Added energy can change how much material is solid or liquid without raising temperature.
By the end: read a heating curve, calculate a melted fraction, and identify pressure and equilibrium assumptions.
Distinguish warming from changing phase
Within one phase and over a modest temperature range, a constant heat-capacity model can approximate warming. During coexistence at a specified pressure, added heat can instead change the phase proportions. For pure water near one atmosphere, ice and liquid can coexist near 0 °C. Adding heat then melts more ice while the equilibrium temperature stays near 0 °C until the solid is gone.
Latent heat per mass is the energy per kilogram for that phase change under specified conditions. It is not a hidden temperature increase. Melting rearranges interactions and increases the available microscopic alternatives. Enthalpy is internal energy plus pressure times volume. The phase-change energy account at fixed pressure includes the appropriate pressure-volume work; heat at constant pressure equals enthalpy change when only pressure-volume work is present.
Explain the energy absorbed during coexistence
Temperature does not uniquely specify the energy of a sample when more than one phase is possible. At a solid-liquid coexistence temperature, two samples can have the same mass, pressure, and temperature but different liquid masses. Their molecular arrangements and enthalpies differ. Adding energy can change those phase proportions instead of raising temperature. This does not suspend energy conservation. It means that a formula based only on temperature change has omitted another way for the sample's state to change.
In ordinary ice, water molecules form an extended structure with constrained relative positions, although they still move and vibrate. Liquid water has a changing network of interactions and allows molecules to rearrange and move past one another. Melting changes those interactions and accessible arrangements. It does not split water molecules into hydrogen and oxygen, nor does it remove every intermolecular attraction. A sketch of lattice positions can distinguish the phases, but its spacing and dot count cannot supply the latent heat numerically.
Melting changes intermolecular arrangement
The same twelve schematic molecule markers appear in each panel. The solid connections indicate a constrained structure, not chemical bonds within water. The liquid drawing omits changing interactions; it does not imply their absence.
At the stated pressure, equilibrium coexistence fixes a common temperature while energy input shifts material toward the liquid phase. To calculate the shift, latent heat of fusion Lf gives an enthalpy difference per kilogram melted. The rounded model value is 334 kJ/kg. Melting 0.2 kg therefore requires 66.8 kJ after that material has reached the melting temperature. This energy changes phase amount. Substituting it into a single-phase heat-capacity formula would incorrectly predict a temperature rise while solid and liquid still coexist.
Why use enthalpy rather than silently calling all constant-pressure heat an internal-energy increase? The sample can change volume as it melts and therefore exchange pressure-volume work. For a closed sample with only that work at fixed external pressure, the first law gives heat input equal to internal-energy change plus pressure times volume change. That sum is the enthalpy change. Ordinary ice contracts on melting near atmospheric pressure, so its expansion-work contribution is negative. The latent-heat account already includes the fixed-pressure contribution rather than requiring an extra correction.
Suppose a 1 kg mixture at the coexistence temperature initially contains 0.3 kg liquid and 0.7 kg ice. Adding 66.8 kJ with no losses melts another 0.2 kg. The final liquid mass is 0.5 kg, so the liquid mass fraction is 0.5. Both initial and final temperatures remain near 0 °C at the stated pressure. The liquid fraction increased by 0.2, but the temperature change is zero. A thermometer alone cannot distinguish these two equilibrium mixtures.
Allocate latent heat to additional liquid mass
66.8 kJ divided by 334 kJ/kg melts 0.2 kg. The bars encode phase masses, not the shape of a container or the volumes occupied by the phases.
A mass strip represents equal masses by equal lengths, even though equal masses of ice and liquid occupy different volumes. For total mass m and liquid mass fraction f, liquid mass is fm and solid mass is the remainder. During coexistence, an increase in fraction by Δf requires heat mLfΔf. This equation assumes pure material, fixed pressure, and equilibrium phase proportions. It does not describe transient pockets at different temperatures within a rapidly heated sample.
Build the heating curve in stages
The model starts with 1 kg of ice at −20 °C. Use rounded constants: ice specific capacity 2.1 kJ per kg per kelvin, latent heat of fusion 334 kJ/kg, and liquid-water capacity 4.18 kJ per kg per kelvin. The first 42 kJ warms the ice to 0 °C. The next 334 kJ melts it, so complete melting occurs after 376 kJ total. Only subsequent heat raises the liquid temperature.
