(z)=-0.6z$ in the upper $50$ m and $-30^{\circ }\mathrm{C}$ below,The two integrals are $525000$ and $6750000\mathrm{kg}\mathrm{K}{\mathrm{m}}^{-2}$, so using the stated ice specific heat capacity,The latent heat of fusion contribution isThereforeThis produces liquid water at $0^{\circ }\mathrm{C}$; warming the meltwater further would require additional heat.

The solar constant measures the incident power per area normal to the sunlight. A spherical Earth of radius $R$ intercepts sunlight over the disc $\pi R^2$, but radiates over its surface $4\pi R^2$. For uniform surface albedo $a$ in the global energy-budget approximation, absorbed power is $(1-a)S\pi R^2$. Treat the surface as a blackbody, assume efficient heat redistribution and neglect internal heating. The Stefan–Boltzmann law and the planetary surface energy balance then giveFor a nonuniform surface the same power balance instead defines an effective planetary equilibrium temperature, $T_E^4=⟨T^4⟩$; its fourth root is not generally the arithmetic mean of the local temperatures. Also, area-averaging the surface albedo is an approximation to the properly insolation-weighted averaging. These are the simplifications implicit in the naked-planet calculation.

Take $R=6.371\times 10^6\mathrm{m}$, so $A_E=4\pi R^2=5.10\times 10^{14}{\mathrm{m}}^2$. The changed area is $A_i=10^{12}{\mathrm{m}}^2$, givingThe initial planetary equilibrium temperature is $T_0=254.8\mathrm{K}$. ThusThe differential estimate $\Delta T\simeq -T_0\Delta a/[4(1-a_0)]$ gives the same result. This is the global equilibrium response of the specified model, not the immediate local Arctic warming or the full atmospheric response.

The exceptional 2007 Arctic sea ice decline is best understood as a weather event acting on an already vulnerable ice cover, with several reinforcing pathways. Warmer air increases downward long-wave heating, lengthens the melt season and delays autumn freezing. Changes in snowfall and the development of melt ponds also change the surface energy balance. Air temperature alone, however, does not identify the source of the observed bottom melt: heat must reach the underside through the ocean.

Warmer Atlantic-origin water supplies a larger subsurface ocean heat content after entering through Fram Strait and the Barents Sea. Its influence on sea ice depends on vertical exchange, because a cold, relatively fresh halocline normally separates it from the ice. Erosion of that halocline, through altered freshwater distributions, wind-driven mixing or changes in circulation, weakens stable density stratification and can increase upward ocean heat flux. A warmer Atlantic layer without this exchange need not produce immediate extra bottom melt. Conversely, warmer Pacific inflow through Bering Strait introduces heat into the shallower western Arctic; advection through the Chukchi Sea provides another route to ice retreat and ocean heat storage.

The Arctic Oscillation changes winds, ice motion and export. Positive phases during earlier years helped remove old, thick ice through Fram Strait and left a younger, thinner pack. Summer 2007 also had a pronounced pressure contrast between the Beaufort region and the Eurasian Arctic, associated with the Arctic dipole anomaly. Its winds promoted ice displacement and warm-air transport. The summer circulation should not be described merely as a particular Arctic Oscillation index: the spatial pattern and the preconditioning from earlier years both matter.

Long-term thinning reduces the latent heat of fusion needed to remove the remaining ice. Replacement of multi-year sea ice by first-year sea ice therefore makes a given summer heat input more effective, while mobile thin ice is more readily redistributed or exported. These dynamical changes can expose open water early enough for substantial solar absorption. Water and melt ponds have lower surface albedo than snow-covered ice, so absorbed short-wave energy rises, increasing upper-ocean heat storage and bottom melt. This ice-ocean albedo feedback also delays refreezing. In the Beaufort Sea in 2007, enhanced solar heating of the newly exposed upper ocean was an especially important explanation of extreme bottom melt. Thus thinning and wind-driven opening can initiate retreat, while solar heating amplifies it; changes in Atlantic or Pacific heat supply can further reinforce the process where circulation permits.

A further positive climate feedback involves terrestrial carbon. Loss of the insulating ice cover permits greater autumn and winter transfer of stored ocean heat to the atmosphere. The resulting regional warming can increase thaw of coastal and inland permafrost. Microbial decomposition of previously frozen organic material then releases carbon dioxide, and oxygen-poor wet environments also release methane. These greenhouse gases strengthen the greenhouse effect, increasing global warming and further thaw. The permafrost carbon feedback provides the causal chain from summer ice retreat to an additional global influence. An Arctic methane feedback involving subsea sources is also possible, but the amount actually reaching the atmosphere depends on oxidation and transport; an abrupt, wholesale release is not a necessary consequence of summer ice loss.

