Example 1
m = 1 kg, Lₜ = 3.34×10⁵ → Q = 3.34×10⁵ J.
Latent Heat of Fusion Calculator
Enter mass and the latent heat of fusion. You get the energy that goes into the phase change itself, with no temperature rise. A kilogram of ice at 334000 J/kg is 334 kJ. Warming that same mass of water by 10 K is only about 42 kJ.
Short page for the same formula: Q = m·Lₜ. Vaporization: Q = m·Lₚ. Heating without a phase change: Q = c·m·ΔT.
Result
Mass m (kg) and Lₜ (J/kg). The result shows up here.
How it works
While something melts, temperature holds still. The energy is not lifting the thermometer. It is breaking the crystal. The formula is Q = m·Lₜ. For ice, Lₜ ≈ 334000 J/kg, so one kilogram needs 334000 J before you have water at 0 °C.
This is the full page for that formula. A shorter copy with the same fields lives on the short URL. Stay here if you want the longer examples and questions.
Type mass in kilograms and Lₜ in joules per kilogram. The result is in joules. School 3.34e5 and 334000 are the same number. A comma and a period both parse.
Ice colder than 0 °C has to be warmed first with Q = c m ΔT (ice c is about 2100). Only then does this melting apply. For 1 kg from −10 °C the sum is about 21 kJ plus 334 kJ, not 334 alone.
Boiling water is much more expensive: Lₚ ≈ 2.26e6 J/kg. The same kilogram, now as boiling water turning into steam, takes about 2.26 MJ. That is a separate page and a different L.
The form needs two fields. The calculator example 1 kg and 334000 gives 334000 J. Half a kilogram of ice is 167000 J. On freezing the size is the same; heat leaves the body instead of entering it.
How to use
- Type the mass that is going to melt, for example 1 kg of ice.
- Type the latent heat Lₜ. For ice, school work usually uses 334000 J/kg or 3.34e5.
- Click Calculate. One kilogram of ice at that constant is 334000 J, the energy for the whole change to water at 0 °C.
- If the ice is still below zero, warm it first on the Q = c m ΔT page and add that result to this melting step.
- Vaporization is on the neighbouring page, with Lₚ. The energy there is several times larger than fusion.
Formula
Q = m · Lt
Unit: J. Temperature stays constant during the change.
Letters in Q = m·Lₜ
While ice melts the temperature holds still, and Q = m·Lₜ pays for the phase change. One kilogram at 334000 J/kg is 334 kJ before any water at 0 °C appears.
- Q
- Phase-change energy in joules. 1 kg of ice at Lₜ = 334000 J/kg gives 334000 J, or 334 kJ.
- m
- Ice mass in kilograms. Half a kilogram at that same constant is 167000 J, not half of the 41860 J water-heating figure.
- Lₜ
- Latent heat of fusion in J/kg. Ice uses 334000; that is not c = 4186 from the Q = c m ΔT card.
Real-life examples
Example 2
m = 0.5 kg, Lₜ = 334000.
Example 3
m = 0.05 kg, ice.
Example 4
m = 2 kg, Lₜ = 334000.
Example 5
m = 0.1 kg.
Example 6
Same |Q|, heat released: m = 1 kg.
Example 7
m = 0.02 kg.
Example 8
m = 5 kg.
Example 9
m = 1 kg, Lₜ = 200000.
Example 10
m = 0.01 kg, ice.
Ways to use this calculator
- You estimate how much energy ice cubes take from a drink before the water itself starts warming.
- You work a problem from ice at −10 °C to water at 20 °C: warm, melt, then c m ΔT again.
Frequently asked questions
How much energy melts 1 kg of ice at Lₜ = 334 kJ/kg?
Q = 334000 J. 0.1 kg is 33400 J. 5 kg of ice is 1.67 MJ, not 50 kJ from mixing this up with c·ΔT.
Does temperature rise while ice melts?
In a clean phase change it holds still until the whole mass has melted.
How is 334 kJ per kilogram of ice different from warming water by 10 K?
Warming 1 kg of water by 10 K is about 42 kJ. Melting the same mass is 334 kJ, roughly eight times more, at a steady 0 °C.
Where is the shorter version of this page?
On latent-heat-of-fusion-q-m-lt. Same formula and fields, less text.
How does this connect to vaporization?
Same idea Q = m L, but Lₚ for water is much larger, about 2.26 MJ/kg.
Is Q negative when water freezes?
In a balance the body gives heat away. Here you compute |Q| = m Lₜ. The sign depends on whether you add or remove energy.
Ice at −10 °C, 1 kg: what first?
First Q = c m ΔT up to 0 °C (ice c ≈ 2100, ΔT = 10 K → 21 kJ), then 334 kJ of melting. Sum ≈ 355 kJ.
What unit do I type for Lₜ?
J/kg. 334 kJ/kg is 334000. The result comes back in joules.
Can I type 3.34e5?
Yes. 3.34e5, 3.34×10⁵ and 334000 are the same ice constant.
How is this different from Q = c m ΔT?
Here temperature holds and the phase changes. There temperature rises or falls and the phase stays the same.
Knowledge sources
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.