Latent heat

Why the thermometer doesn't move. Heat supplied to a solid at its melting point does not speed up molecular vibration (which is what a rising temperature actually measures); instead it goes entirely into overcoming the intermolecular forces holding the solid's ordered lattice together. Once every particle has broken free into the looser liquid arrangement, any further heat resumes raising the temperature. The same logic applies at the boiling point, only now the bonds must be broken almost completely, separating molecules so far apart that they become a gas.

Specific latent heat. The specific latent heat $L$ of a substance is the heat required to change the state of unit mass of it, at constant temperature:

where: $Q$ is heat supplied or released (J), $m$ is mass (kg or g), and $L$ is specific latent heat (J/kg or J/g), a fixed property of the substance and the particular change of state. Two named values are prescribed by the syllabus. Specific latent heat of fusion, $L_f$, is the heat to convert unit mass of solid to liquid at the melting point (or released on freezing); for ice, $L_f \approx 336\ \text{J/g} = 3.36\times10^5\ \text{J/kg}$. Specific latent heat of vaporisation, $L_v$, is the heat to convert unit mass of liquid to vapour at the boiling point (or released on condensation); for steam, $L_v \approx 2268\ \text{J/g} \approx 2.27\times10^6\ \text{J/kg}$. Notice $L_v$ is nearly seven times $L_f$ for water — vaporisation separates molecules completely, while fusion only loosens their fixed positions, so vaporisation is by far the costlier change.

Melting point vs boiling point. The melting point is the fixed temperature at which a solid turns to liquid; impurities lower it, and increased pressure lowers ice's melting point specifically. The boiling point is the fixed temperature at which a liquid turns to vapour throughout its bulk; increased pressure raises it, which is exactly why a pressure cooker — sealed, so pressure builds — cooks food faster than an open pan.

Evaporation vs boiling. Evaporation happens only at a liquid's exposed surface, at any temperature, and proceeds slowly and gradually. Boiling happens throughout the entire liquid, but only at one fixed temperature — the boiling point. Treating the two as interchangeable is a common and penalised mix-up.

The heating curve. Heat a solid steadily and plot temperature against time: the graph climbs (solid warming), then runs perfectly flat at the melting point while latent heat of fusion is absorbed and the solid liquefies, then climbs again (liquid warming), then runs flat again at the boiling point while latent heat of vaporisation is absorbed, and finally climbs once more as vapour is superheated. The two flat plateaus, at fixed temperatures despite continuous heat input, are the graph's unmistakable signature of a change of state.

Steam at 100°C of mass 10 g is passed into 200 g of water at 20°C in a calorimeter of negligible heat capacity. Find the final temperature of the water, assuming the steam fully condenses and no heat is lost to the surroundings, taking $L_v(\text{steam}) = 2268\ \text{J/g}$ and $c(\text{water}) = 4.2\ \text{J/g}\,\cdot\,\text{K}$.

Step 1. The steam first condenses to water at 100°C (releasing latent heat), and that condensed water then cools from 100°C down to the final temperature $T$ (releasing sensible heat), so the heat released has two parts:

Step 2. The original 200 g of water simply warms from 20°C to $T$, a purely sensible-heat gain:

Step 3. With no heat lost to surroundings, heat lost equals heat gained, so equate the two expressions and solve for $T$:

Step 4. Only 10 g of steam, against 200 g of water twenty times its mass, still pushes the final temperature up from 20°C to about 49.5°C — the huge latent heat released on condensation dominates the mixture, which is exactly why steam burns are far more severe than hot-water burns at the same starting temperature.

Competitive stretch

If the same 10 g had arrived as boiling water at 100°C instead of steam, only the sensible-heat term would apply: $10 \times 4.2 \times (100-T) = 840(T-20)$ gives $T \approx 21.8°\text{C}$ — dramatically cooler than the steam case, isolating just how much of the 49.5°C rise above came from the latent heat of condensation alone.

• Forgetting the condensation or melting step in mixed problems — when steam meets water, students often apply only $Q=mc\Delta T$ to the steam and skip the much larger $Q=mL_v$ term for condensation; always check whether a state change happens before or after the temperature change.
• Assuming latent heat causes a temperature rise — by definition temperature stays constant throughout melting or boiling, and all the heat goes into changing state, exactly what the flat plateau on a heating curve shows.
• Mixing up fusion and vaporisation values — $L_f(\text{ice}) \approx 336\ \text{J/g}$ is much smaller than $L_v(\text{steam}) \approx 2268\ \text{J/g}$, because vaporisation must separate molecules completely while fusion only loosens their fixed arrangement.
• Treating evaporation and boiling as the same process — evaporation is a slow surface-only process at any temperature, while boiling occurs throughout the liquid at one fixed temperature; do not use the two words interchangeably.

Ask the tutor: Want to try a trickier combined problem, like ice starting below 0°C being converted all the way to steam — ask for a step-by-step walkthrough.