For the bottom-melt estimate, use ${\rho }_i=917\mathrm{kg}{\mathrm{m}}^{-3}$ and the given $L_f=336000\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Assume the ice at the melting interface is already at its local freezing temperature; there is then no separate sensible-heating term. Per square metre the latent heat requirement isJuly through September has $92$ days, so the mean ocean heat flux to the ice isUsing ${\rho }_i\simeq 900\mathrm{kg}{\mathrm{m}}^{-3}$ instead gives about $76\mathrm{W}{\mathrm{m}}^{-2}$, within the intended accuracy. Around the date of this examination, the estimated net anthropogenic radiative forcing since preindustrial times was about $1.6\mathrm{W}{\mathrm{m}}^{-2}$; doubling carbon dioxide gives about $3.7\mathrm{W}{\mathrm{m}}^{-2}$. The calculated bottom flux is tens of times larger. There is no energy-budget contradiction: it is a local, summer ocean-to-ice heat transfer, supplied by absorbed sunshine, stored heat and transport, whereas radiative forcing is a global perturbation to the radiation budget. Redistribution and local absorption can produce much larger regional fluxes than the global forcing.

For an already ice-free $100\mathrm{m}$ water column, take ${\rho }_w=1025\mathrm{kg}{\mathrm{m}}^{-3}$ and liquid-water specific heat capacity $c_w\simeq 4000\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Assuming all the same energy remains in that uniformly mixed column,If the water initially lies near the seawater freezing point, about $-1.8{}^{\circ }\mathrm{C}$, its final temperature is approximately $\boxed{-0.3{}^{\circ }\mathrm{C}}$. An absolute final temperature requires that additional initial condition; the energy calculation itself determines only the temperature increase. The given ice specific heat capacity is not the appropriate value for heating liquid seawater.

The Maykut-Untersteiner model solves a seasonally forced, one-dimensional sea ice thermodynamics problem with a snow layer above saline ice. Let $z$ increase downward from the upper surface, with snow–ice interface $z=h_s$ and ice bottom $z=h_s+h_i$. Write $q=-k{T}_z$ for conductive heat flux, positive downward, and $I(z)$ for downward shortwave radiation penetrating the material. Local energy conservation givesFor $I=I_0{e}^{-\kappa z}$, the radiation term is $+\kappa I$: absorption heats the interior. Here $c_{\mathrm{eff}}$ is the derivative of specific enthalpy with temperature at prescribed ice salinity, and includes the latent heat of changing brine fraction. Using only pure-ice specific heat capacity would omit this storage. A usual approximation has ${\rho }_i{c}_{\mathrm{eff}}={\rho }_i{c}_{i,0}+a{S}_i/T_C^2$ and $k_i=k_{i,0}+b{S}_i/T_C$, with $T_C<0$ in degrees Celsius and positive, unit-dependent constants $a,b$. Thus warmer saline ice has larger effective heat capacity and lower thermal conductivity. Layerwise constant $k_i$ gives the usual finite-layer heat equation; full differentiation retains its spatial variation. Snow uses its own density and thermal conductivity, and its stronger extinction can make shortwave penetration negligible beneath it.

Define the net atmospheric input to the upper surface bywhere $\alpha$ is albedo, $H_{\uparrow }$ is turbulent sensible heat loss and $E_{\uparrow }$ is turbulent latent heat loss. Their signs can reverse. The outgoing longwave term uses the Stefan–Boltzmann law. Subtract the penetrating shortwave $I_0$, because it heats the interior rather than the infinitesimal surface. If $M_0\ge 0$ is surface ablation speed and ${\rho }_0$ the density of the material currently exposed, the upper boundary obeysBelow the melting point, $M_0=0$ and the flux balance determines $T_0$. Once it reaches $T_m$, the positive excess produces surface ablation; snow melts first, and only afterwards does that ablation remove ice. At the snow–ice interface, absent interfacial phase change,At the base, $T_i=T_f(S_w)$ and the Stefan condition isHere $G_b$ is signed bottom growth speed, and $F_o>0$ is ocean heat flux toward the ice. An upward conductive loss $-q_b>0$ freezes ice; an excess $F_o$ melts it. For a prescribed upward turbulent ocean heat flux, the model needs no separate ocean temperature solution. A water-side closure could write $F_o={\rho }_w{c}_{p,w}{K}_w{T}_{w,z}$ at the boundary in this downward coordinate. Snowfall is an independent mass input: with snowfall mass flux $P_s$, ${\dot{h}}_s=P_s/{\rho }_s-M_s$ in a constant-density approximation. The ice thickness obeys ${\dot{h}}_i=G_b-M_i$. Compaction or refreezing, if added, requires corresponding mass and energy terms rather than counting snowfall directly as ice growth. These boundary conditions and interior equations determine temperature, surface melt and basal growth together.