At 209 kJ total input, subtract the initial 42 kJ of warming. The remaining 167 kJ is half the melting requirement, so the sample is half liquid by mass at 0 °C. At 585 kJ, subtract 376 kJ; the remaining 209 kJ warms 1 kg of water by 50 K. The horizontal coordinate is energy input, not time. The model stops below boiling and ignores evaporation and heat loss.
Allocate each energy increment to the correct stage
The initial ice is colder than its melting temperature, so added energy first warms solid material. For 1 kg and specific heat capacity 2.1 kJ per kilogram per kelvin, a 20 K rise requires 42 kJ. The whole-sample capacity is therefore 2.1 kJ/K. Specific capacity and whole-sample capacity happen to have the same numerical value here because the mass is one kilogram, but their units differ. Doubling the mass doubles the whole-sample capacity while leaving the assumed specific capacity unchanged.
Once warming is complete, the next 334 kJ melts the entire kilogram. The cumulative input at the end of melting is 42 plus 334, or 376 kJ. This distinction between stage input and cumulative input prevents a common error. At a total input of 300 kJ, melting has received only 258 kJ. The liquid mass is 258/334, about 0.7725 kg, so the mass fraction is about 77.25%. The remaining solid mass is about 0.2275 kg at the same coexistence temperature.
For an input beyond 376 kJ, subtract that threshold before calculating liquid warming. A total input of 459.6 kJ leaves 83.6 kJ for liquid water. Its whole-sample capacity is 4.18 kJ/K, giving a temperature rise of 83.6/4.18 = 20 K. The final temperature is 20 °C. Dividing the entire 459.6 kJ by liquid capacity would count warming and melting energy as if it all raised the temperature of water that was liquid from the beginning.
Separate warming and melting energy
The narrow first segment is 42 kJ, not the full 376 kJ needed to finish melting. Liquid warming begins only after the cumulative input passes 376 kJ.
The inverse question is equally useful. What input produces a specified fraction or temperature? To obtain 60% liquid from the initial cold ice, first supply 42 kJ, then 0.6 times 334 = 200.4 kJ. Total input is 242.4 kJ. To obtain liquid at 20 °C, supply the complete 376 kJ through melting, then 83.6 kJ for warming. A target of 0 °C is different: many inputs between 42 and 376 kJ reach that temperature with different phase fractions.
Read the heating graph as a map of equilibrium states against cumulative energy. The solid segment rises more steeply than the liquid segment because the ice capacity is smaller in this rounded model. The slope is temperature change divided by energy change, the reciprocal of whole-sample capacity. Along the plateau, energy continues to increase while temperature does not. A horizontal segment is therefore not zero heat input. It indicates a changing phase fraction that the temperature coordinate alone cannot show.
Try resolving an ambiguous observation without using an unavailable control. On paper, compare inputs of 75.4 kJ and 342.6 kJ. After subtracting 42 kJ, the melting inputs are 33.4 kJ and 300.6 kJ. These correspond to liquid fractions 0.1 and 0.9. Both samples read 0 °C, yet their enthalpies differ by 267.2 kJ. The mass bars show the missing state information. Temperature equality establishes thermal equilibrium between them, not equal enthalpy or equal amounts of liquid.
Temperature alone leaves the phase fraction unknown
Each state starts from the same 1 kg of ice at −20 °C. Inputs differ by 267.2 kJ, entirely assigned to melting an additional 0.8 kg. Gold is ice; blue is liquid.
Work through the mechanism
Within one phase
c is appropriate to the heating constraint and is treated as constant. This formula does not include a phase transition.
During melting at fixed pressure
m melted is the mass that changes phase. Lf is about 334 kJ/kg in the rounded water model. Heat used to reach the melting point must be accounted for separately.
Add energy to ice, through melting, into liquid
ice and liquid; temperature 0 °C; liquid 50% by mass. Reference: 209 kJ, 0 °C, 50% liquid.
The curve uses 1 kg at approximately one atmosphere and rounded constant material properties. The fixed ring marks half melted at 209 kJ. The point is the selected state. Inputs replace total added heat, not elapsed time; no heat loss, evaporation, gradients, or boiling is modeled.
Total added heat
Temperature
Liquid mass fraction
0 kJ
−20 °C
0
42 kJ
0 °C
0
209 kJ
0 °C
0.5
376 kJ
0 °C
1
585 kJ
50 °C
1
Rounded constants and quasi-equilibrium at approximately one atmosphere, with negligible loss.
