Latent Heat Calculator

Physics

Or start from a worked scenario:

Substance

Presets carry melting and boiling points, latent heats and per-phase specific heats.

Mass unit

Mass m

Start temperature T₁

End temperature T₂

Temperature unit

Start phase

Needed when the temperature sits exactly on a transition point.

End phase

Molar mass M (g/mol)

Optional — enables kJ/mol and the Trouton check.

Pressure (bar)

L_v and the boiling point both move with pressure.

Decimal places

Show the step-by-step working

MODEL — latent pressure dependence

Water (liquid)'s latent heats are tabulated at 100 °C and 1 atm. L_v is strongly pressure-dependent — water's falls from 2500.9 kJ/kg at 0 °C to 1939.8 kJ/kg at 200 °C and to zero at the critical point — so treat any result at another pressure as an estimate. L_f barely moves with pressure and is safe to treat as constant.

Heating curve

Total energy: 3094600.000 J

Latent share: 83.82 %

Latent energy: 2594000.000 J

Sensible energy: 500600.000 J

Drawn to scale: energy runs left to right, temperature bottom to top. The shaded blocks are latent plateaus, where the line is flat because energy is going into the phase change rather than the thermometer.

Latent against sensible

#	Segment	Type	Q	Cumulative	% of total
1	Solid -20 °C → 0 °C	sensible	41800.000 J	41800.000	1.35 %
2	Melt at 0 °C	latent	334000.000 J	375800.000	10.79 %
3	Liquid 0 °C → 100 °C	sensible	418600.000 J	794400.000	13.53 %
4	Vaporise at 100 °C	latent	2260000.000 J	3054400.000	73.03 %
5	Gas 100 °C → 120 °C	sensible	40200.000 J	3094600.000	1.30 %

The substance starts as a solid and ends as a gas. Per-phase specific heats used: solid 2090, liquid 4186, gas 2010 J/(kg·K).

Phase map and reference values

Solid

Below the melting point. Absorbing L_f melts it; releasing L_f freezes it back.

Liquid

Between melting and boiling. Absorbing L_v boils it; releasing L_v condenses it.

Gas

Above the boiling point. Deposition takes a gas straight to a solid, releasing L_s.

Property	Water (liquid)
Melting point	0 °C
Boiling point	100 °C
Fusion (melt / freeze) `L_f`	334,000 J/kg
Vaporisation (boil / condense) `L_v`	2,260,000 J/kg
Sublimation (sublime / deposit) `L_s`	2,834,100 J/kg
Molar enthalpy ΔH = L · M	6.017 kJ/mol
Trouton ratio ΔH_vap / T_b	109.110 J/(mol·K) — 24.0 % from the 88 J/(mol·K) rule

MODEL — L_s equals L_f + L_v only at a common temperature

Adding the tabulated L_f and L_v gives 2,594,000 J/kg against a measured sublimation value of 2,834,100 J/kg, a -8.47 % gap. That is not rounding: the two latent heats were measured at different temperatures, and L_v falls as temperature rises.

How this substance compares

Water (liquid)

L_f 334 kJ/kg · L_v 2260 kJ/kg

Ethanol

L_f 108 kJ/kg · L_v 841 kJ/kg

Ammonia (liquid)

L_f 332 kJ/kg · L_v 1371 kJ/kg

Lead

L_f 23 kJ/kg · L_v 858 kJ/kg

Aluminium

L_f 397 kJ/kg · L_v 10900 kJ/kg

Upper bar is the latent heat of fusion, lower bar the latent heat of vaporisation, both against the same scale. Water’s vaporisation value is the outlier that shapes the entire heating curve.

How L_v collapses as temperature rises

Temperature	Estimated L_v (kJ/kg)
0 °C	2543.7
50 °C	2408.7
100 °C	2260.0
150 °C	2093.4
200 °C	1901.7
300 °C	1374.0
373.946 °C (critical point)	0.0

Estimated with the Watson correlation, not looked up. At the critical point liquid and vapour become indistinguishable, so the latent heat is exactly zero — there is no transition left to pay for.