Under periodically repeated annual forcing, annual sea-ice thermodynamic equilibrium is a periodic annual state, not constant thickness:Winter accretion then balances net annual ablation. “Equilibrium thickness” normally denotes a mean or a specified point in this annual cycle. Different forcing can instead eliminate a perennial-ice solution.

Increasing ocean heat flux reduces bottom growth directly in the Stefan condition, lowering equilibrium thickness and eventually removing perennial ice. The stabilizing winter response is visible in the approximate conductive fluxwhen ice thins, conductive loss and winter regrowth increase. More snow lowers this flux and inhibits winter growth. However, snow also raises albedo and delays summer exposure of dark ice. The competing snow insulation and summer ice protection effects mean an annual snow sensitivity cannot be deduced from winter insulation alone. Under the original forcing, the Maykut-Untersteiner model gives an almost flat annual-thickness response below about $0.7\mathrm{m}$ annual snow depth, followed by strong thickening as snow persists into summer. This particular transition is not a universal threshold for other forcing or snowfall timing. Increasing summer albedo reduces absorbed solar energy, reduces surface and internal melting, and increases equilibrium thickness; sufficiently low albedo removes perennial ice. The original thermodynamic sensitivity calculations establish these three responses.

The three curves are original qualitative sketches, not a new numerical integration or digitization. The snow marker refers to that historical forcing; the unnumbered loss boundaries and ordinate indicate trends rather than calibrated universal thresholds.

For the summer retreat discussed around the time of this examination, the proposed mechanisms act as follows.

These processes overlap: circulation causes export, export changes thickness and open-water fraction, and those changes alter solar absorption. Without specified regional budgets there is no defensible unique percentage partition. For enhanced Beaufort summer basal melt, upper-ocean solar heating and western-Arctic heat pathways are more directly relevant than invoking unmodified heat from a deep, isolated Atlantic layer. ice-mass-balance observations of extreme Beaufort melt identify increased solar absorption in open water as an important source.

Take ${\rho }_i=900\mathrm{kg}{\mathrm{m}}^{-3}$ as an explicit ice-density approximation and use the supplied $L_f=336000\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. The latent heat for $1.5\mathrm{m}$ basal ablation isJune–September contains about $122$ days, soUsing $917\mathrm{kg}{\mathrm{m}}^{-3}$ would give $43.8\mathrm{W}{\mathrm{m}}^{-2}$. This is the net flux used for melting: the gross ocean heat flux can also supply conductive losses and changes in cold content. It greatly exceeds an annual central-pack background of a few $\mathrm{W}{\mathrm{m}}^{-2}$, but is compatible with a summer upper ocean absorbing solar radiation at tens to hundreds of $\mathrm{W}{\mathrm{m}}^{-2}$. For scale, an added open-water fraction $0.3$, a shortwave input $200\mathrm{W}{\mathrm{m}}^{-2}$ and an ice–water albedo contrast $0.6$ produce $0.3\times 200\times 0.6=36\mathrm{W}{\mathrm{m}}^{-2}$ additional area-mean absorption before losses. This is an illustrative budget, not a measurement. Wind mixing and circulation can deliver stored solar or advected heat to the underside.

The ice-albedo feedback closes the warming loop: retreat exposes darker water; extra solar absorption warms the upper ocean; basal and lateral melt increase retreat. The stored heat also delays autumn freeze-up, exposing water to atmospheric heat transfer for longer. Its later release warms the atmosphere, but must not be counted as a second source in addition to the solar energy that originally entered the ocean. In winter, greater heat loss from open water and rapid growth of thin ice partly oppose the summer positive feedback.

A convective chimney is a localized, deeply mixed region within a stratified ocean, formed when surface buoyancy loss allows convection to penetrate far below the surrounding mixed layer. The chimney describes the mixed water column and its circulation; instantaneous downward transfer occurs through many smaller plumes, accompanied by compensating motion and lateral exchange, rather than through one permanently uniform downward pipe. Open-ocean convection occurs in the Labrador and Irminger Seas, the Greenland Sea, the northwestern Mediterranean, and episodically in the open Weddell Sea under a large polynya. Dense-water formation on a continental shelf is related but geometrically distinct from deep open-ocean chimney convection.

