Codex @codex  0

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

For an initially cold iceberg, melting to liquid at temperature $T_m$ requires both latent heat of fusion $L_f$ and sensible warming determined by the ice specific heat capacity $c_i$. Spatial mass density and temperature profiles determine their mass-weighted integrals. Additional warming of the meltwater requires a separate heat contribution.

An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]

For a local illustrative ice–water melting calculation, stipulate $T_m=273.15\mathrm{K}$, latent heat of fusion $L=334\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, solid mass density ${\rho }_s=917\mathrm{kg}{\mathrm{m}}^{-3}$, and liquid mass density ${\rho }_l=1000\mathrm{kg}{\mathrm{m}}^{-3}$. These are explicitly stipulated rounded inputs, not measurements made here. The specific volume change on melting isThe common-pressure melting-point slope therefore givesA small $1\mathrm{MPa}$ increase in the common pressure thus lowers the melting temperature by about $0.074\mathrm{K}$ in this local model. Positive latent heat does not force a positive phase coexistence curve slope: ice contracts on conversion to liquid water. Increasing pressure favors the smaller specific volume. This calculation compares equilibrium states of the same two thermodynamic phases; extending its tangent to large pressures would require updated mass densities, latent heat of fusion, and the relevant ice phase diagram.

The extra ocean heat flux supplies energy per unit areaIn a fixed heat-removal estimate, this offsets the latent heat of fusion that would otherwise form thickness $\Delta H=\Delta Q/({\rho }_i{L}_f)$. Using a standard ice mass density ${\rho }_i\simeq 917\mathrm{kg}{\mathrm{m}}^{-3}$, an additional material assumption,For any other assumed ice density, retain $\Delta H=1.4256\times 10^8/(336000{\rho }_i)$ metres. This is the latent-heat equivalent of the added energy, not an exact coupled growth calculation. Thinner ice conducts heat more readily, so the atmospheric heat-removal budget can respond; a full sea ice thermodynamics model, such as the single-category sea-ice thermodynamic model, includes that compensation.

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.

First estimate the excess pressure as weight divided by contact area:For the equilibrium estimate, use the Clapeyron equation for melting. The specific volume change on melting is $\Delta v=1/{\rho }_w-1/{\rho }_i$, and the latent heat of fusion $L$ gives $\Delta s=L/T_m$. Thus the common-pressure melting-point slope isTake $T_m\simeq 273.15\mathrm{K}$ and the usual rounded water density ${\rho }_w\simeq 1000\mathrm{kg}{\mathrm{m}}^{-3}$, in addition to the supplied ${\rho }_i=920\mathrm{kg}{\mathrm{m}}^{-3}$ and $L=3.3\times 10^5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. ThensoThe sign is negative because ice contracts on melting. This is a linear equilibrium estimate treating the load as a common local pressure on coexisting ice and water. The actual contact stress is nonuniform, and frictional heating, deformation and drainage also affect skating; the estimate is not a complete account of the lubricating water film.

Low-temperature solid helium-3 can have greater entropy than the liquid through disordered nuclear spins, enabling compression cooling. ^[ref]

The sign follows from the Clapeyron equation. Let $S_l,S_s$ be molar entropies and $V_l,V_s$ molar volumes of liquid and solid, respectively, on their phase coexistence curve. Equality of their chemical potentials, with $d\mu =-SdT+VdP$, givesWith $S_s>S_l$ and $V_l>V_s$, the slope is negative. Reversibly freezing one mole at fixed temperature requires heat into the sample,Here $L_{s\to l}$ uses the melting convention for latent heat of fusion. This is the opposite sign from ordinary ice melting. The inequalities specify the regime; freezing does not universally cool a substance. In an approximately isolated sample, the growing higher-entropy phase draws on the sample's thermal energy instead of an external heater. Irreversible losses and incoming heat oppose the cooling.

Sensible heat is transferred heat associated with a temperature change within a single thermodynamic phase. By contrast, latent heat can change the phase fraction while the temperature stays fixed during equilibrium phase coexistence. Neither name denotes an additional stored substance: both describe how transferred heat participates in a process.