Latent Heat Calculator – Q = mL, Heating Curves and Phase Change

Put a thermometer in a pan of melting ice and it will not move. Energy is pouring in, the ice is visibly disappearing, and the reading sits at 0 °C the whole time. That stubborn plateau is latent heat— the energy a substance absorbs or releases while it changes state, at constant temperature. This latent heat calculator works Q = m · L for whichever quantity you leave blank, and walks full heating curves that alternate latent plateaus with ordinary sensible heating.

Latent heat against sensible heat

Sensible heat moves the thermometer: Q = m · c · ΔT. Latent heat moves the phase: Q = m · L, with ΔT = 0. Melting a kilogram of ice costs 334 000 J; boiling a kilogram of water costs 2 260 000 J. Neither raises the temperature by a single degree. The specific latent heat of fusion L_f covers melting and freezing, the latent heat of vaporisation L_v covers boiling and condensing, and the latent heat of sublimation L_scovers solids that go straight to gas, like dry ice at −78.46 °C.

The heating curve, drawn to scale

Take one kilogram of water from −20 °C ice to 120 °C steam and the journey splits into five stages: warming the ice costs 41 800 J, melting it costs 334 000 J, warming the water costs 418 600 J, boiling it costs 2 260 000 J, and superheating the steam costs 40 200 J. The total is 3 094 600 J, of which 83.82 % is latent and the vaporisation plateau alone is 73.03 %. Boiling water away costs 5.4 times as much energy as heating it from freezing to boiling in the first place, and L_v is 6.77 times L_f. For water, latent heat is not a footnote to thermodynamics — it is most of the bill.

Why steam burns are so much worse

Steam and boiling water read the same 100 °C, yet the burns are not comparable. Steam has to condense before it can cool, releasing the full latent heat on your skin. A hundred grams of steam delivers 226 000 J condensing plus 26 372 Jcooling to 37 °C — 252 372 J in total, against 26 372 J for the same mass of boiling water. That is 9.57 times the energy from an identical thermometer reading.

Partial phase changes give a state, not a temperature

Spend a fixed energy budget and it may run out mid-plateau. Give 500 g of −10 °C ice exactly 100 000 J: warming it to 0 °C uses 10 450 J, and the remaining 89 550 J melts only 53.62 %of it. The answer is 268.114 g of water sitting alongside 231.886 g of ice, both at 0 °C. Reporting a single temperature there would hide the physics completely, so the calculator returns a state with a melt fraction. The same logic explains an ice bath: the plateau absorbs energy without letting the temperature move.

Melting, freezing and running costs

Because Q = m · L rearranges to m = Q / L, a megajoule vaporises 442 g of water but melts 2.994 kg of ice — the ratio is exactly L_v / L_f. Feed the same equation a heater and you get time: a 2 kW kettle at 85 % efficiency needs 1994.12 sto boil 1.5 kg of water dry, over six times longer than the 313.95 s it took to heat that water to boiling. Run it in reverse for refrigeration and freezing 2 kg of 20 °C water into −18 °C ice removes 910 680 J, of which 73.35 % is the freezing plateau rather than the cooling.

Latent heat of vaporisation is not a constant

Water’s L_vis 2500.9 kJ/kg at 0 °C, 2256.4 kJ/kg at 100 °C, 1939.8 kJ/kg at 200 °C, and exactly zero at the critical point of 373.946 °C, where liquid and vapour become indistinguishable. Pressure cookers, autoclaves and boilers all sit well off the one-atmosphere figure. The latent heat of fusion barely moves with pressure, so melting calculations stay reliable.

