Latent heat: melting, boiling and the heating curve: WASSCE Physics

Put a pot of water on the fire and watch the thermometer. It climbs steadily to 100 °C — and then stops, even though the fire still burns. Where is all that heat going? Into changing water to steam. That hidden heat is latent heat.

Learning objectives

By the end of this lesson you should be able to:

describe the changes of state and interpret a heating curve
define latent heat, specific latent heat of fusion and of vaporisation
use $Q = mL$ together with $Q = mc\Delta\theta$
explain latent heat using the kinetic theory

Changes of state

Change	Name	Heat
solid → liquid	melting (fusion)	absorbed
liquid → solid	freezing (solidification)	given out
liquid → gas	boiling / evaporation (vaporisation)	absorbed
gas → liquid	condensation	given out
solid → gas directly	sublimation (e.g. iodine, camphor, naphthalene)	absorbed

The heating curve

Temperature against time as ice at minus 10 degrees Celsius is heated: the temperature rises to 0, stays flat while the ice melts, rises to 100, stays flat while the water boils, then rises again as steam

Figure 1. Heating curve of water. WASSCE Tutor diagram.

Ice warms from −10 °C to 0 °C (temperature rises).
At 0 °C the ice melts: the temperature stays constant until all the ice has melted.
Water warms from 0 °C to 100 °C.
At 100 °C the water boils: the temperature stays constant until all the water has become steam.
Steam warms above 100 °C.

The flat sections show that heat is being absorbed without a change in temperature. This heat is latent ("hidden") heat.

Latent heat

Definition

Latent heat is the heat absorbed or released when a substance changes state without a change in temperature.

Specific latent heat of fusion $L_f$ : the quantity of heat needed to change 1 kg of a substance from solid to liquid at its melting point, without a change in temperature.

Specific latent heat of vaporisation $L_v$ : the quantity of heat needed to change 1 kg of a substance from liquid to vapour at its boiling point, without a change in temperature.

Q = mL \quad (L \text{ in J kg}^{-1})

\begin{gathered}Q = mL \\[6pt] (L \text{ in J kg}^{-1})\end{gathered}

Substance	Value
Ice (fusion)	$3.36 \times 10^{5}$ J kg⁻¹ (336 kJ kg⁻¹)
Water (vaporisation)	$2.26 \times 10^{6}$ J kg⁻¹ (2260 kJ kg⁻¹)

It takes almost seven times as much energy to boil away water as to melt the same mass of ice.

Kinetic explanation

During melting, the energy supplied does not speed the particles up (temperature unchanged); it breaks the bonds holding them in fixed positions.
During boiling, the energy separates the particles completely, overcoming their attraction and pushing back the atmosphere as the vapour expands. This needs far more energy, which is why $L_v > L_f$ .

Worked examples

Worked example

How much heat is needed to melt 2 kg of ice at 0 °C? ( $L_f = 3.36 \times 10^{5}$ J kg⁻¹)

Q = mL_f = 2 \times 3.36 \times 10^{5} = 6.72 \times 10^{5}\ \text{J}

\begin{aligned}Q &= mL_f \\ &= 2 \times 3.36 \times 10^{5} \\ &= 6.72 \times 10^{5}\ \text{J}\end{aligned}

Worked example

Find the heat needed to change 0.5 kg of ice at −10 °C completely into steam at 100 °C. ( $c_{ice} = 2100$ , $c_{water} = 4200$ J kg⁻¹ K⁻¹, $L_f = 3.36 \times 10^{5}$ , $L_v = 2.26 \times 10^{6}$ J kg⁻¹)

Stage	Heat
Warm ice −10 → 0 °C	$0.5 \times 2100 \times 10 = 10\,500$ J
Melt ice at 0 °C	$0.5 \times 3.36 \times 10^{5} = 168\,000$ J
Warm water 0 → 100 °C	$0.5 \times 4200 \times 100 = 210\,000$ J
Boil water at 100 °C	$0.5 \times 2.26 \times 10^{6} = 1\,130\,000$ J
Total	1 518 500 J ≈ 1.52 MJ

Boiling takes three-quarters of the total.

Worked example

Ice at 0 °C is added to 0.4 kg of water at 30 °C until the water just reaches 0 °C with all the ice melted. What mass of ice was added?

Heat lost by water $= 0.4 \times 4200 \times 30 = 50\,400$ J.

Heat gained by ice (melting only) $= m \times 3.36 \times 10^{5}$ .

m = \dfrac{50\,400}{3.36 \times 10^{5}} = 0.15\ \text{kg}

\begin{aligned}m &= \dfrac{50\,400}{3.36 \times 10^{5}} \\ &= 0.15\ \text{kg}\end{aligned}

In Sierra Leone

A scald from steam is worse than one from boiling water at the same temperature. Each kilogram of steam that condenses on the skin releases 2.26 MJ of latent heat before it even starts to cool as water.

Key points

Temperature stays constant during a change of state; the heat goes into breaking bonds (latent heat).
$Q = mL$ : $L_f$ for melting, $L_v$ for boiling.
For ice at 0 °C, $L_f \approx 3.36 \times 10^{5}$ J kg⁻¹; for water at 100 °C, $L_v \approx 2.26 \times 10^{6}$ J kg⁻¹.
Multi-stage problems: add $mc\Delta\theta$ for each temperature change and $mL$ for each change of state.

WASSCE examination focus

WAEC tests this

Define the specific latent heats precisely (include "at constant temperature" / "at its melting/boiling point"), interpret or sketch a heating or cooling curve, and calculate the total heat for a multi-stage change — the ice-to-steam calculation is a favourite. Explain why steam scalds are more severe than boiling-water scalds.

Quick check

Why does the temperature stay constant while ice melts?
How much heat is released when 0.2 kg of steam condenses at 100 °C? ( $L_v = 2.26 \times 10^{6}$ J kg⁻¹)
Name a substance that sublimes.

Answers: 1. The heat supplied breaks the bonds between the particles instead of increasing their kinetic energy. 2. $0.2 \times 2.26 \times 10^{6} = 4.52 \times 10^{5}$ J. 3. For example: iodine, camphor or naphthalene.

Latent heat: melting, boiling and the heating curve

Learning objectives

Changes of state

The heating curve

Latent heat

Kinetic explanation

Worked examples

Key points

WASSCE examination focus

Quick check

Keep going with change of state

More lessons in Change of State