Observed Greenland Sea convective chimneys include compact, approximately circular, weakly stratified cores with temperature and salinity much more uniform vertically than in the ambient ocean. A mapped example was about $20\mathrm{km}$ across, extended to about $2500\mathrm{m}$ in winter, and rotated anticyclonically. In summer a fresh, low-density cap and shallower lateral intrusions could isolate the deep core, while winter buoyancy loss removed the cap and renewed deep contact. Such cores could persist through several seasons, so presence of a deep homogeneous core is not proof that vigorous convection is occurring continuously. These particular dimensions and seasonal structure come from successive hydrographic surveys of Greenland Sea chimneys.

Formation is favoured by the cyclonic Greenland gyre, which domes density surfaces and can reduce the depth of the stratified barrier. Cold, dry wind outbreaks produce strong sensible, latent and longwave heat loss. Where frazil ice and pancake ice form in the Odden ice tongue, ice growth removes relatively fresh material while brine rejection increases the density of the remaining water. Ice advection can keep exposing water for further freezing and separate local brine release from later freshwater return. This combination of preconditioning, cooling and sustained salt input can make a formerly buoyant surface layer dense enough to convect.

Conversely, a fresh surface cap supplied by Arctic freshwater and ice melt strengthens density stratification. Warmer atmospheric outbreaks reduce surface heat loss and ice production; circulation changes affecting the cold northwesterly winds and the Odden can reduce the brine source. Atlantic advection supplies both heat and salt, with competing effects on density, so temperature alone is not an adequate diagnostic. The historical decline in Odden ice formation and changes in wind regimes offer a mechanism for less favourable chimney production around the period considered, but ice-driven formation is a physical hypothesis, not evidence that every observed chimney requires ice or that every local change follows a single climate index. The Greenland Sea salt-flux model explicitly includes ice production and advection.

For the brine rejection mechanism, let ${\dot{h}}_i>0$ be ice growth and let $s_w,s_i$ be water and ice salt mass fractions. Newly formed ice removes mass ${\rho }_i{\dot{h}}_i$ per area per time. Compared with removing water at concentration $s_w$, it leaves an excess salt flux in the liquid ofThe units are $\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. With the exam's approximation $s=10^{-3}S$ for salinity $S$ in psu, the corresponding mixed-layer salinity tendency is, to first order,If ${\beta }_S=(1/{\rho }_w)(\partial \rho /\partial S)>0$, the surface buoyancy input is negative, with scale $-g{\beta }_S({\rho }_i/{\rho }_w){\dot{h}}_i(S_w-S_i)$. The enhanced density produces downward saline plumes that deepen mixing if this buoyancy loss overcomes the available density stratification. This explains why exporting new ice while retaining its rejected salt can strengthen overturning, whereas later local melt reverses the freshwater effect.

For the numerical comparison, the cover gives the specific heat capacity of ice, not of seawater. A seawater value must therefore be supplied explicitly. Take $c_{p,w}=4000\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and use the stipulated ${\rho }_w=1025\mathrm{kg}{\mathrm{m}}^{-3}$. A $100\mathrm{m}$ column has initial water mass per areaCooling it by $1.8\mathrm{K}$ removesThe specified thermal density relation givesThis assumes the whole column mixes and neglects entrainment and advected heat. Using the supplied ice heat capacity $2100$ for liquid seawater would change $Q$ incorrectly.

At the freezing point the same heat extraction produces a pure-ice mass per areaHere $f_i$ is the fraction of the original water mass removed into ice. With the explicit ice-density approximation ${\rho }_i=900\mathrm{kg}{\mathrm{m}}^{-3}$, the resulting ice thickness isThe water-equivalent thickness uses a reference water density: with fresh water at $1000\mathrm{kg}{\mathrm{m}}^{-3}$, it is $m_i/1000=\boxed{2.196\mathrm{m}\simeq 2.2\mathrm{m}}$. The depth of the original seawater removed is instead $m_i/1025=\boxed{2.143\mathrm{m}}$. These three quantities are different; $2.2\mathrm{m}$ is not the physical ice thickness for the stated density approximation.

Let the initial salinity be $S_0$. Complete exclusion of salt from the ice leaves all the initial salt in a liquid mass $M_1=M_0-m_i$. Conservation of salt during freezing givesThe initial salinity is not printed, so a unique numerical salinity increase requires another explicit assumption. For $S_0=35\mathrm{psu}$,The total density increase after cooling followed by freezing is then about $0.9464\mathrm{kg}{\mathrm{m}}^{-3}$. The factor $1/(1-f_i)$ is required because salt concentration refers to the remaining liquid mass. Using ice thickness divided by the original water depth as the removed mass fraction would violate this accounting.

