On the changing-state page we met that strange flat stretch where energy pours in but the temperature refuses to move — melting and boiling. That “hidden” energy has a name: latent heat. Now we put a number on it with Q = mL, and finish the job of accounting for every joule on a heating curve.
📚 What you need to know
Latent heat is the energy absorbed or released during a change of state, at constant temperature.
Specific latent heatL = the energy per kilogram to change a substance’s state (no temperature change). Units: J kg−1.
Q = mL.
Two kinds: latent heat of fusionLf (melting / freezing) and of vaporisationLv (boiling / condensing).
Lv > Lf — vaporisation fully separates the particles; melting only loosens the lattice. For water Lf = 3.34 × 105 J kg−1, Lv = 2.26 × 106 J kg−1.
On a heating curve: sloped parts use Q = mcΔT (temperature changing); flat plateaus use Q = mL (state changing).
The energy hiding in a change of state
When a substance melts or boils, energy keeps flowing in but the thermometer sits still. That energy hasn’t vanished — it’s going into potential energy, prising particles apart against the forces holding them, rather than speeding them up. Because it produces no temperature change, it’s called latent (“hidden”) heat. The energy per kilogram to change state is the specific latent heat:
Energy for a change of stateQ = mL
with Q the energy (J), m the mass (kg) and L the specific latent heat (J kg−1).
“Latent” literally means hidden, and that’s the perfect word: the energy is real and considerable, but it’s stored invisibly as potential energy between the particles. A thermometer can’t see it — which is exactly why the temperature reads constant right through a change of state.
Two latent heats: fusion and vaporisation
There are two, because there are two “up the ladder” changes. The latent heat of fusionLf covers solid⇆liquid (melting and freezing); the latent heat of vaporisationLv covers liquid⇆gas (boiling and condensing). And vaporisation always needs far more:
Melting only loosens the lattice — particles stay close. Boiling has to pull them completely apart, which costs much more energy, so Lv > Lf.
For water, vaporisation needs about seven times the energy of fusion per kilogram — 2260 kJ kg−1 versus 334 kJ kg−1.
This is the number behind the steam-burn from a few pages back. When steam condenses on your skin it hands over that whole huge Lv — about 2260 kJ for every kilogram — before the water even starts to cool. That’s why a steam burn is so much worse than one from boiling water at the very same temperature.
The full heating curve
Now the two equations work together. Follow a substance from cold solid to hot gas and you alternate between heating (temperature rising) and changing state (temperature flat). Each sloped section is a Q = mcΔT calculation; each plateau is a Q = mL calculation.
The complete heating curve: red slopes are heating (mcΔT), blue plateaus are changes of state (mLf then mLv). The boiling plateau is much longer because Lv is so large.
For a “how much energy to go from ice to steam” question, don’t look for one formula — there isn’t one. Break the journey into stages, use mcΔT for every slope and mL for every plateau, work out each separately, then add them up. That method handles any heating-curve problem they throw at you.
Worked examples
WE 1
How much energy is needed to melt 0.50 kg of ice at 0 °C into water at 0 °C? (Lf = 3.34 × 105 J kg−1)
Change of state at constant temperature → use Q = mLf= 0.50 × 3.34 × 105Q = 1.67 × 105 J = 167 kJNo temperature change — every joule goes into breaking the ice lattice (potential energy).
WE 2
How much energy is needed to boil away 0.20 kg of water at 100 °C into steam? (Lv = 2.26 × 106 J kg−1)
Change of state at constant temperature → use Q = mLv= 0.20 × 2.26 × 106Q = 4.52 × 105 J = 452 kJFar more than melting the same mass — vaporisation has to separate the particles completely.
WE 3
How much total energy turns 0.10 kg of ice at 0 °C completely into steam at 100 °C? (cwater = 4200, Lf = 3.34 × 105, Lv = 2.26 × 106)
Three stages — melt, heat, boil:1) melt at 0 °C: Q1 = mLf = 0.10 × 3.34 × 105 = 33 400 J2) heat 0→100 °C: Q2 = mcΔT = 0.10 × 4200 × 100 = 42 000 J3) boil at 100 °C: Q3 = mLv = 0.10 × 2.26 × 106 = 226 000 JQ = 33 400 + 42 000 + 226 000Q ≈ 301 000 J = 301 kJBoiling alone is about 75% of the total — the vaporisation step dwarfs the other two.
🔧 Latent heat & heating-curve calculations
Which equation? Constant temperature → Q = mL. Changing temperature → Q = mcΔT.
Which L? Fusion (melt / freeze) or vaporisation (boil / condense).
Units: mass in kg, L in J kg−1 → Q in J.
Multi-stage (e.g. ice → steam)? Split into steps and add: mcΔT for slopes, mL for plateaus.
Reverse changes (freezing, condensing) release the same energy they absorbed.
mass m
× L
energy Q change of state
at constant temperature
no ΔT
Quick recap: Latent heat is the energy of a change of state at constant temperature, Q = mL. Fusion Lf covers melting/freezing; vaporisation Lv covers boiling/condensing, and Lv > Lf. On a heating curve, use mcΔT for the sloped (heating) parts and mL for the flat (state-change) plateaus — and add the stages for a full ice-to-steam calculation.
💡 Top tips
Constant T → mL; changing T → mcΔT. Choosing the wrong one is the classic slip.
Pick the right L: fusion for melting/freezing, vaporisation for boiling/condensing.
Add the stages. An ice-to-steam problem is separate mL and mcΔT steps summed — there’s no single formula.
Lv is always bigger. Separating particles costs more than loosening them.
Reverse changes release energy. Freezing gives out mLf; condensing gives out mLv.
⚠ Common mistakes
Using Q = mcΔT during a phase change — there’s no ΔT, so use Q = mL
Using Lf for a boiling change (it needs Lv), or vice versa
Forgetting a stage in ice→steam problems — especially the 0 → 100 °C heating in the middle
Leaving mass in grams — convert to kilograms first
Thinking latent heat raises the temperature — it doesn’t; the energy is potential, so temperature stays constant
You can now account for every joule on a heating curve — the sloped mcΔT stretches and the flat mL plateaus. That completes the “how much energy” story for temperature and state. The natural next question is how that energy actually travels from one place to another. Coming up: the three methods of thermal energy transfer, starting with thermal conduction.
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