Explore the plateau and its two boundaries
Begin with the initial-ice preset, then select the preset where ice reaches its melting point. The added energy changes from zero to 42 kJ and the temperature rises from −20 °C to 0 °C. Liquid fraction remains zero. Now select the half-melted preset at 209 kJ. The additional 167 kJ changes the liquid fraction to one half without changing temperature. The plotted point moves horizontally because this graph's horizontal coordinate is energy input. Neither displacement nor playback time represents a heating duration.
Compare the half-melted state with the all-melted preset at 376 kJ. Their temperatures match, but another 167 kJ has changed the remaining half kilogram into liquid. Select the liquid-at-50 °C preset next. Its 585 kJ total input exceeds the melting threshold by 209 kJ. Dividing that excess by 4.18 kJ/K gives 50 K. These successive differences identify the energy assigned to each stage. Subtracting consecutive cumulative inputs is often easier to interpret than treating every preset as a separate formula.
Use nearby slider settings to examine the corners rather than only the widely separated presets. At 41 kJ, the ice temperature is −20 plus 41/2.1, about −0.476 °C, with no liquid. At 43 kJ, the sample is at 0 °C and 1/334, about 0.299%, is liquid by mass. One kilojoule on either side of the threshold has different effects because the available equilibrium state changes. The corner is a change in response, not a jump in the temperature itself.
Repeat that comparison around complete melting. At 375 kJ, liquid fraction is 333/334, about 99.701%, at 0 °C. At 377 kJ, all material is liquid and the extra kilojoule warms it to about 0.239 °C. These four neighborhood values are available as integer slider settings. Predict each output before moving the slider, then identify whether the energy changes temperature or mass fraction. The model makes idealized sharp boundaries; its numerical continuity does not imply that real samples equilibrate instantly.
Inspect both ends of the plateau
Both horizontal scales span 4 kJ, but the temperature scales differ. Near 42 kJ the warming solid reaches coexistence; near 376 kJ the last solid disappears and liquid warming begins.
Next change the mass on paper while retaining the same rounded properties and starting temperature. For 2 kg of ice, warming to 0 °C requires 84 kJ, and complete melting requires another 668 kJ. The cumulative threshold is 752 kJ. At 209 kJ total input, only 125 kJ remains for melting. It melts 125/334, about 0.3743 kg, which is only about 18.71% of the two-kilogram sample. Reusing the one-kilogram graph without rescaling would give the wrong fraction.
Energy per kilogram provides the scaling rule. The one-kilogram half-melted state needs 209 kJ/kg, so a two-kilogram sample needs 418 kJ to reach that same temperature and fraction. Its liquid mass is then one kilogram, rather than half a kilogram. The interface does not change mass; this comparison is a separate calculation. It distinguishes an intensive state description, temperature and phase fraction, from extensive quantities such as total enthalpy difference and liquid mass. Matching intensive quantities does not require matching total energy input.
Scale energy and phase masses together
Both bars are 50% liquid by mass at 0 °C. The second bar is twice as long because it contains twice the mass. The same energy per kilogram gives the same temperature and fraction under the model assumptions.
Pressure, gradients, and metastability matter
Boiling occurs when a liquid’s equilibrium vapor pressure matches the surrounding pressure. Evaporation can occur below the boiling temperature. Lower ambient pressure generally lowers water’s boiling temperature, so 100 °C is not a universal boiling point. A pressure-temperature phase diagram locates phase stability and coexistence; a heating curve instead follows one process and can include changing proportions at a fixed coexistence temperature.
Real samples may have temperature gradients, dissolved substances, or delayed nucleation. Supercooling and superheating are metastable states, not evidence that equilibrium phase rules have ceased to apply. Water also has an unusual solid-liquid density relationship: ordinary ice is less dense than liquid water. The model’s liquid fraction is a mass fraction, not a claim about equal solid and liquid volumes.
Connect coexistence with pressure and entropy
A heating curve follows a process at a selected pressure; a pressure-temperature phase diagram answers a different question. Its coexistence line identifies combinations where two phases can be in equilibrium. It does not specify how much of each phase is present. At one point on the solid-liquid line, a sample could contain little liquid or mostly liquid. The phase fractions require another account, such as the added-energy balance. One phase-diagram point can correspond to a whole heating-curve plateau.
