Melting, Boiling, and the Hidden Heat of Phase Changes

The plateau that puzzles everyone

Take a block of ice from the freezer and heat it steadily on a stove, tracking its temperature. At first the temperature climbs — cold ice warms toward 0 °C. Then something strange happens: while the ice melts to water, the temperature stops rising and sits at 0 °C, even though you keep pouring in heat. Once fully melted, the water warms up to 100 °C — and stalls again while it boils away to steam. Two flat plateaus interrupt the climb. Where does all that heat go if the temperature is not changing?

Latent heat: energy that breaks bonds

During a phase change the added heat does not speed the molecules up — it does the work of pulling them apart, loosening the bonds that hold a solid rigid or a liquid together. Since molecular speed (and hence temperature) stays fixed while this rearrangement happens, the thermometer flatlines. The heat hidden in this process is latent heat — 'latent' meaning hidden, because it changes the state without changing the temperature.

Q = m \, L

Heat to change the phase of a mass m. L is the latent heat: L_f for fusion (melting/freezing), L_v for vaporization (boiling/condensing). No ΔT appears — temperature is constant.

For water the numbers are striking: the latent heat of fusion is L_f ≈ 334 kJ/kg, but the latent heat of vaporization is L_v ≈ 2260 kJ/kg — nearly seven times larger. Boiling away a kilogram of water takes far more energy than melting a kilogram of ice, because turning liquid into free-flying vapour means tearing the molecules completely apart. It is also why a steam burn is so much worse than hot-water scald: condensing steam dumps that huge L_v straight into your skin.

Reading the heating curve as five stages

The heating curve is really the two equations from this track stitched together. On the sloped segments, only temperature changes, so heat follows Q = mc\,\Delta T — and the slope is gentle where c is large (liquid water climbs slowly). On the flat segments, only the phase changes, so heat follows Q = mL. To carry ice at −20 °C all the way to steam at 120 °C you add up five separate pieces.

Q_{\text{total}} = m c_{\text{ice}}\Delta T_1 + m L_f + m c_{\text{water}}\Delta T_2 + m L_v + m c_{\text{steam}}\Delta T_3

Warming ice, melting it, warming the water, boiling it, then warming the steam: three mcΔT ramps and two mL plateaus, added together.

Evaporation, boiling, and why sweating cools

Water does not need to reach 100 °C to become vapour. At any temperature the fastest molecules at the surface can break free — that is evaporation. Because the escaping molecules carry away above-average energy, the liquid left behind is a little cooler. This is evaporative cooling: sweat evaporating from your skin drains latent heat and chills you; a dog pants; a wet cloth on a fevered brow cools it. The body's radiator runs on latent heat.

Worked example: melting and warming ice

How much heat turns 0.20 kg of ice at 0 °C into water at 25 °C? Use L_f = 334 kJ/kg = 334000 J/kg and c(water) = 4186 J·kg⁻¹·K⁻¹.
Step 1 — melt the ice at 0 °C (a flat plateau, use Q = mL): Q₁ = (0.20)(334000) = 66800 J.
Step 2 — warm the resulting water from 0 to 25 °C (a slope, use Q = mcΔT): Q₂ = (0.20)(4186)(25) = 20930 J.
Total Q = Q₁ + Q₂ = 66800 + 20930 ≈ 87700 J ≈ 88 kJ. Notice that just melting the ice took more than three times the heat of warming it 25 degrees — latent heat is the big spender.