Heating and changes of state: revision notes and practice

What you will learn

• Choose between specific heat capacity and latent heat equations.
• Calculate separate stages of warming and changing state.
• Explain temperature plateaus and evaluate a heating practical.

Key rules

Temperature changes

E = mcΔT. Use mass in kg, c in J/(kg °C) and the temperature difference, not the final temperature.

Changes of state

E = mL. Specific latent heat L is in J/kg. Melting or boiling a pure substance at constant pressure can occur at constant temperature.

Combine stages

Calculate warming energy and change-of-state energy separately, then add. Include cooling/heating losses only if the question supplies a model.

Worked examples

Example 1: Warm 0.5 kg by 10 °C with c = 400 J/(kg °C). Find E.

1. E = mcΔT.
2. 0.5 × 400 × 10 = 2000.

2000 J

Example 2: Melt 0.2 kg at its melting point, with L = 330 000 J/kg.

1. Temperature is not rising in this stage.
2. E = mL = 0.2 × 330 000.

66 000 J

A practice question

A 2 kg sample warms from 17 °C to 30 °C without changing state. Its specific heat capacity is 1000 J/(kg °C). Calculate the energy absorbed.

1. Temperature and energy describe different things

Temperature indicates how hot a substance is and is related to the average kinetic energy of its particles. Energy transferred by heating can change the temperature, the arrangement of particles, or both during a sequence of processes. Two objects at the same temperature do not necessarily have the same thermal energy, because their masses and substances can differ.

Use a clear system description: identify the sample, its starting state and its final state. A solid warming without melting requires a different equation from a solid melting at its melting temperature. Avoid choosing an equation merely because it contains a familiar number. Ask whether temperature changes or a change of state is taking place.

2. Specific heat capacity links energy to temperature rise

Specific heat capacity is the energy needed to raise the temperature of one kilogram of a substance by one degree Celsius without a state change. For a mass m, multiply mass by specific heat capacity c and temperature change. Heating 0.50 kilograms with c = 400 joules per kilogram per degree Celsius through 20 degrees Celsius requires 4000 joules.

Temperature change means final temperature minus initial temperature. From 18 to 38 degrees Celsius, use 20 degrees, not 38. A temperature difference in Celsius has the same numerical size as a difference in kelvin. Convert grams to kilograms before substituting, and use the units given for c consistently. A large c means more energy for the same mass and rise.

E = mcΔT; E in J, m in kg, c in J/(kg °C), ΔT in °C

3. Rearrange and check a warming calculation

If energy, mass and temperature rise are known, calculate c = E/(mΔT). A 2-kilogram sample absorbing 12 000 joules while warming by 15 degrees Celsius has c = 12 000/(2 × 15) = 400 joules per kilogram per degree Celsius. Keep the entire product mΔT in the denominator; dividing by mass and then multiplying by the rise gives the wrong relationship.

The ideal calculation assumes that the stated energy is absorbed by the sample and that c stays approximately constant over the temperature interval. If a heater supplies energy electrically, use E = Pt with power in watts and time in seconds. Supplied energy may also warm the heater, container and surroundings, so it is not always equal to energy absorbed by the sample.

4. Latent heat describes a change of state

Specific latent heat is the energy needed per kilogram for a change of state without a temperature change. Latent heat of fusion concerns melting or freezing; latent heat of vaporisation concerns boiling or condensing. Multiply mass by the appropriate L. Melting 0.20 kilograms with L = 330 000 joules per kilogram requires 66 000 joules when the sample is already at its melting temperature.

During a pure substance’s melting or boiling at constant pressure, transferred energy changes particle arrangements and the potential energy associated with interactions. Average particle kinetic energy, and therefore temperature, stays approximately constant during the transition. Do not describe this as breaking the chemical bonds inside every molecule. Melting changes the state while retaining the substance’s chemical identity.