Pressure matters because changing phase can change volume. Increasing pressure favors the phase with smaller volume for the same amount of matter. Near ordinary melting conditions, liquid water is denser than ice, so increasing pressure lowers the equilibrium melting temperature. Many substances show the opposite trend because their solids are denser than their liquids. A schematic line shows this contrast without predicting a numerical shift. Quantifying it requires property data and a model for that pressure range.
Pressure-temperature position does not encode phase amount
Axes show directions of increasing absolute pressure and temperature without numerical scales. Phase labels identify the lower-temperature solid and higher-temperature liquid sides of each line. No numerical melting-point shift is implied.
The entropy difference between coexisting phases also has a direct energy interpretation. Imagine melting one kilogram reversibly at the coexistence temperature, using an infinitesimally warmer source. With the rounded latent heat, sample entropy increases by 334,000 J divided by 273.15 K, about 1,222.77 J/K. Use kelvin here even though the heating graph labels its vertical axis in Celsius. Dividing by 0 °C would be meaningless. The positive entropy change belongs to the sample; reversibility requires the surroundings to lose an equal amount.
Contrast that reference with direct heat transfer from an ideal reservoir at 300 K to the melting sample at 273.15 K. Keep the transferred energy at 334 kJ and allow no other exchanges. The sample's endpoint entropy increase is still about 1,222.77 J/K because its initial and final equilibrium states are unchanged. The reservoir loses 334,000/300, about 1,113.33 J/K. Total entropy therefore rises by about 109.44 J/K. The finite temperature difference generates entropy even though the sample remains on its temperature plateau.
Melting from a warmer reservoir generates total entropy
The sample endpoint change uses a reversible melting reference at 273.15 K. Actual heat comes from 300 K. Signed bar lengths use 0.22 drawing units per J/K; the positive total is the difference, not an extra heat input.
Freezing reverses the sample's phase and entropy changes, but it does not reverse every possible heat-transfer route. Reversibly freezing one kilogram at coexistence releases 334 kJ and lowers sample entropy by about 1,222.77 J/K. That negative subsystem change is allowed if the surroundings receive sufficient entropy. Sending the released energy to a colder reservoir can give a positive isolated-total change. The second law concerns the complete boundary, not a rule that each material must always become more liquid or increase its own entropy.
Finally, distinguish equilibrium states from delayed transitions. A pure liquid can remain temporarily below its equilibrium freezing temperature if the transition lacks an effective starting structure. This metastable state is not the two-phase equilibrium mixture assumed by the plateau calculation. Impurities and spatial temperature differences can also change the observed relationship. The model assigns equilibrium states and predicts no waiting time. Use it to reason about energy partitions and mass fractions, not to infer nucleation rates or certify a material's thermal behavior.
Exercises
Finish a partial melting calculation
Starting from the same 1 kg of ice at −20 °C, add 125.5 kJ. Find the temperature and liquid fraction.
Hint
First pay the 42 kJ needed to reach the melting temperature.
Worked answer
Energy use
Amount
Warm ice to 0 °C
42 kJ
Melt one quarter
83.5 kJ
Total
125.5 kJ
Final state
0 °C, 25% liquid by mass
The plateau still absorbs energy.
After warming, 83.5 kJ remains for melting. That is one quarter of 334 kJ, so the sample contains 0.25 kg liquid and 0.75 kg ice at 0 °C. Applying the liquid-water heat capacity to the entire 125.5 kJ would skip both warming the ice and the phase transition.
Interpret two samples at the same temperature
Two 1 kg equilibrium samples at 0 °C and one atmosphere contain respectively 25% and 75% liquid by mass. Does equal temperature imply equal enthalpy? Find the difference using the model constants.
Hint
The second sample has melted an additional half kilogram.
Worked answer
Comparison
Value
Additional liquid mass
0.5 kg
Latent heat per mass
334 kJ/kg
Enthalpy difference
167 kJ
Equal temperature can coexist with a large phase-energy difference.
No. The 75% liquid sample has 0.5 kg more melted material. Its enthalpy is higher by about 167 kJ. Equal temperature fixes thermal equilibrium, not the phase proportions or total energy-related state variables.
Sources and further reading
These optional references support the stated laws and model assumptions. The numerical examples and diagrams are authored teaching models, not experimental measurements or engineering specifications. Selected relevant passages were reviewed, not entire textbooks or a comprehensive research literature.