Latent Heat Calculator
Find the heat absorbed or released when a mass melts, freezes, boils, condenses or sublimes, using Q = mL with built-in values for water, ethanol and aluminium.
Latent Heat Calculator
Background.
Melting a kilogram of ice at 0 °C takes 334 kJ of heat, enough to warm the same kilogram of liquid water from 0 °C to almost 80 °C, yet a thermometer standing in the melting ice does not move. Boiling a kilogram of water away at 100 °C takes 2,257 kJ, more than five times the 418.6 kJ needed to bring it from freezing to the boil. That hidden energy, taken in or given out while a substance changes phase at a fixed temperature, is latent heat. It is why ice cools a drink so well, why a pan of boiling water sits at 100 °C with the burner on full, and why a steam burn is worse than a splash of boiling water.
The arithmetic is a single multiplication, Q = m × L: the mass that changes phase times the specific latent heat for that substance and transition. Water, ethanol, aluminium and nitrogen have built-in values at their normal transition temperatures under standard atmospheric pressure: 334 and 2,257 kJ/kg for water, 108 and 838 kJ/kg for ethanol, 397 kJ/kg for melting aluminium and 199.2 kJ/kg for boiling nitrogen. For anything else, including melting nitrogen, any sublimation or a material from a supplier's data sheet, you enter L yourself in kJ/kg. Say whether heat is absorbed (melting, boiling, sublimation) or released (freezing, condensation, deposition) and the answer carries a plus or minus sign to match.
What Q = mL leaves out is any change of temperature. Warming ice from freezer temperature up to 0 °C, or heating the meltwater afterwards, is sensible heat, worked out from the specific heat capacity, and a full heating or cooling load adds the two kinds together. For water the latent part is usually the bigger share, about 85% of the energy needed to turn ice at −10 °C into steam, which is why ice banks, steam plant and evaporative coolers are designed around it.
What is latent heat calculator?
Specific latent heat, L, is the energy needed to change the phase of one kilogram of a substance without changing its temperature, measured in joules per kilogram or, more conveniently, kilojoules per kilogram. The word latent comes from the Latin for hidden: Joseph Black, working in Glasgow in the 1760s, noticed that a mixture of ice and water stays at 0 °C while heat keeps flowing in, and gave the name to heat that vanishes into a change of state instead of showing on a thermometer. Each pair of phases has its own value: fusion between solid and liquid, vaporisation between liquid and gas, and sublimation straight from solid to gas. The same amount comes back out in the reverse direction, so freezing releases exactly what melting absorbed. At constant pressure, L is the same quantity chemistry tables list as the specific enthalpy of the transition.
How to use this calculator.
- Enter the Mass of substance and choose kilograms or grams under Mass unit. Count only the material that actually changes phase: if half of a 2 kg block melts, enter 1 kg.
- Pick the Phase transition: fusion for melting or freezing, vaporisation for boiling, evaporation or condensation, and sublimation for a solid turning straight into gas.
- Set the Direction of transition. Heat absorbed gives a positive result; heat released (freezing, condensation, deposition) gives the same size of answer with a minus sign.
- Choose the Substance. Water and ethanol have built-in fusion and vaporisation values, aluminium has a melting value and nitrogen a boiling value; Custom, any sublimation, melting nitrogen and boiling aluminium all need a number in the next field.
- Type a Custom specific latent heat in kJ/kg when you have one from a data sheet or steam table. Leave it at 0 to use the built-in figure; anything above zero replaces it, even for water.
- Read the heat in joules, kilojoules and megajoules, and check the specific latent heat used, shown in J/kg, to confirm which value went into the sum.
The formula.
Q = m × L, with m in kilograms and L in joules per kilogram, gives Q in joules. A mass entered in grams is divided by 1,000 first, and a custom L typed in kJ/kg is multiplied by 1,000. The sign records the direction of flow: positive when the substance takes heat in (melting, boiling, sublimation), negative when it gives heat out (freezing, condensation, deposition). The size is the same either way, because at the transition temperature the change can run in both directions with the same energy.
The relationship is a straight proportion, so doubling the mass doubles the heat, and the choice of L sets the scale. Vaporisation values dwarf fusion values because boiling has to pull molecules completely apart and also push the surrounding atmosphere back as the vapour expands. At 100 °C a kilogram of steam fills about 1.672 m³, against roughly 0.001 m³ as liquid, a 1,600-fold expansion. Pushing back 101,325 Pa through that volume accounts for about 169 kJ of water's 2,257 kJ/kg; the remaining 2,088 kJ goes into separating the molecules.
The built-in figures assume a pure substance changing phase at its normal transition temperature under one standard atmosphere, with the whole mass changing phase. Vaporisation is the value most sensitive to conditions: for water it is 2,501 kJ/kg at 0 °C, 2,257 kJ/kg at 100 °C, and it falls to zero at the critical point, 373.9 °C and 22.064 MPa, where liquid and vapour can no longer be told apart. Solutions such as brine or syrup freeze gradually over a range of temperatures as the remaining liquid grows more concentrated, so a single L only approximates them.
A worked example.
Take 350 g of ice at 0 °C, a generous handful of cubes, and ask how much heat it must absorb to melt. Enter 0.35 kg, fusion, heat absorbed and water, and the calculator applies L = 334,000 J/kg. Q = 0.35 × 334,000 = 116,900 J, which is 116.9 kJ or 0.1169 MJ. It is positive because melting draws heat in. In a drink, that heat comes out of the liquid around the ice. Dividing 116,900 J by water's specific heat of about 4,186 J/(kg·K) and a 25 K drop shows that, through melting alone, the ice could in principle bring about 1.12 kg of water from 25 °C down to 0 °C. Cubes straight from a freezer at −18 °C do a little more: warming 0.35 kg of ice through 18 K at about 2.1 kJ/(kg·K) soaks up another 13.23 kJ before any melting starts, roughly 11% on top of the latent figure.
Frequently asked questions.
Why does a pan of boiling water stay at 100 °C with the burner on full?
Why is the latent heat of vaporisation so much larger than that of fusion?
Why does choosing sublimation, or melting nitrogen, ask me for a value?
Does the latent heat of vaporisation change with temperature?
How much heat turns 1 kg of ice at −10 °C into steam at 100 °C?
How much cooling does a block of ice actually store?
Why do growers spray water on crops on a frosty night?
Can I use it for salt water, syrup or other mixtures?
How do I turn a molar enthalpy from a chemistry table into the value used here?
Why does supercooled water freeze only partly when it finally freezes?
References& sources.
- [1]Black, J. (1803). Lectures on the Elements of Chemistry, delivered in the University of Edinburgh. Robison, J. (ed.). Edinburgh: Mundell and Son.
- [2]NIST Chemistry WebBook. Water. National Institute of Standards and Technology.
- [3]NIST Chemistry WebBook. Nitrogen, fluid saturation properties at 1 atm. National Institute of Standards and Technology.
- [4]Halliday, D., Resnick, R., and Walker, J. (2013). Fundamentals of Physics, 10th ed. John Wiley & Sons. ISBN 978-1-118-23072-5
- [5]Callen, H.B. (1985). Thermodynamics and an Introduction to Thermostatistics, 2nd ed. John Wiley & Sons. ISBN 978-0-471-86256-7
- [6]ASHRAE (2021). ASHRAE Handbook-Fundamentals, Chapter 1: Psychrometrics. American Society of Heating, Refrigerating and Air-Conditioning Engineers.
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- Q = mL
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