Energy f or a change o f state = mass × speci f ic latent heat E = m L…

The idea

E = m L has no temperature term, as temperature is constant during a change of state. Boiling 0.25 kg of water needs E = 0.25 × 2 260 000 = 565 000 J. On heating and cooling graphs, sloping parts belong to ΔE = m c Δθ and flat plateaus belong to E = m L. Heating ice at -10 °C to steam at 100 °C involves four stages: warming ice, melting, warming water, and boiling. Cooling graphs show the reverse, with plateaus for condensing and freezing. Two standard practical errors exist. If thermal energy escapes to the surroundings, less energy reaches the water, making the calculated L too high. If water evaporates before boiling, the measured mass change is too large for the energy supplied, making L too low.

What earns the marks

• Choosing E = m L for a change of state and ΔE = m c Δθ for a temperature change.
• Correct substitution with the mass that actually changed state, in kilograms.
• The answer in joules.
• Stating that thermal energy transferred to the surroundings makes the experimental value of L too high.
• Stating that water evaporating before boiling makes the measured mass change too large, so the value of L is too low.

Common mistakes and how to put them right

• Multiplying by a temperature change in E = m L, which has no temperature term.

• Reading the melting point from a sloping section of the graph instead of the plateau.

• Using the total mass of the sample when only part of it changed state.

• Adding the two stages of a graph question together but using c for both, or L for both.

Exam tips

• Label each section of a heating or cooling graph with the equation you will use before you calculate anything.
• When asked about an error, say whether the value of L is too high or too low, and give the reason in the same sentence.
• Check the mass: it is the mass that changed state, often a mass loss read from a balance, not the starting mass.