The latent heat calculator above answers the question that a temperature-based heat calculation cannot: how much energy does it take to turn a solid into a liquid, or a liquid into a gas, when the temperature never moves? That energy is real, it is often much larger than the energy needed to warm the same material up to the transition point, and it is invisible on a thermometer. The tool relates mass, latent heat and energy through the single relation Q = mL, solves for whichever of the three you are missing, and turns the answer into a time using a heating power you supply.
Arb Digital builds free calculators that each own one job and finish it. This page owns the phase change. The boundary against the site's specific heat calculator is exactly the one that trips people up in practice: that tool handles heating and cooling within a phase, where temperature changes and Q = mcΔT applies, while this one handles the flat step at the transition, where temperature does not change and Q = mL applies instead. A complete heating problem usually needs both, run in sequence.
What This Latent Heat Calculator Does
In its default mode it takes a mass and a latent heat value and returns the energy in kilojoules, watt-hours and BTU, plus the time a heater of the power you name would take to supply it. Switch the solve-for selector and it will instead tell you how much mass a given quantity of energy can melt or boil, or back out the latent heat implied by a measured energy and mass — which is how the constant is determined experimentally in the first place.
The phase selector matters more than it looks. Fusion and vaporisation are different transitions with very different constants for the same substance, and for water the gap is nearly a factor of seven. Selecting a substance fills the latent heat field with the standard value for the transition you picked, so switching from melting to boiling with water selected changes 334 into 2,256 kJ/kg automatically. If your material is not in the list, or you are working at a pressure where the published value does not apply, choose Custom value and the field stops being overwritten.
Every result is reported in three energy units at once. Kilojoules are the coherent SI answer, watt-hours make the electricity cost obvious, and BTU is what refrigeration and HVAC specifications are still written in. Having all three on screen removes the conversion step that introduces most of the arithmetic errors in this kind of problem.
How to Use It
- Pick the transition first. Fusion covers melting and freezing; vaporisation covers boiling and condensing. The same numbers apply in both directions — freezing releases exactly what melting absorbs.
- Choose the substance or type your own latent heat. The presets carry standard values at atmospheric pressure. Anything unusual, and you should be sourcing the constant yourself.
- Enter the mass in whatever unit you have it. Grams, kilograms, pounds and tonnes are all accepted and converted internally to kilograms.
- Set the heating power if you want a duration. This is the power actually reaching the material, not the nameplate rating of the appliance — see the efficiency section below.
- Read the hero for the quantity you solved for and the grid for the same answer expressed three ways plus the time estimate.
The Formula: How Latent Heat Is Calculated
The relation is Q = mL, where Q is the heat transferred, m is the mass changing phase and L is the latent heat of the substance for that particular transition. OpenStax University Physics Volume 2, section 1.5 on phase changes, writes it as Q = mLf for melting and freezing and Q = mLv for vaporisation and condensation, and notes that these constants are called latent — hidden — because in a phase change the energy enters or leaves the system without causing any temperature change at all.
Work the default values. Two kilograms of ice at exactly 0 °C, melting to water at 0 °C, needs Q = 2 × 334 = 668 kJ. That is 185.6 watt-hours, or 633.1 BTU. A 1,500 W heater delivering all of its output into the ice would take 668,000 J ÷ 1,500 W = 445.3 seconds, which is 7 minutes 25 seconds. Nothing about the temperature changes during those seven minutes.
The same source's table gives the constants the presets use: water at 334 and 2,256 kJ/kg, ethanol at 104 and 854, ammonia at 332 and 1,370, nitrogen at 25.5 and 201, mercury at 11.8 and 272, lead at 24.5 and 871, copper at 134 and 5,069, and aluminium at 380 and 11,400. Note that in every case the vaporisation figure is the larger of the two, which is a general result rather than a coincidence about these particular materials.
Why Boiling Costs So Much More Than Melting
Melting only has to loosen a solid's lattice enough for the molecules to slide past one another. They stay in contact, and the substance barely changes volume. Boiling has to pull the molecules completely apart and push back the atmosphere to make room for the vapour, which occupies roughly a thousand times the volume of the liquid it came from. The energy bill reflects that difference, and for water the ratio is 2,256 to 334 — 6.75 times as much heat per kilogram.
This is why a pan of water takes a few minutes to reach boiling and then far longer to boil dry, and it is the reason sweating cools a body so effectively. It also explains why steam burns are so much worse than hot-water burns at the same temperature: condensing steam dumps 2,256 kJ per kilogram into the skin before the resulting water has cooled by a single degree. The boiling point calculator tells you the temperature at which this transition happens at a given pressure; this page tells you what it costs once you get there.
A Complete Heating Problem Needs Both Equations
Taking ice from −18 °C to steam at 100 °C is five separate calculations, not one. Warm the ice to 0 °C with Q = mcΔT using the specific heat of ice. Melt it with Q = mLf. Warm the water from 0 to 100 °C with mcΔT using the specific heat of water. Boil it with mLv. Superheat the steam if you need to, with mcΔT again using the specific heat of steam.
