Changes of state and specific latent heat — AQA GCSE Physics
Test yourself on Changes of state and specific latent heat with AQA GCSE practice questions.
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Changes of state and specific latent heat explained
The two quantities answer different questions.
Read the full explanation
Specific heat capacity is the energy per kilogram to change the temperature by one degree with no change of state, unit J/kg °C, and it governs the sloping parts of a heating graph. Specific latent heat is the energy per kilogram to change state with no change of temperature, unit J/kg, and it governs the flat parts. Read the question for the giveaway words: warms, cools or rises from one temperature to another means c; melts, freezes, boils, condenses or evaporates means L. A problem that says ice at 0 °C is turned into water at 20 °C needs both, E = m L first at the plateau, then ΔE = m c Δθ up the slope.
Your focus
- Decide from the wording of a question whether ΔE = m c Δθ or E = m L is required, and justify the choice.
- Compare specific heat capacity and specific latent heat in terms of what changes, what stays constant, and the units of each.
- Calculate the total energy needed to take ice at 0 °C to water at 20 °C by combining both equations in the correct order.
Changes of state and specific latent heat exam tips
Quick Revision Summary (Key Takeaway)
Specific latent heat is the energy required to change the state of one kilogram of a substance with no change in temperature. During a change of state, energy input alters intermolecular bonds and increases internal potential energy rather than kinetic energy, keeping the temperature constant.
Topic Overview
Changes of state and specific latent heat explore how internal energy alters when substances transition between solids, liquids, and gases. Unlike temperature changes governed by specific heat capacity, state changes occur at fixed temperatures as energy is directed into breaking or forming intermolecular bonds.
This topic connects the particle model of matter directly to quantitative thermodynamics in AQA GCSE Physics. Mastering these principles allows students to interpret heating and cooling curves accurately and solve multi-step energy conservation problems in real-world thermal systems.
Key Concepts
- →Internal energy is the total energy stored inside a system by the particles, comprising the sum of their kinetic and potential energy stores.
- →During a change of state, mass is conserved and the change is purely physical, meaning the material recovers its original properties if the change is reversed.
- →Specific latent heat (L) is defined as the energy required to change the state of 1 kg of a substance with no change in temperature, governed by E = m x L.
- →Specific latent heat of fusion applies to transitions between solid and liquid, whereas specific latent heat of vaporisation applies to transitions between liquid and gas.
- →Heating and cooling curves feature flat horizontal plateaus during phase changes because thermal energy input or removal changes potential energy, not average kinetic energy.
Marking Points
- one mark for specific heat capacity involving a temperature change with no change of state
- one mark for specific latent heat involving a change of state with no temperature change
- one mark for contrasting the units, J/kg °C against J/kg
- one mark for matching each quantity to the sloping or flat section of a heating graph
Examiner Tips
- 💡Underline the verb in the question: melts and boils point to L, warms and cools point to c.
- 💡State both differences when asked to distinguish them, what changes and what stays the same, not just the units.
- 💡In a multi-stage calculation, set the work out as separate labelled stages so an examiner can award each part.
- 💡Always convert time to seconds and mass to kilograms before applying the formula E = m x L or E = P x t.
- 💡In explanation questions, explicitly distinguish between kinetic energy (linked directly to temperature) and potential energy (linked to bond breaking and intermolecular separation).
- 💡On heating curve graphs, label the sloped sections as changes in temperature (E = m x c x delta theta) and the flat horizontal plateaus as changes of state (E = m x L).
Common Mistakes
- using ΔE = m c Δθ across the plateau of a heating graph, where the temperature is constant
- inserting a temperature change into E = m L
- quoting J/kg °C as the unit of specific latent heat
- assuming a substance sitting at its melting point is receiving no energy because the thermometer reading is steady
- Believing that thermal energy input always leads to a temperature rise; during a phase change, temperature remains constant because kinetic energy does not increase.
- Treating boiling and evaporation as identical; evaporation occurs only at the surface at any temperature below boiling, whereas boiling occurs throughout the liquid at a defined boiling point.
- Assuming mass is lost during boiling or evaporation; mass is strictly conserved in all physical state changes, though gas particles may escape an open system.
Revision Plan
- 1Day 1: Review particle arrangements and define internal energy in terms of kinetic and potential energy stores.
- 2Day 2: Practise sketching and annotating heating and cooling curves, marking clearly where E = mc delta theta and E = mL apply.
- 3Day 3: Memorise the definitions and units of specific latent heat of fusion versus vaporisation, practicing formula rearrangements for m and L.
- 4Day 4: Solve combined exam-style calculation questions involving electrical power (P = E / t) and specific latent heat (E = m x L).
Exam Question Types
- 📋Calculation questions: Rearranging and calculating values using E = m x L, often combined with electrical energy supplied (E = P x t or E = V x I x t).
- 📋Graph interpretation: Identifying states of matter, melting points, boiling points, and explaining why specific sections of a heating or cooling curve are horizontal.
- 📋Extended response (4-6 marks): Describing an experiment to measure the specific latent heat of fusion of ice or vaporisation of water, including apparatus, measurements, and sources of error.
Command Word Expectations (AQA)
Give reasons based on physics principles. For state changes, you must link particle separation and intermolecular forces to changes in internal potential energy rather than kinetic energy.
Show all working stages: state the equation, substitute unrounded values with correct units (kg, J, s), and evaluate the final answer to an appropriate number of significant figures.
State what happens or outline the steps of a method clearly without needing to provide theoretical reasons for why it happens.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A 100 W electric immersion heater is placed into 0.15 kg of pure liquid water already at 100 °C. The heater is switched on for 5.0 minutes. Calculate the mass of water converted into steam during this time. The specific latent heat of vaporisation of water is 2.26 x 10^6 J/kg.
- 1.Step 1: Calculate the total energy supplied by the heater using E = P x t. Time must be in seconds: t = 5.0 min x 60 s = 300 s. Energy E = 100 W x 300 s = 30,000 J.
- 2.Step 2: Recall and rearrange the specific latent heat equation E = m x L to solve for mass: m = E / L.
- 3.Step 3: Substitute values into the rearranged equation: m = 30,000 J / (2.26 x 10^6 J/kg) = 0.01327... kg.
- 4.Step 4: Round to an appropriate number of significant figures (2 s.f. based on given data): m = 0.013 kg (or 13 g).
Question: An ice cube of mass 0.040 kg at 0 °C absorbs 13,360 J of thermal energy from its surroundings. Determine whether all the ice melts, and calculate the specific latent heat of fusion of water if this energy is exactly sufficient to melt it completely.
- 1.Step 1: State the relationship between thermal energy, mass, and specific latent heat of fusion: E = m x Lf.
- 2.Step 2: Rearrange the equation to make specific latent heat (Lf) the subject: Lf = E / m.
- 3.Step 3: Substitute the provided values: Lf = 13,360 J / 0.040 kg.
- 4.Step 4: Calculate the numerical result and supply standard units: Lf = 334,000 J/kg (3.34 x 10^5 J/kg).