Specific Latent Heat Revision Notes | A Level AQA Physics

1. The Puzzle of the Flat Line

Picture a beaker of ice on a hotplate. You supply energy continuously, yet the thermometer reads 0 °C for several minutes without budging. The ice is absorbing energy — so where is it going? This is the central puzzle of specific latent heat, and once you crack it, a whole class of exam questions becomes straightforward.

The answer lies in what energy actually does at a molecular level. During a phase transition, supplied energy breaks intermolecular bonds rather than increasing the kinetic energy of molecules. Because temperature is a measure of average molecular kinetic energy, the temperature stays flat until the transition is complete.

2. Defining Specific Latent Heat

There are two named values you need to know. The specific latent heat of fusion ($l_f$) applies to the solid-liquid transition — melting or freezing. The specific latent heat of vaporisation ($l_v$) applies to the liquid-gas transition — boiling or condensing. Vaporisation values are always larger than fusion values for the same substance, because completely separating molecules from a liquid requires far more energy than merely loosening them from a solid lattice.

For water: $l_f \approx 3.34 \times 10^5$ J kg⁻¹ and $l_v \approx 2.26 \times 10^6$ J kg⁻¹. Notice that vaporisation requires roughly seven times more energy per kilogram than fusion — a fact that explains why steam burns are so much more severe than water burns at the same temperature.

3. The Core Equation: $Q = ml$

We need a way to calculate the energy transferred during a change of state. The relationship is direct: double the mass and you need double the energy; choose a substance with a higher specific latent heat and you need more energy per kilogram. That proportionality gives us the equation at the heart of this topic.

The equation rearranges to $l = Q/m$ (useful when you measure energy supplied and mass in an experiment) and $m = Q/l$ (useful when you know the energy available and want to find how much mass changes state). Keep track of which transition you are dealing with — fusion or vaporisation — and use the corresponding value of $l$.

4. Molecular Interpretation of Latent Heat

Zoom in to the molecular scale and the flat-line temperature behaviour makes complete sense. In a solid, molecules vibrate about fixed positions held by strong intermolecular forces. As you supply energy at the melting point, that energy increases the potential energy of the molecules — stretching and eventually breaking the bonds that hold the lattice together — rather than increasing their kinetic energy.

Because temperature measures average molecular kinetic energy, and kinetic energy is not increasing during the transition, the temperature stays constant. Only once all the bonds associated with that transition have been broken does further energy input go back into raising kinetic energy, and the temperature climbs again.

Vaporisation requires more energy than fusion because molecules in a liquid still experience significant intermolecular attraction. To escape into the gas phase, they must overcome those attractions almost entirely. In a solid-to-liquid transition, molecules only need enough energy to move past fixed lattice positions — the intermolecular forces are weakened but not eliminated.

5. Heating Curves and Identifying Phase Transitions

A heating curve plots temperature against time (or energy supplied) for a substance heated at a constant rate. Knowing how to read one is a core exam skill. The curve has sloping regions where temperature rises within a single phase, and flat regions (plateaux) where a phase transition is occurring.

The two flat sections correspond to melting (at 0 °C for water) and boiling (at 100 °C). The length of each plateau is proportional to the specific latent heat for that transition — a longer plateau means more energy is needed per kilogram. The steepness of the sloping sections depends on the specific heat capacity of each phase.

Notice that the boiling plateau is longer than the melting plateau for water. This reflects the fact that $l_v > l_f$ — vaporisation demands more energy per kilogram than fusion, so at the same heating rate it takes longer to complete.

6. Worked Calculations Using $Q = ml$

Exam questions on this topic follow a reliable pattern: you are given two of the three quantities ($Q$, $m$, $l$) and asked to find the third. The method is always the same — identify the phase transition, select the correct value of $l$, substitute, and check units. Here is a full walkthrough.

