Latent Heat Calculator

Latent heat is the energy a phase change hides

Heat a block of ice and the thermometer climbs steadily until it reaches 0 °C, and then it stops. It stays at 0 °C while the ice turns to water, and only starts climbing again once the last of the solid has gone. The energy has not stopped arriving. It is going into pulling the crystal lattice apart instead of into faster molecular motion, and temperature only measures the motion.

That hidden energy is the latent heat, and for a given substance it is proportional to how much of it changes state. Melting one gram of ice takes 334 joules, so melting a hundred grams takes 33.4 kilojoules. There is no temperature term in the equation at all, because there is no temperature change to put in it.

Fusion and vaporisation are not the same size

Every substance has two latent heats, and they differ by more than most people expect. Melting only has to loosen a rigid lattice into a liquid whose molecules still touch each other. Boiling has to separate them completely, pushing back the atmosphere as the vapour expands into it. For water the two figures are 334 and 2260 joules per gram, a factor of about 6.8.

This is why a pan of water reaches boiling point in a few minutes and then takes a long time to boil dry, and it is why steam burns are so much worse than hot water burns. A gram of steam condensing on skin delivers 2260 joules before it has cooled by a single degree.

Worked example

Melting 100 g of ice already at 0 °C, with a specific latent heat of fusion of 334 J/g:

• Q = m × L = 100 × 334 = 33,400 J, or 33.4 kJ.
• The same 33.4 kJ put into 100 g of liquid water would raise it by 33400 / (100 × 4.184) = 79.8 K, taking it from freezing to nearly boiling.
• A 2 kW kettle element supplies that in 33400 / 2000 = 16.7 s, which is the same order as the time it takes to bring the resulting water to the boil.

That second figure is the one worth remembering. The energy to melt ice is roughly the energy to heat the same mass of water by 80 degrees, which is why a drink with ice in it stays cold for so long: the ice is absorbing far more heat than its temperature suggests it should.

Heating curves need both equations

A problem that takes a substance across a phase boundary cannot be done in one step. Ice at minus 20 °C becoming steam at 120 °C is five stages: warm the ice with Q = mcΔT, melt it with Q = mL, warm the water, boil it, then warm the steam. Each stage uses a different constant, and the two phase changes together contribute more than all three heating stages combined.

On a graph of temperature against energy supplied, the phase changes appear as flat plateaus and the heating stages as sloped lines. The slope of each line is inversely proportional to the specific heat capacity of that phase, which is why the ice and steam sections are steeper than the liquid water section: water has an unusually high specific heat, about twice that of ice.

Common mistakes

• Putting a temperature change into Q = mL. There is no ΔT in this equation. If the temperature is changing, you are in the wrong stage of the problem and want mcΔT instead.
• Using one equation across a phase boundary. Applying a single mcΔT from minus 20 °C to 120 °C ignores both plateaus and understates the answer by a large factor.
• Mixing J/g and J/kg. Tables quote water as 334 J/g or 334,000 J/kg, and both are correct. Pairing the J/kg figure with a mass in grams gives an answer a thousand times too large.
• Confusing fusion with vaporisation. They differ by nearly a factor of seven for water, so using the wrong one is not a small error.
• Assuming the substance is at its transition temperature. Q = mL only covers the change of state. Ice at minus 10 °C needs warming to 0 °C first, and that energy is separate.

Converting units first? Use the mass and energy conversion tables.