Sublimation and the molar view

Dry ice never melts at atmospheric pressure: it sublimes straight to gas, absorbing 571 000 J/kg. Hess’s law says L_s = L_f + L_v, but only when both are measured at the same temperature — using water’s familiar 100 °C figure makes the sum 8.6 % too low. Multiply any latent heat by the molar mass and you get the molar enthalpy: 40.71 kJ/molfor water’s vaporisation. Divide that by the boiling point in kelvin and Trouton’s rulesays normal liquids land near 88 J/(mol·K). Water reaches 109.1, a 24 % excess caused by hydrogen bonding — the same bonding that makes its latent heat so large in the first place.

Where the model stops

Mixing results assume a perfectly insulated container with no evaporation from an open surface. Supercooling and superheating are not modelled: the calculator assumes the transition happens exactly at the melting or boiling point, though real water can sit below 0 °C without freezing. Boiling points themselves move with pressure — water boils near 93 °C at 2000 m altitude — so a plateau drawn at 100 °C is a sea-level plateau. Every one of these limits is flagged in the results as a MODEL warning, distinct from the harmless ARITHMETIC residue of floating-point maths.

Is the Latent Heat Calculator free?

Yes, Latent Heat Calculator is totally free :)

Can I use the Latent Heat Calculator offline?

Yes, you can install the webapp as PWA.

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How does this latent heat calculator work?

It works Q = m·L, the energy a substance absorbs or releases at constant temperature while it changes state, and rearranges it for whichever quantity you leave blank. For a temperature span it walks a full heating curve instead, alternating sloped sensible segments (Q = m·c·ΔT) with flat latent plateaus (Q = m·L) and totalling them stage by stage. Every input is normalised to SI at full double precision and rounded only when it is printed.

What is the difference between latent heat and sensible heat?

Sensible heat moves the thermometer; latent heat moves the phase while the thermometer stands still. Melting 1 kg of ice at 0 °C absorbs 334 000 J without the temperature changing at all, and boiling 1 kg of water at 100 °C absorbs 2 260 000 J the same way. Taking 1 kg of water from −20 °C ice to 120 °C steam costs 3 094 600 J in total, and 83.82 % of that happens at constant temperature — latent heat is the larger half of thermodynamics for most everyday materials.

Why is a steam burn so much worse than a boiling-water burn?

Because steam must condense before it can cool, and condensing releases the full latent heat of vaporisation. 100 g of 100 °C steam landing on 37 °C skin delivers 226 000 J condensing plus 26 372 J cooling — 252 372 J altogether. The same 100 g of 100 °C water delivers only the 26 372 J of cooling. That is 9.57 times the energy from the same thermometer reading.

What happens when the energy runs out part-way through melting?

You get a two-phase mixture, not a temperature. Giving 500 g of −10 °C ice exactly 100 000 J warms it to 0 °C using 10 450 J and then melts 53.62 % of it with the remaining 89 550 J, leaving 268.114 g of water and 231.886 g of ice, both at 0 °C. The calculator reports that as a state with a melt fraction, because printing a single temperature there would hide the physics entirely.

Is the latent heat of vaporisation a fixed value?

No, and this is the biggest limitation of any latent-heat table. Water's L_v is 2500.9 kJ/kg at 0 °C, 2256.4 kJ/kg at 100 °C, 1939.8 kJ/kg at 200 °C, and exactly zero at the critical point of 373.946 °C, where liquid and vapour become indistinguishable and no phase transition exists. The latent heat of fusion barely moves with pressure, so melting calculations are safe, but any boiling calculation away from 1 atm should be treated as an estimate.

Can I add the latent heats of fusion and vaporisation to get sublimation?

Only when both are taken at the same temperature. Ice at 0 °C has L_f = 333.55 kJ/kg and L_v = 2500.9 kJ/kg, which sum to 2834.45 kJ/kg against a measured sublimation value of 2834.1 kJ/kg — a 0.01 % match. Using the familiar 100 °C figure of 2256.4 kJ/kg instead gives 2589.95 kJ/kg, which is 8.6 % low. The calculator flags that mismatch rather than letting the sum look exact.