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HomeAQA GCSE PhysicsInternal energy and specific latent heat
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Internal energy and specific latent heat

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What you'll learn

This revision guide covers internal energy and specific latent heat as required for AQA GCSE Physics. You'll understand how heating affects particle energy stores, why temperature remains constant during state changes, and how to calculate the energy required for melting, boiling, freezing and condensing substances.

Key terms and definitions

Internal energy — the total kinetic energy and potential energy of all the particles in a system

Specific latent heat — the energy required to change the state of 1 kg of a substance without changing its temperature

Specific latent heat of fusion — the energy required to change 1 kg of a substance from solid to liquid (or released when 1 kg changes from liquid to solid) at constant temperature

Specific latent heat of vaporisation — the energy required to change 1 kg of a substance from liquid to gas (or released when 1 kg changes from gas to liquid) at constant temperature

State change — a physical change where a substance changes between solid, liquid and gas without changing its chemical composition

Kinetic energy store — the energy store associated with the movement of particles

Potential energy store — the energy store associated with the position of particles relative to each other (affected by forces of attraction between particles)

System — a defined object or group of objects that we're studying

Core concepts

Internal energy and particle behaviour

Internal energy is the total energy stored by all particles within a system. This energy exists in two forms:

  • Kinetic energy of particles moving and vibrating
  • Potential energy from the forces of attraction between particles

When you heat a substance, you transfer energy to it. This energy increases the internal energy of the system in one or both of these ways:

  • Increases the kinetic energy of particles (they move faster) — temperature rises
  • Increases the potential energy of particles (they move further apart, weakening attractive forces) — no temperature change

The temperature of a substance is directly related to the average kinetic energy of its particles. When particles move faster, the temperature increases. When particles move slower, the temperature decreases.

Changes of state and energy

When a substance changes state (melts, boils, freezes, condenses, sublimates), bonds between particles are either broken or formed:

Solid to liquid (melting):

  • Energy is supplied to break some bonds between particles
  • Particles vibrate more and can move past each other
  • Internal energy increases (potential energy increases)
  • Temperature remains constant during melting

Liquid to gas (boiling/evaporating):

  • Energy is supplied to break remaining bonds between particles
  • Particles move freely and rapidly
  • Internal energy increases significantly (potential energy increases)
  • Temperature remains constant during boiling

Gas to liquid (condensing):

  • Energy is released as bonds form between particles
  • Internal energy decreases (potential energy decreases)
  • Temperature remains constant during condensing

Liquid to solid (freezing):

  • Energy is released as more bonds form between particles
  • Particles can only vibrate in fixed positions
  • Internal energy decreases (potential energy decreases)
  • Temperature remains constant during freezing

Why temperature stays constant during state changes

This is a crucial concept that students often misunderstand. When you heat ice at 0°C, the temperature does not rise above 0°C until all the ice has melted. Here's why:

The energy supplied during a state change is used entirely to break bonds between particles (increasing potential energy). None of this energy increases the kinetic energy of particles, so the average kinetic energy — and therefore the temperature — remains constant.

Once all bonds are broken and the state change is complete, any additional energy then increases the kinetic energy of particles, causing the temperature to rise.

Heating and cooling curves

A heating curve shows how temperature changes when a substance is heated continuously at a constant rate:

Flat sections represent state changes:

  • Temperature constant
  • Energy used to break bonds (increase potential energy)
  • Duration of flat section depends on specific latent heat

Sloped sections represent heating within one state:

  • Temperature increases
  • Energy increases kinetic energy of particles
  • Gradient depends on specific heat capacity

The longer the flat section, the more energy is required for that state change. For most substances, the flat section for boiling is longer than for melting because more energy is required to separate particles completely (liquid to gas) than to allow them to move past each other (solid to liquid).

Calculating energy changes during state changes

The energy required to change the state of a substance is calculated using:

Energy = mass × specific latent heat

E = m L

Where:

  • E = energy transferred (joules, J)
  • m = mass (kilograms, kg)
  • L = specific latent heat (joules per kilogram, J/kg)

Use L_f for specific latent heat of fusion (melting/freezing)

Use L_v for specific latent heat of vaporisation (boiling/condensing)

Key values to know:

For water:

  • Specific latent heat of fusion: 334,000 J/kg (or 334 kJ/kg)
  • Specific latent heat of vaporisation: 2,260,000 J/kg (or 2260 kJ/kg)

The specific latent heat of vaporisation is much larger than the specific latent heat of fusion. This is because:

  • Melting only weakens bonds enough for particles to move past each other
  • Boiling completely separates particles against attractive forces
  • Much more energy is needed to overcome all attractive forces

Energy conservation during state changes

State changes are reversible physical processes. The energy:

  • Released when freezing = energy absorbed when melting (for same mass)
  • Released when condensing = energy absorbed when boiling (for same mass)

This is important in many real-world contexts:

Caribbean/tropical applications:

  • Ice melting in drinks absorbs energy, cooling the drink
  • Sweat evaporating from skin absorbs energy, cooling the body (essential in hot climates)
  • Condensation on cold water pipes releases energy, warming surroundings