For this assumed salinity, the freezing-to-cooling density efficiency shows that equal heat extraction at freezing produces about $0.7664/0.18=\boxed{4.26}$ times the density increase of the initial cooling. At low temperature the thermal expansion response is weak, whereas brine rejection changes salinity efficiently. This supports strong freezing-driven convection where water is already near its freezing point. It does not mean freezing always creates deep chimneys: the initial density stratification, freshwater supply, mixing depth and later melting remain decisive. The freezing point itself decreases as salinity rises; keeping it at $-1.8^{\circ }\mathrm{C}$ is the exam's approximation. Allowing that small additional cooling and a temperature-dependent equation of state would require a refined heat budget.

Important other sites of freezing-enhanced downward transfer include the Ross and Weddell Antarctic shelves and their coastal polynyas, Adélie Land and Cape Darnley, Arctic shelves such as the Barents and Kara Seas, and the Sea of Okhotsk. On shelves the saline dense water can cascade down a continental slope and feed deeper waters; the same salt-removal balance operates without requiring an isolated open-ocean chimney. A later direct observational example is bottom-water formation associated with Cape Darnley sea-ice production.

Atmospheric warming of the Greenland Sea can reduce surface cooling and ice formation, while increased freshwater input can inhibit convection. If these changes weaken the Atlantic meridional overturning circulation and its northward heat transport, their ocean-mediated tendency is relative cooling, or reduced warming, of northwestern Europe compared with otherwise identical atmospheric warming. This is not a prediction of net European cooling under all emissions scenarios, nor does one chimney determine the whole overturning circulation: other convection regions, winds, freshwater pathways and coupled ocean adjustment also matter. The physical conclusion is the sign of the heat-transport response, not a specified century-scale temperature decrement.

Use the well-mixed natural ventilation model for a distributed floor source. Let the building height be $H_b$, its volume be $\mathcal{V}$, its temperature excess be $\theta$, and the exterior mass density be $\rho$. Let ${\alpha }_T$ be the coefficient of thermal expansion, so reduced gravity is $g^{\prime }=g{\alpha }_T\theta$. Convert the total heat input $Q$ to the temperature-volume flux $\mathcal{H}=Q/(\rho c_p)$, where $c_p$ is specific heat capacity. If the quoted heat flux is per floor area, first multiply it by that area.

For effective opening area for pressure-driven ventilation, define the equal-opening hydraulic coefficient $K_A$ such that the signed throughflow $q$ obeys $q|q|=K_A\Pi$, with $\Pi$ the total driving pressure divided by mass density. In the effective-area convention of question 6, each opening has discharge $A\sqrt{\delta p/\rho }$, so $K_A=A^2/2$. The usual physical-area orifice law instead gives $K_A=C_d^2{A}^2$; that prefactor does not affect the stability conclusions.

Set $W=\Delta P/\rho >0$ and $G_T=g{\alpha }_T{H}_b$. Positive $q$ denotes entry at the floor and exit at the roof. hydrostatic pressure and wind combine to giveBecause the inflowing air is at exterior temperature in either direction, the heat budget isThe well-mixed ventilation temperature balance at steady state is thereforeAll roots must satisfy $\theta >0$ and their appropriate flow-direction inequality; squaring a pressure relation without imposing those conditions could add spurious branches.

The assisting case has a unique positive solution and it is stable. For opposing wind, define ${\theta }_w=W/G_T$. There is always one solution with $\theta >{\theta }_w$, a buoyancy-dominated upward flow. Below ${\theta }_w$, the heat-removal curve is $R(\theta )=\sqrt{K_A}\theta \sqrt{W-G_T\theta }$. It is zero at both endpoints, and its maximum occurs at $\theta =2{\theta }_w/3$:For $0<\mathcal{H}<R_{\max }$ there are two downward-flow roots as well as the upward-flow root. Equivalently, setting $y=G_T\theta /W$ and $\epsilon =G_T\mathcal{H}/(\sqrt{K_A}{W}^{3/2})$, the equilibrium curve isFor linear stability analysis of the opposing-wind ventilation bistability, linearize the heat budget: the eigenvalue is $-R^{\prime }(\theta )/\mathcal{V}$. The cooler downward root has $0<\theta <2{\theta }_w/3$ and $R^{\prime }>0$, so it is stable. The warmer downward root has $2{\theta }_w/3<\theta <{\theta }_w$ and $R^{\prime }<0$, so it is unstable and separates the two basins. The upward root has $R^{\prime }>0$ and is stable. At $\mathcal{H}=R_{\max }$ the two downward roots merge at a saddle-node bifurcation; above that value only the upward equilibrium remains.

