Latent Heat Calculator
Heat to melt or boil water from mass with fusion and vapor values.
What it does
Phase changes eat heat without temperature moves. Melting a kilo of ice needs 334 kilojoules, boiling it needs 2260. Mass times those constants gives the total.
The tool above takes water mass and change type and returns heat, kWh, and small mass. The defaults 1 kg to steam match the worked example, so the page opens showing 2260 kJ. Switch to imperial for pounds and BTU.
Fusion uses 334 kJ/kg (144 BTU/lb), vaporization 2260 kJ/kg (970 BTU/lb) for water near standard pressure. Kilowatt hours divide kJ by 3600 or BTU by 3412. Here 2260 kJ is 0.63 kWh.
Use this for ice melt loads, kettle energy, and steam budgets. Sensible heating to the change point needs mc delta T on top. Pressure shifts constants slightly.
This page shows the constants, one full hand checked example at 1 kg to steam, a reference table, and practice checks that run through the tool in both unit systems.
How to use
- Pick Metric (kg, kJ) or Imperial (lb, BTU).
- Type water mass.
- Pick Melt ice or Boil water.
- Read Heat needed as the main result.
- Read kWh for energy bills.
- Add sensible heat to reach the change point separately.
Formula
- m
- Water mass.
- L
- 334 fusion, 2260 vapor kJ/kg.
- Q
- Heat in kJ or BTU.
Water near 0 C melt and 100 C boil at 1 atm. Salt and altitude shift points and constants slightly.
Worked example
Example: 1 kg to steam
Boil off one kilo of water at 100 C. Find heat.
- Write m = 1 kg, L = 2260 kJ/kg.
- Multiply: 1 x 2260 = 2260 kJ.
- kWh: 2260 / 3600 = 0.63 kWh.
- Small mass: 1000 g.
- Imperial mirror 2.2 lb gives 2134 BTU.
- Add sensible heat from start temp separately.
Reference table
| Change | kJ | BTU/lb mirror |
|---|---|---|
| Melt | 334 kJ | 144 BTU/lb |
| Boil | 2260 kJ | 970 BTU/lb |
Limits
- Water only near 1 atm.
- No sensible heat here. Add mc delta T.
- Constants approximate with temp and purity.
- No superheat past steam here.
Common mistakes
- Using fusion for boiling and undercounting 7x.
- Mixing kg with pounds in one product.
- Forgetting kJ to kWh divide by 3600.
- Ignoring sensible heat to reach 0 or 100 C.
- Applying water constants to other fluids.
Try it yourself
Quick checks. Answers sit tucked under each question.
2 kg to melt?
Heat 668 kJ, 0.19 kWh, mass 2000 g.
0.5 kg to steam?
Heat 1130 kJ, 0.31 kWh, mass 500 g.
Questions
1 kg to steam?
2260 kJ.
Melt?
334 kJ/kg here.
kWh?
kJ over 3600. Here 0.63 kWh.
BTU?
970 per lb to steam here.
Sensible?
Separate mc delta T to reach change point.
Other fluids?
Need their own L constants.