Worked example: Calculating Energy to Melt Ice

Calculate the energy required to completely melt 0.25 kg of ice at 0 °C. The specific latent heat of fusion of water is $3.34 \times 10^5$ J kg⁻¹. [2 marks]

1. $Q = ml$ — This is a change of state at constant temperature, so $Q = ml$ applies. [M1] for selecting this equation.
2. $m = 0.25$ kg, $l = 3.34 \times 10^5$ J kg⁻¹ — We are dealing with melting (solid to liquid), so we use the latent heat of fusion. Both values are already in SI units — no conversion needed.
3. $Q = 0.25 \times 3.34 \times 10^5$ — [M1] for correct substitution with consistent units.
4. $Q = 0.25 \times 334000 = 83500$ J — Expanding: $0.25 \times 334000 = 83500$. Check: $0.25 \times 300000 = 75000$ and $0.25 \times 34000 = 8500$, so $75000 + 8500 = 83500$. Consistent.
5. $Q = 8.35 \times 10^4$ J — [A1] for correct answer with units. Expressing in standard form is good practice for large values.

Answer: $Q = 8.35 \times 10^4$ J (or 83 500 J)

Worked example: Finding Specific Latent Heat from Experimental Data

A student supplies 1.13 × 10⁶ J of energy to 0.50 kg of water already at 100 °C, causing it to boil away completely. Calculate the specific latent heat of vaporisation of water from this data. [3 marks]

1. $Q = ml \Rightarrow l = \dfrac{Q}{m}$ — [M1] for rearranging correctly. We are solving for $l$, so divide both sides by $m$.
2. $Q = 1.13 \times 10^6$ J, $m = 0.50$ kg — The substance is already at 100 °C, so all the energy goes into the phase transition — no $Q = mc\Delta T$ component is needed here.
3. $l = \dfrac{1.13 \times 10^6}{0.50}$ — [M1] for correct substitution.
4. $l = \dfrac{1130000}{0.50} = 1130000 \times 2 = 2.26 \times 10^6$ J kg⁻¹ — Dividing by 0.50 is the same as multiplying by 2. $1130000 \times 2 = 2260000 = 2.26 \times 10^6$.
5. $l = 2.26 \times 10^6$ J kg⁻¹ — [A1] for correct answer with correct units. This matches the accepted value for the specific latent heat of vaporisation of water — a useful sanity check.

Answer: $l = 2.26 \times 10^6$ J kg⁻¹

7. Experimental Determination of Specific Latent Heat

AQA expects you to understand how specific latent heat can be measured experimentally. The standard approach uses an electrical heater of known power, allowing you to calculate the energy supplied ($Q = Pt$, where $P$ is power in watts and $t$ is time in seconds). You then measure the mass that changes state and apply $l = Q/m$.

The main source of systematic error in this experiment is heat exchange with the surroundings. Some energy from the heater warms the air and container rather than melting ice, so the measured value of $l$ tends to be an overestimate. To reduce this, use well-insulated containers and work quickly. A control experiment — running without the heater to measure how much ice melts from room-temperature heat alone — allows a correction to be applied.

8. Exam Technique and Mark-Winning Strategies

This topic has a small number of high-frequency exam traps. Knowing them in advance is worth marks. The guidance below covers the most common errors seen in AQA mark schemes, along with the language that secures full credit.

For extended-response questions asking you to evaluate an experimental method, structure your answer as: identify the source of error → explain the direction of the effect (overestimate or underestimate) → suggest a specific improvement → justify why the improvement works. This four-part chain reliably picks up all available marks.

Specific latent heat — summarised in $Q = ml$ — quantifies the energy needed to change the state of a substance at constant temperature, with the molecular explanation rooted in changes to intermolecular potential energy rather than kinetic energy. Mastering the distinction between fusion and vaporisation, and knowing when to switch between $Q = ml$ and $Q = mc\Delta T$, covers the vast majority of marks available on this topic.

During a change of state, the thermal energy transferred is given by Q = ml, where l is the specific latent heat. This equation applies when the temperature stays constant and the energy transferred changes the state rather than the kinetic energy of the particles.