E = mL; E in J, m in kg, L in J/kg

5. Combine the stages of a heating curve

A heating curve may have sloping sections and plateaus. A sloping section indicates rising temperature within one state; use mcΔT for that stage. A plateau represents a change of state; use mL for that stage. To calculate a complete process, work out each stage separately and add the energies, rather than trying to put the total temperature range into one equation.

For example, warming 0.10 kilograms through 10 degrees with c = 2000 needs 2000 joules. Melting it with L = 300 000 then needs 30 000 joules, so the combined requirement is 32 000 joules. If a question starts at the melting temperature, omit the preliminary warming stage. If only part of the sample melts, use the mass that actually changes state for mL.

6. Cooling reverses the transfer direction

A freezing or condensing substance transfers energy to the surroundings. The magnitude for a complete state change is still mL, but the sample loses rather than gains energy. While cooling within one state, the temperature decreases; use the temperature drop if asked for the positive magnitude of energy released. Explain the direction separately if the question does not use signed energies.

Evaporation can occur at a liquid surface below its boiling point. More energetic particles escape, leaving a lower average kinetic energy among the remaining liquid particles, so the liquid can cool. Boiling occurs throughout the liquid under the stated conditions. Distinguishing these processes prevents treating every loss of liquid as a temperature plateau at the boiling point.

7. Evaluate a supervised heating experiment

Measure the sample’s mass, record a stable starting temperature, supply a measured electrical energy, then record the temperature rise. Use a probe with good thermal contact and insulate the sample to reduce transfer to the surroundings. Follow the teacher’s method for heaters and hot materials. Repeat sensible measurements and compare results rather than relying on one uncertain temperature reading.

If all supplied energy is treated as absorbed by the sample despite losses, the calculated c is usually too large: the measured temperature rise is smaller than it would be without those losses. Insulation reduces this systematic problem. A poorly positioned probe can give an unrepresentative temperature. Name the measurement affected and the resulting bias; human error alone does not explain the effect.

8. Link energy requirements to heater power

Useful heater power tells you how quickly energy reaches the sample. Once the full energy requirement is known, calculate time using t = E/P. A process requiring 36 000 joules with constant useful power of 600 watts takes 60 seconds in the ideal model. If the supplied power is twice as large and losses are unchanged, the ideal time is halved.

Distinguish useful power reaching the sample from electrical input power when efficiency is supplied. A heater with 800-watt input and efficiency 0.75 provides 600 watts of useful heating power in that model. Do not multiply by efficiency again if the question already states useful power. For a multi-stage process, use the sum of the warming and state-change energies before dividing by constant useful power.

Check your understanding

How to earn the marks

• Use the temperature difference rather than final temperature.
• Match c or L to the relevant substance and stage.
• Add separate warming and state-change energies.
• State assumptions about losses and describe specific practical limitations.

A common mistake

Energy can be transferred during a temperature plateau. It changes particle arrangements rather than average kinetic energy.

Course context and official specifications

This guide covers selected skills. Some extensions are Higher content; choose your official tier separately from practice difficulty in the app. These are original questions and explanations.

• AQA GCSE Physics — 4.1.1.3; 4.3.2.2–4.3.2.3

  E = mcΔT and E = mL calculations are available at both tiers. This bank uses supplied material constants and selected heating or phase-change examples.

• Edexcel GCSE Physics — 14.6–14.11

  E = mcΔT and E = mL calculations are available at both tiers. The bank includes supplied constants and selected thermal examples, not the full particle-model topic.

• OCR Gateway GCSE Physics A — P1.2c–P1.2f; PM1.2i–PM1.2ii

  The heating and latent-heat equations are available at both tiers. Supplied constants and selected examples do not represent full P1.2 coverage.

• OCR Twenty First Century GCSE Physics B — P6.1.3–P6.1.6

  E = mcΔT and E = mL calculations are available at both tiers. This selected bank uses supplied constants and does not cover the entire matter chapter.

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