For ventilation switching under wind and heating changes, increasing wind pressure or decreasing heat input both reduce $\epsilon$, so both can create the pair of wind-driven equilibria. They nevertheless have different physical and transient effects. On the hot upward branch, implicit differentiation of $\mathcal{H}=\sqrt{K_A}\theta \sqrt{G_T\theta -W}$ shows that increasing $W$ increases $\theta$, while decreasing $\mathcal{H}$ decreases $\theta$. Both reduce the upward ventilation rate, but only a wind change shifts ${\theta }_w$.

Under slow parameter variation, a state follows its current stable branch while that branch persists; the upward branch does not disappear when wind increases or heating decreases. A sufficiently large abrupt wind increase can raise ${\theta }_w$ beyond the current temperature and place that state below the new unstable threshold, causing flow reversal and cooling towards the wind-dominated equilibrium. A jump insufficient to cross its basin boundary returns to the upward equilibrium.

At fixed $W$, reducing a still-positive heat input cannot force an initially upward-flow state through ${\theta }_w$: at that boundary the temperature balance has $\dot{\theta }=\mathcal{H}/\mathcal{V}>0$. Thus heat reduction alone does not cause that reversal in this model. Conversely, decreasing wind or increasing heat from the stable downward branch eventually destroys it at the fold and forces a switch to the upward branch, giving history dependence. Wind strengthening and heat reduction have the same effect on the steady dimensionless control parameter, but not the same temperature change or switching dynamics. These stability statements use quasi-steady opening flow and a single well-mixed temperature; other stratification or airflow-inertia models require their own stability analysis.

Use a planar saline Stefan problem with upward coordinate $z$, the initial contact at $z=0$, and phase boundaries $z=a(t)$ and $z=b(t)$ enclosing pure ice, where $a<b$. Fresh liquid occupies $z>b$ and brine occupies $z<a$. We assume negligible bulk flow, equal constant mass density $\rho$, specific heat capacity $c_p$ and thermal diffusivity $\kappa$ in all regions, and zero salt content and salt transport in the ice. Equal densities remove phase-change volume flow; suppressing convection isolates the molecular-transport mechanism. Different material properties would change the numerical coefficients. Set $K=\rho c_p\kappa$ for the common thermal conductivity and $\Delta T=T_m-T_{\infty }>0$. Denote the brine salt diffusion coefficient by $D$.

Use a linear ice liquidus, $T_L(C)=T_m-mC$, with $m>0$ and $C$ the salt mass fraction. Assume the initial brine is a stable liquid, $T_{\infty }>T_L(C_0)$, and all concentrations considered are below the eutectic composition. Thus $R=m{C}_0/\Delta T>1$. These assumptions exclude independent nucleation in a supercooled brine or an eutectic system event, neither of which is specified by the initial data alone. Start with a negligible seed of ice and use local phase equilibrium without interfacial kinetics or curvature. Temperature differences may be in Celsius, since only differences enter this calculation.

WriteThe paper calls $S$ a Stefan number; it is the latent-to-sensible heat ratio, reciprocal to another common Stefan number convention. At the lower phase boundary, define $T_i=T_L(C_i)$ and letThe liquidus relation is $\vartheta =Rc$.

A similarity solution gives an explicit calculation of the two positions. Putand use $u=z/(2\sqrt{\kappa t})$, $v=z/(2\sqrt{Dt})$. The temperatures and salinities areThese error function and complementary error function profiles satisfy the heat equation and the salt diffusion equation, their far-field conditions, and the interfacial temperature conditions. The fresh-water temperature is constant because the interface and the far field are both at $T_m$.

Salt conservation at the moving lower phase boundary requiresFor upward motion this is dilution by melting salt-free ice, rather than salt rejection by freezing brine. Substitution gives the dilution function for a melting saline Stefan front:For $A>0$, $0<c<1$, so the interfacial brine is fresher than the remote liquid.

The Stefan condition at the upper boundary is $\rho L\dot{b}=K{T}_{s,z}(b)$, since the fresh liquid has no temperature gradient. At the lower boundary it isThese signs follow from the jump in enthalpy: the upper front freezes liquid while a positive $\dot{a}$ melts solid. With $E=\mathrm{erf}B-\mathrm{erf}\delta$, they reduce toTogether with $\delta =ϵA$, these equations determine $A,B,\vartheta$, and hence both positions, without discarding the salt-diffusion correction.

For the stated large latent-to-sensible heat ratio, the layer is thin relative to the thermal diffusion length. Expanding the thermal equations for $B,\delta \ll 1$ givesConsequently a useful leading calculation iswhere the positive $A$ is obtained fromThe left side decreases with $A\ge 0$ while the right side increases, and their values at zero and infinity guarantee a unique positive root for large $S$. Thus this leading calculation includes the translation of the lower boundary as well as the increasing thickness.