UK/temperate applications:

  • De-icing roads with salt lowers freezing point, requiring more energy to be removed
  • Steam burns are more severe than boiling water burns because steam releases latent heat when condensing on skin
  • Frost formation (deposition) releases latent heat to surroundings

Worked examples

Example 1: Melting ice

Question: Calculate the energy required to melt 0.50 kg of ice at 0°C. The specific latent heat of fusion of ice is 334,000 J/kg. [3 marks]

Solution:

Step 1: Write down the equation E = m L [1 mark]

Step 2: Substitute values E = 0.50 kg × 334,000 J/kg [1 mark]

Step 3: Calculate and include unit E = 167,000 J (or 167 kJ) [1 mark]

Mark scheme notes: Award marks for correct equation, correct substitution with units, and correct answer with unit. Alternative forms like 1.67 × 10⁵ J are acceptable.

Example 2: Boiling water

Question: A kettle boils 1.5 kg of water. Calculate the energy transferred to change the water at 100°C to steam at 100°C. The specific latent heat of vaporisation of water is 2,260,000 J/kg. Give your answer in kJ. [4 marks]

Solution:

Step 1: Identify which latent heat to use Boiling uses latent heat of vaporisation [1 mark]

Step 2: Write equation and substitute E = m L_v E = 1.5 kg × 2,260,000 J/kg [1 mark]

Step 3: Calculate E = 3,390,000 J [1 mark]

Step 4: Convert to kJ E = 3,390 kJ (or 3390 kJ or 3.39 MJ) [1 mark]

Mark scheme notes: The question specifies the answer should be in kJ, so conversion is required for full marks.

Example 3: Combined heating and state change

Question: A student heats 2.0 kg of ice at 0°C until it becomes water at 0°C. The specific latent heat of fusion is 334,000 J/kg.

(a) Calculate the energy required. [3 marks] (b) Explain why the temperature does not change during melting. [2 marks]

Solution:

(a) E = m L_f [1 mark] E = 2.0 kg × 334,000 J/kg [1 mark] E = 668,000 J or 668 kJ [1 mark]

(b) The energy supplied is used to break bonds between water molecules [1 mark]. This increases the potential energy of the particles but not their kinetic energy, so temperature (which depends on average kinetic energy) remains constant [1 mark].

Mark scheme notes: For part (b), both points about potential energy increasing AND kinetic energy not changing are needed for full marks. Saying "energy used to change state" without explanation is insufficient.

Common mistakes and how to avoid them

  • Using mass in grams instead of kilograms — Always convert mass to kg before using E = mL. If mass is given as 500 g, convert to 0.5 kg. Using grams will make your answer 1000 times too small.

  • Confusing fusion and vaporisation — Fusion refers to melting/freezing (solid ↔ liquid). Vaporisation refers to boiling/condensing (liquid ↔ gas). Check the states in the question carefully and use the correct specific latent heat value.

  • Stating that temperature increases during a state change — Temperature remains constant during state changes at constant pressure. The energy supplied increases potential energy (by breaking bonds) not kinetic energy, so temperature doesn't change.

  • Forgetting that energy is released during freezing and condensing — When substances change from gas to liquid or liquid to solid, energy is transferred to the surroundings. The same equation E = mL applies, but energy is released rather than absorbed.

  • Confusing specific heat capacity with specific latent heat — Specific heat capacity relates to temperature change within one state. Specific latent heat relates to state changes at constant temperature. Read questions carefully to identify which applies.

  • Not including units in final answers — Always include the unit (J or kJ) in your final answer. In calculations worth 3-4 marks, the unit is often worth 1 mark. Without it, you lose marks even if the number is correct.

Exam technique for "Internal energy and specific latent heat"

  • "Calculate" questions — Show the equation, substitution with units, and final answer with unit. For 3-mark calculations, these three steps typically earn one mark each. Don't skip working out.

  • "Explain" questions about temperature during state changes — Must mention both that energy breaks/forms bonds (changing potential energy) AND that kinetic energy doesn't change (so temperature constant). One-sentence answers rarely score full marks.

  • Units matter — For specific latent heat, the unit is J/kg. For energy, accept J, kJ, or MJ but convert if the question specifies. Always check whether mass is given in kg or g.

  • Command word "Describe" — State what happens in sequence without necessarily explaining why. For heating curves, describe flat sections (state changes at constant temperature) and sloped sections (temperature increases while heating one state).

Quick revision summary

Internal energy is the total kinetic and potential energy of all particles in a system. When substances change state at constant temperature, energy supplied breaks or forms bonds (changing potential energy) without changing kinetic energy. Specific latent heat is the energy needed to change 1 kg of a substance's state without temperature change. Use E = mL for calculations, with L_f for fusion (melting/freezing) and L_v for vaporisation (boiling/condensing). Water's latent heat of vaporisation (2,260,000 J/kg) is much larger than its latent heat of fusion (334,000 J/kg).

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