For one kilogram of water the two flat steps dominate. Warming the liquid across its whole 100-degree range costs about 419 kJ; boiling it at the top costs 2,256 kJ. The plateau is over five times the ramp. Any energy estimate that skips the latent terms will be badly low whenever a phase change is involved, and this is the single most common failure in back-of-envelope thermal work.
Run the ramps on the specific heat calculator and the plateaus here, then add the results. The energy converter will put the total into whatever unit your specification uses, and the specific heat converter handles capacity values quoted in imperial units.
Heater Power Is Not the Power Reaching the Material
The time figure this tool reports assumes every watt you enter arrives in the substance. Real systems never manage that. An electric kettle is unusually good because the element sits in the water, but a pan on a hob loses a large share of the burner's output around the sides of the pan, and an open container loses heat to the room continuously while you are heating it.
The practical way to handle this is to enter the power actually delivered rather than the appliance rating — if you believe a gas ring puts 40 per cent of its heat into the pan, enter 40 per cent of its rating. The result is then an estimate rather than a bound. The time figure is also a floor in another sense: it assumes the material is already sitting exactly at its transition temperature, so any warming still to be done comes on top.
For a system losing heat to its surroundings while you heat it, the loss is not constant either, because it grows with the temperature difference. The Newton's law of cooling calculator models that decay, and the heat transfer calculator handles conduction through a wall or vessel.
Latent Heat Is Pressure-Dependent, and the Presets Are Not
The constants in the substance list are values at one atmosphere. Latent heat of vaporisation falls as pressure rises, because at higher pressure the liquid and vapour states are less different from one another. At the critical point the two become indistinguishable and the latent heat of vaporisation reaches zero. For water that point is far above ordinary conditions, but the trend matters well below it: steam plant working at high pressure sees a materially lower latent heat than the atmospheric figure.
The NIST Chemistry WebBook entry for water carries the phase change data — boiling point, triple point and critical point among them — for anyone who needs to work outside atmospheric conditions. If you are doing refrigeration or steam work, take the latent heat from a property table at your actual pressure and enter it in the custom field rather than accepting the preset.
The energy unit underneath all of this is the joule, defined in the SI Brochure published by the BIPM. Kilocalories and BTU are historical conveniences layered on top, and mixing them mid-calculation is a reliable way to be wrong by a factor of four or of a thousand.
Arb Digital builds free tools like this one because useful pages earn attention. If you want tools, calculators or content built for your own audience, we can help.
Browse All Free Tools Talk to Arb DigitalCommon Mistakes to Avoid
- Using Q = mcΔT across a phase change — there is no ΔT to use, and the specific heat equation returns zero for a step that actually costs the most energy in the whole problem.
- Mixing up fusion and vaporisation constants — for water they differ by a factor of nearly seven, so the wrong one is not a small error.
- Assuming the substance starts at its transition temperature — if it does not, you owe the warming energy first, and this calculator does not include it.
- Taking an atmospheric latent heat into high-pressure work — the vaporisation constant falls as pressure rises and vanishes entirely at the critical point.
- Entering nameplate heater power — the time figure assumes every watt lands in the material, which no real heating arrangement achieves.
Related Free Tools From Arb Digital
Pair this page with the specific heat calculator for the temperature ramps either side of the transition, and the specific heat converter if your capacity figures are in imperial units. Find the transition temperature itself with the boiling point calculator, rescale results with the energy converter, and model the surroundings with the Newton's law of cooling calculator or the heat transfer calculator. For thermodynamic state functions rather than raw heat, the enthalpy calculator is the next step. Everything Arb Digital has built is listed on the free online tools hub.
Frequently Asked Questions
It is the energy a substance absorbs or releases when it changes phase, without its temperature changing at all. Melting ice sits at zero degrees for the whole time it is melting, yet it is absorbing heat the entire time. That absorbed energy is the latent heat.
Specific heat governs warming and cooling within a single phase, where energy produces a temperature change and Q equals mcΔT. Latent heat governs the transition between phases, where energy produces no temperature change and Q equals mL. A full heating problem needs both, applied in sequence.
Melting only loosens a lattice enough for molecules to move past one another. Boiling separates them completely and pushes back the atmosphere to make room for a vapour roughly a thousand times the volume of the liquid. For water the ratio is about 6.75 to one.
Yes. The latent heat of fusion is a property of the transition, not of its direction, so freezing a kilogram of water releases the same 334 kilojoules that melting it absorbs. The tool reports magnitude, and you supply the sign from the direction of your process.
No. They are values at one atmosphere. Latent heat of vaporisation falls as pressure rises and reaches zero at the critical point, so high-pressure steam or refrigeration work needs a value read from a property table at the actual pressure, entered in the custom field.
It is a floor. It assumes every watt you enter reaches the material and that the material is already at its transition temperature. Real heating loses part of its output to the surroundings, so enter the power you believe is actually delivered rather than the appliance rating.
Yes, as long as you supply the latent heat of sublimation yourself through the custom field. The equation Q equals mL is identical; only the constant differs, and it is roughly the sum of the fusion and vaporisation values for the same substance.
This tool is provided for educational and estimating use. It performs an idealised Q equals mL calculation and does not model heat losses, pressure effects, mixtures or the rate at which heat can actually be moved into a material, so treat its output as a physics result rather than a process design.