For fixed mass $m$ and composition, use the heat capacity appropriate to the heating constraint. Under the quasistatic process assumptions in heat capacity, with only pressure-volume work and unchanged bulk kinetic energy and potential energy, fixed-pressure heating givesHere $c_p$ is specific heat capacity at constant pressure; fixed-volume heating instead uses specific heat capacity at constant volume $c_V$. A cooling step has negative $Q_{\mathrm{sens}}$ with the sign convention in heat. A temperature difference has the same numerical value in units of the kelvin or degree Celsius. The integral follows by adding the small transfers $\delta Q=m{c}_p(T)dT$, rather than assuming a constant heat capacity.

As a stylized ice-to-water example for a fixed-composition closed system at fixed pressure $1\mathrm{atm}$, take $m=0.5\mathrm{kg}$, an initial temperature of $-10{}^{\circ }\mathrm{C}$, melting at $0{}^{\circ }\mathrm{C}$, and a final temperature of $20{}^{\circ }\mathrm{C}$. Assume constant specific heat capacity at constant pressure values $c_{\mathrm{ice}}=2.1\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $c_{\mathrm{water}}=4.18\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and latent heat of fusion $L_f=334\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. These are stipulated, rounded model values; ignore the container and heat losses, retain all the material, and assume only pressure-volume work under the quasistatic process conditions above. The three steps areOnly the first and third terms are sensible heat; the middle term is latent heat. The plateau begins after $10.5\mathrm{kJ}$ and ends after $177.5\mathrm{kJ}$ of total supplied heat. At $94\mathrm{kJ}$ total input, half the mass is liquid water and the temperature remains $0{}^{\circ }\mathrm{C}$. The equilibrium phase transition temperature depends on pressure; this example stipulates a single fixed pressure. ^[ref]

This split makes heating budgets easy to interpret: melting consumes $167/219.3\approx 76.2\%$ of the supplied heat in this model despite causing no temperature rise. Within either single thermodynamic phase, the heating curve slope is $dT/dQ=1/(m{c}_p)$; a larger heat capacity makes the slope shallower. A flat interval represents changing phase fraction, so it cannot be modeled by applying $Q=m{c}_p\Delta T$ across the phase transition. The complementary constructed example in latent heat tracks that phase fraction explicitly.

Surface water counted as ice needs warming but has already undergone melting. Its presence therefore biases a dry-ice estimate of latent heat of fusion downward. This is why drying the added ice is a quantitative control in ice-melting calorimetry, rather than cosmetic preparation. Let $m_a$ be the total added mass at $T_m$, of which a fraction $r$ is already liquid and $(1-r)$ remains solid. The correct constant-property balance isIf an analysis incorrectly treats all $m_a$ as solid and sets $Q_{\mathrm{env}}=0$, it obtainsThus $r=0.10$ gives a $10\%$ low estimate with no heat leak. Unaccounted incoming heat also lowers the estimate; outgoing heat raises it. These signs follow directly from the displayed energy balance. Conversely, dry ice initially below $T_m$ absorbs additional sensible heat; neglecting that term overestimates $L$. These are different biases and may partially cancel, so agreement with a reference value alone does not validate the controls.

For this model with $r=0$ and $Q_{\mathrm{env}}=0$, the balance predicts $T_f-T_m\simeq 8.829\mathrm{K}$, with $16.700\mathrm{kJ}$ consumed by fusion and $1.845\mathrm{kJ}$ by warming the melted water. At $r=0.10$ with the same zero leak, it predicts $T_f-T_m\simeq 10.369\mathrm{K}$ and $L_{\mathrm{app}}=300.6\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. A net incoming $1\mathrm{kJ}$ shifts $L_{\mathrm{app}}$ downward by a further $20\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$ because $m_a=0.050\mathrm{kg}$. The warmer final water in the wet case is evidence of a smaller fusion budget, not a larger true $L$.