How this instrument works
Latent heat is energy a substance takes on to change phase while its temperature sits perfectly still. Joseph Black pinned it down at Glasgow around 1761, prompted by an everyday oddity: snow banks linger for weeks in air well above freezing. Were melting simply a matter of arriving at 0 °C, he argued, every thaw would come as a catastrophic flood. Energy must therefore be vanishing somewhere invisible — hidden, or latent — spent prising a crystal lattice apart rather than hurrying molecules along faster. James Watt, then mending instruments across that same university, took the idea into steam: knowing what condensation gives back is why he stopped cooling and reheating one cylinder and bolted on a separate condenser instead.
Magnitudes here tend to startle people. Water asks 334 kJ/kg to melt and roughly 2260 kJ/kg to boil at atmospheric pressure, nearly seven times more, because tearing molecules entirely free of one another costs far more than merely letting them slide. Ammonia surrenders about 1370 kJ per kilogram evaporated, R-134a only 217, liquid nitrogen 199. Refrigeration engineers live inside those figures, since an evaporator's entire capacity is latent heat per kilogram multiplied by circulating mass flow. So does anyone who perspires: evaporating sweat off warm skin carries away close to 2430 kJ/kg, which is why still humid air feels punishing and a breeze rescues you.
Q = m·L rests on two assumptions, both of which slip in practice. It expects your whole mass to cross that boundary, so half-melted ice belongs in this sheet as its melted fraction only. And it treats L as belonging to some substance outright, when L really belongs to that substance under stated conditions. Water's heat of vaporisation reads 2257 kJ/kg boiling at 100 °C, climbs to 2442 kJ/kg for evaporation at 25 °C, and shrinks to exactly zero at its critical point — 374 °C and 22.1 MPa — where liquid and vapour cease to be distinguishable things. Careful tables print a temperature and pressure beside every value for precisely that reason.
- Put however much material genuinely changes phase into Mass; that field accepts grams, tonnes and pounds as readily as kilograms.
- Type your published figure into Specific latent heat (J/kg): 334000 melts ice, 2260000 boils water, 199000 evaporates liquid nitrogen.
- Pick fusion values for melting or freezing and vaporisation values for boiling or condensing — one substance carries different numbers for each.
- Read Heat required in whichever unit fits your work — joules or kilojoules on a bench, kilocalories against food data, watt-hours for electrical sizing.
- Freezing and condensing hand back precisely what melting and boiling took, so read an identical answer as energy your equipment must remove.
Worked example — melting one kilogram of ice
A kilogram of ice sits in a cool box, already sitting at 0 °C, and you want to know what melting it will cost. Mass = 1, Specific latent heat (J/kg) = 334000. One multiplication finishes it: Q = 1 × 334000 = 334,000 J, which that output menu reads back as 334 kJ, 79.8 kcal or 92.8 Wh.
Set it beside ordinary warming and its scale lands properly. Those same 334 kJ, poured into a kilogram of liquid water instead, would carry it from freezing all the way to 80 °C. Ice is not merely cold, then — it is a buffer, holding a drink at 0 °C for as long as any crystal survives. A 100 W element would need 56 minutes to see this kilogram off, which is roughly how long a bag of cubes lasts in a warm car.
That same number hides inside an industrial unit still in daily use. One ton of refrigeration was defined as melting a short ton of ice over 24 hours: 907 kg × 334 kJ/kg, spread across a day, works out at 3.5 kW — the 12,000 BTU/h stamped on air-conditioning nameplates the world over.
Questions
What separates latent heat from sensible heat?
Sensible heat shifts temperature; latent heat shifts phase. A thermometer registers one and stays blind to the other, which is exactly where that name originated. Warming ice from −10 °C up to 0 °C is sensible, needing Q = m·c·ΔT. Melting it at 0 °C is latent, needing Q = m·L, with your thermometer motionless from start to finish. Any journey crossing a phase boundary wants both terms, worked as separate stages and added.
Do I want a fusion value or a vaporisation value?
Fusion for anything crossing between solid and liquid, vaporisation for anything crossing between liquid and gas. Those differ by a factor near seven for water — 334 kJ/kg against 2260 kJ/kg — so confusing them is no subtle slip. Sublimation, solid straight to vapour, costs both stages summed at that same low temperature — 334 plus roughly 2501, so near 2835 kJ/kg for ice at 0 °C. Note that second term is not 2260, which belongs to boiling at 100 °C. Freeze-drying pays that bill, and so does frost quietly vanishing on a cold dry morning.
Does freezing release energy or absorb it?
Release, in exactly what melting absorbs — a single value of L serves both directions. Orchard growers turn this to account: spraying water over budding fruit trees on a frosty night lets it freeze, and every kilogram converting to ice dumps 334 kJ back into those buds, pegging them at 0 °C rather than letting them plunge below. This sheet hands you a magnitude and leaves direction to context — you know whether energy enters your sample or leaves it.
Why does my substance show different latent heats in different tables?
Because L belongs to a transition, not to a substance on its own. Water vaporises at 2257 kJ/kg boiling under one atmosphere at 100 °C, yet 2442 kJ/kg evaporating at 25 °C, and that figure slides steadily toward zero as pressure climbs to critical. Reputable data states a temperature and pressure alongside each entry. Confirm your quoted conditions resemble your real ones before leaning on a fourth digit.
Which units do published latent heats arrive in?
SI works in joules per kilogram, so that field wants 334000 rather than 334. Textbooks usually print kJ/kg, meaning you multiply by 1000 first. Chemistry sources favour molar enthalpy in kJ/mol, where water's 6.01 kJ/mol becomes 334 kJ/kg once divided by 0.018 kg/mol. Older engineering data comes as BTU/lb, in which ice fusion reads 143.4; multiplying by 2326 returns joules per kilogram.
Why does boiling cost so much more than melting?
Melting only loosens a lattice, while boiling dismantles it outright. Molecules in liquid water still cling to neighbours through hydrogen bonds, and fusion buys nothing more than freedom to slide past one another — 334 kJ/kg covers that. Vaporisation must break every bond left and shove back an atmosphere to make room for vapour around 1600 times bulkier, and 2260 kJ/kg is where such work goes. Steam scalds far worse than boiling water for identical reasons: condensing against skin, it returns all of that at once.