If the scale separation also obeys $ϵSA\ll 1$, the familiar simpler result iswithThis last simplification needs the logarithmic refinement $ϵS\sqrt{\log (RS)}\ll 1$. The algebraic ordering $ϵ\ll S^{-1}$ alone should not be used to discard an arbitrarily large logarithmic correction; the preceding coupled equations remain the appropriate calculation when that refinement is unavailable. This distinction is captured by large latent heat in a freezing and melting ice layer.

The illustration uses $S=40$, $ϵ=2.5\times 10^{-4}$ and $R=2$. Solving the full equations gives $A=1.73707$, $B=0.0143053$ and $\vartheta =0.0158766$. It shows the ice layer between freshwater and cold brine, with both fronts advancing upward and a strongly diluted lower liquid boundary.

The phase diagram shows the spatial path from remote brine to the lower interface: temperature first changes over $\sqrt{\kappa t}$ at almost unchanged salinity, then salinity changes over its much smaller diffusion scale at nearly constant temperature. The liquid path ends on the liquidus at $(C_i,T_i)$; the solid has $C=0$ and its temperature rises from $T_i$ to $T_m$. The eutectic position is schematic, with $C_E=2C_0$ in the illustration, and is not used in the calculation. A point initially in the fresh layer freezes when $b$ reaches it, cools in the ice, and later melts when $a$ reaches it. The inset indicates this temporal path in the opposite direction through the solid branch and into the liquid.

The mechanism is simultaneous upper freezing and lower melting, with net growth of the ice thickness. Cold brine accepts most of the latent heat released by upper freezing. Salt reaching the lower contact lowers its equilibrium melting temperature, and melting freshens that liquid until its liquidus is close to $T_m$. The heat delivered through the ice supplies the much smaller melting demand there. The remote brine remains liquid even though it is below the pure-water melting temperature, because it lies above its own saline liquidus. Thus “cold” does not by itself decide which phase is stable. These conclusions apply while the layers remain effectively deep and the stated no-convection, pure-ice and phase-equilibrium assumptions hold.

Use seawater specific heat capacity $c_w\simeq 4.0\times 10^3\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, which is an additional standard material approximation: the paper supplies the ice heat capacity, not the seawater heat capacity. Relative to freezing, the assumed uniform column has $\Delta T=7-(-1.8)=8.8\mathrm{K}$. Its ocean heat content per horizontal area isThis is heat above the reference freezing state, not the absolute internal energy of seawater.

If all this heat reaches ice already at its fusion temperature, divide by the supplied latent heat:For a chosen ice mass density ${\rho }_i=917\mathrm{kg}{\mathrm{m}}^{-3}$, this corresponds to about $5.86\mathrm{m}$ of ice. Density was not specified for this conversion, so the mass per area is the result that needs no further ice-density assumption.

Cold ice must first be warmed. If its initial temperature is $T_i<T_m$, the corresponding ideal melt mass is $Q/[L_f+c_i(T_m-T_i)]$, using the supplied ice specific heat capacity $c_i=2100\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. No initial ice temperature or salinity-dependent fusion temperature is supplied. The latent-only result is therefore an upper bound; actual melting is smaller when sensible warming, ocean or atmospheric losses and incomplete heat transfer matter.

Let $z$ measure distance inward from the surface. While $a\ll R$, curvature is negligible and this is a planar two-phase Stefan problem. Use a common reference mass density $\rho$, specific heat capacity $c_p$, thermal conductivity $k$ and thermal diffusivity $\kappa =k/(\rho c_p)$; retain the mass density differences only in the subsequent buoyancy estimates. Define $\Delta T=T_m-T_s>0$, $\Theta =\overline{T}-T_m\ge 0$, and $\vartheta =\Theta /\Delta T$. The paper's parameter $S=L/(c_p\Delta T)$ is the latent-to-sensible heat ratio, reciprocal to the more usual Stefan number.

For superheated two-phase Stefan similarity, solving $T_t=\kappa T_{zz}$ with $T(0)=T_s$, $T(a)=T_m$ and $T(\infty )=\overline{T}$ gives the Neumann solution of the Stefan problem. With $\eta =z/(2\sqrt{\kappa t})$,The solid expression holds for $0<z<a$ and the liquid expression for $z>a$. The interface gradients inserted in the Stefan condition determine the positive constant $\lambda$:The right-hand bracket decreases from infinity and crosses zero, while the equivalent left-hand side $S\sqrt{\pi }\lambda e^{{\lambda }^2}$ increases, proving uniqueness. The liquid-side flux subtracts from the roof cooling: it is heat brought to the interface by the superheated interior.

