Latent heat calculator
Heat to melt, freeze, boil or condense, and the whole ice-to-steam curve.
Melts at 32 °F (L_f = 334 kJ/kg), boils at 212 °F (L_v = 2,256 kJ/kg).
Specific heats in J/(kg·°C), from OpenStax Table 1.3 where it lists them
What is latent heat and how do I calculate it?
Latent heat is the energy a substance absorbs or gives off while it changes phase without changing temperature: Q = m × L_f for melting or freezing and Q = m × L_v for boiling or condensing (OpenStax University Physics, Volume 2, section 1.5). For water at 1 atm L_f = 334 kJ/kg (79.8 kcal/kg) and L_v = 2,256 kJ/kg (539 kcal/kg), so melting one pound of ice takes 0.4536 kg × 334 kJ/kg = 151.5 kJ, or 143.6 BTU, and boiling one pound of water takes about 970 BTU. Freezing and condensing release the same amounts.
How much heat does it take to turn ice at −20 °C into steam at 120 °C?
Five steps, each with its own formula. For 1 kg with the OpenStax values: warm the ice, 1 × 2,090 × 20 = 41.8 kJ; melt it, 334 kJ; warm the water from 0 to 100 °C, 1 × 4,186 × 100 = 418.6 kJ; boil it, 2,256 kJ; warm the steam, 1 × 2,020 × 20 = 40.4 kJ. The total is 3,090.8 kJ, and boiling alone is almost three quarters of it. The Heating curve tab lists every step with a running total; its pre-filled example is 1 lb from −4 °F to 248 °F, which comes to about 1,329 BTU.
Why doesn’t the temperature change while ice melts or water boils?
The energy goes into loosening the bonds between molecules, not into making them move faster, so there is no temperature rise until the change is complete (OpenStax University Physics, section 1.5). That is why a drink with ice in it stays at 0 °C until the last piece melts. Evaporation also works below the boiling point and needs a little more energy: at body temperature (37 °C) the heat of vaporization of water is 2,430 kJ/kg, which is why sweating cools you. In OpenStax Example 1.12 a person at rest gives off 120 W, and evaporating 2.96 g of sweat per minute carries all of it away.
How much ice does it take to cool a drink?
Use the Ice in water tab. OpenStax Example 1.9: three ice cubes of 6.0 g each, at 0 °C, are dropped into 0.25 kg of soda at 20 °C, and the water and ice reach about 13 °C (13.3 °C unrounded), assuming the soda has the water’s specific heat and no heat enters from outside. The ice takes 0.018 × 334,000 = 6,012 J just to melt, which is about 29% of the 20,930 J the soda gives up in cooling to 0 °C. If there is more ice than the water can melt, the result stays at 0 °C and the calculator says how much ice is left.
Where do the numbers come from, and how do they differ from the specific heat calculator?
Latent heats come from OpenStax Table 1.4 (values at the normal melting and boiling points at 1 atm; helium is left out because it has no solid phase at atmospheric pressure), specific heats from Table 1.3: ice 2,090 J/(kg·°C) averaged from −50 to 0 °C, water 4,186 at 15 °C and steam 2,020 at constant pressure. The table lists the specific heat of other phases only for a few substances; where one is missing, type your own. The specific heat calculator uses NIST values at 25 °C (4.1813 J/(g·°C) for water), so the two can differ by a fraction of a percent. Water’s latent heat also varies slightly with pressure, and real ice contains air.
Sources: OpenStax University Physics, Volume 2, sections 1.4 to 1.6 (Tables 1.3 and 1.4, Examples 1.9 and 1.12); OpenStax College Physics 2e, section 14.3 (heating curve of ice). Related: specific heat calculator, heat conduction calculator, thermal expansion calculator and energy converter.