enthalpy of vaporization
/ EN-thal-pee /
Think about how long a pot of water takes to boil completely dry compared with how long it takes to merely reach a boil — it lingers and lingers. That stubborn extra energy demand is the enthalpy of vaporization: the heat you must pour in to turn a given amount of liquid into vapor at its boiling point, without raising the temperature at all.
It is the latent heat of boiling, written as a thermodynamic quantity, usually per mole and measured at constant pressure (which is what 'enthalpy' signals). The energy goes entirely into pulling molecules apart against the attractions holding the liquid together. So the size of this number is a direct readout of how strongly a liquid's molecules cling: water's is unusually large because of its hydrogen bonds, which is why water is such a good coolant and moderates Earth's climate.
Enthalpy of vaporization shows up everywhere it counts. It sets how much cooling you get from sweating or from a sprinkler in a heatwave; it determines the energy bill of distillation columns separating crude oil; and it is the key quantity inside the Clausius–Clapeyron equation. Its partner, the enthalpy of fusion (for melting), is almost always much smaller, since boiling tears molecules fully apart while melting only loosens them.
Vaporizing one mole of water (18 grams) at 100 °C demands about 40.7 kilojoules — far more than the 6.0 kilojoules it takes to melt that same mole of ice, because boiling fully separates the molecules.
Boiling costs much more energy than melting, molecule for molecule.
Enthalpy of vaporization shrinks as temperature rises and falls to zero at the critical point, where liquid and gas become identical and there is nothing left to separate. The value quoted is usually the one at the normal boiling point unless stated otherwise.