For large $S$, $\lambda \ll 1$. Keeping the superheat term while using $\mathrm{erf}\lambda \simeq 2\lambda /\sqrt{\pi }$ and $\mathrm{erfc}\lambda \simeq 1$ yields $S{\lambda }^2+(\vartheta /\sqrt{\pi })\lambda \simeq 1/2$. Hence the small-interface-parameter growth with liquid superheat approximation isThe solid temperature is approximately linear across its thin crust, while the liquid thermal boundary layer has thickness of order $\sqrt{\kappa t}$. For fixed superheat and $S\gg 1$, $A\simeq \sqrt{2/S}$; for $\vartheta \gg \sqrt{S}$, $A\simeq \sqrt{\pi }/\vartheta$. This retains the strong suppression of growth by a very hot interior.

To estimate overturn, take local gravitational acceleration $g$ at the roof. The liquid's mass density contrast is $\Delta {\rho }_l={\rho }_f\alpha \Theta$, while the solid contrast is $\Delta {\rho }_s={\rho }_s-{\rho }_f$. A liquid thermal layer of thickness $\delta$ has Rayleigh number $\Delta {\rho }_{l}g{\delta }^3/({\mu }_l\kappa )$. As $\delta \sim \sqrt{\kappa t}$, it reaches its critical value after a time scaling asIn viscous overturn of a dense crust, an unstably stratified viscous crust has a Rayleigh-Taylor instability growth time of order ${\mu }_s/(\Delta {\rho }_{s}ga)$. Setting this equal to its age, with $a=A\sqrt{\kappa t}$, givesThese are formation-to-overturn times. A specified critical liquid Rayleigh number ${\mathrm{Ra}}_c$ multiplies ${\tau }_l$ by ${\mathrm{Ra}}_c^{2/3}$; logarithmic amplification and geometrical constants also affect the crust estimate. They are omitted for the requested order-of-magnitude scaling.

The given numbers yield $S\simeq 2.54$, $\Delta {\rho }_s=500\mathrm{kg}{\mathrm{m}}^{-3}$ and $\Delta {\rho }_l=100\vartheta \mathrm{kg}{\mathrm{m}}^{-3}$. ThusLiquid thermal overturn is enormously faster than crustal overturn throughout the stated range. Even a critical-Rayleigh prefactor of order $10^2$ does not change that ordering. Since $S\simeq 2.54$ is not asymptotically large, low-superheat coefficients from the permitted approximation are only approximate. At the upper end $\alpha \Theta =1.25$, so the linear thermal-density approximation also ceases to be quantitatively credible; that endpoint is a formal scaling estimate under the supplied constant-property model. At a common age the instantaneous viscous instability-time ratio instead scales as $5A10^{-18}/\vartheta$; it should not be confused with the above formation-time ratio.

For a boundary-layer renewal model, the liquid layer reaches ${\delta }_l\sim \sqrt{\kappa {\tau }_l}$. Its cycle-averaged thermal flux is consequentlyIf a critical Rayleigh number is retained, insert a factor ${\mathrm{Ra}}_c^{-1/3}$. Averaging the diffusive $t^{-1/2}$ flux over a cycle only changes an order-one coefficient. The crust reachesso its average solid-volume flux per roof area is the crust-removal fluxThe corresponding mass flux is ${\rho }_s{J}_s$ and latent-heat flux is ${\rho }_{s}L{J}_s$. This independent-cycle estimate is a scale for a crust that reaches overturn thickness and is then renewed. Because the liquid renews much more rapidly, in a hot interior its average heat flux can limit actual crust growth: the consistent interfacial budget is $\rho L\dot{a}\simeq k\Delta T/a-F_l$. A stalled crust would not realize repeated solid cycles. In a superheated interior, detached solid may also remelt; permanent core deposition is most directly justified in the late regime requested next.

At late times $\Theta =0$, liquid thermal buoyancy vanishes and $A^2\kappa \simeq 2\kappa /S=2k\Delta T/(\rho L)$. ThusAssuming thin renewed crust, efficient sinking and constant asteroid radius, conservation of mass gives $4\pi b^2\dot{b}\simeq 4\pi R^2{J}_0$ under the permitted equal-density approximation. Therefore the asteroidal core growth by crustal sinking law isFor a self-gravitating approximately uniform sphere, $g\simeq 4\pi G\rho R/3$, so $J_0\propto R^{1/3}$ and a zero-seed core scales as $b\propto R^{7/9}(t-t_0{)}^{1/3}$. This law applies while the thin-crust renewal model remains valid and liquid remains available; it cannot continue beyond complete solidification. The specific numerical lifetime cannot be inferred without the absolute solid viscosity and conductivity.

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.