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6.5 Specific Latent Heat

GCSE Physics · AQA · Chapter 6: Molecules and Matter

A free GCSE Physics lesson for AQA (8463) from OBA School: latent heat, specific latent heat of fusion and vaporisation, E = m L, measuring L for ice and water. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • Explain what latent heat is and why the temperature does not change during a change of state
  • Define specific latent heat of fusion and specific latent heat of vaporisation
  • Use E = m L, rearranged where needed, and combine it with ΔE = m c Δθ and E = P t
  • Describe how to measure the specific latent heat of fusion of ice and of vaporisation of water
  • Evaluate these experiments and explain why results can be too high

Key words

  • Latent heat: Energy transferred to or from a substance when it changes state, with no change in temperature.
  • Specific latent heat (L): The energy needed to change the state of 1 kg of a substance with no change in temperature. Unit: J/kg.
  • Fusion: Melting (solid to liquid). LF is the specific latent heat of fusion.
  • Vaporisation: Boiling (liquid to gas). LV is the specific latent heat of vaporisation.
  • Joulemeter: A meter that records the total energy transferred by an electric current, in joules.

Latent heat

When a substance is at its melting or boiling point, energy supplied to it changes its state rather than its temperature. The particles use the energy to break away from each other, so their potential energy increases. This energy is called latent heat (latent means “hidden”).

  • Melting and freezing involve the latent heat of fusion. Energy is taken in when a solid melts; the same amount is released to the surroundings when the liquid freezes, as the particles bond into a rigid structure.
  • Boiling and condensing involve the latent heat of vaporisation. Energy is taken in when a liquid boils; the same amount is released when the vapour condenses and the particles form new bonds.

Exam tip: A scald from steam is worse than one from boiling water at the same temperature. As the steam condenses on skin, it releases its latent heat of vaporisation as well.

Specific latent heat and E = m L

The specific latent heat of a substance is the energy needed to change the state of 1 kg of it without changing its temperature. Its unit is J/kg.

E = m L

E = energy for the change of state, in joules (J)

m = mass that changes state, in kilograms (kg)

L = specific latent heat, in joules per kilogram (J/kg)

Rearranged: m = E / L and L = E / m

  • water; LF in J/kg (melting) 3.34 × 10⁵; LV in J/kg (boiling) 2.26 × 10⁶
  • ethanol; LF in J/kg (melting) 1.08 × 10⁵; LV in J/kg (boiling) 8.5 × 10⁵
  • lead; LF in J/kg (melting) 2.3 × 10⁴; LV in J/kg (boiling) 8.6 × 10⁵

Worked example: melting ice for a cool box

How much energy is needed to melt 0.25 kg of ice at 0 °C?
E = m LF = 0.25 × 334 000

Answer: 83 500 J (8.35 × 10⁴ J)

Worked example: boiling water away

What mass of water at 100 °C is turned to steam by 452 kJ of energy?
E = 452 kJ = 452 000 J; m = E ÷ LV = 452 000 ÷ 2 260 000

Answer: 0.20 kg

Multi-stage problems

If the temperature changes and the state changes, work out each stage separately and add them. Use ΔE = m c Δθ for a temperature change and E = m L for a change of state. Use E = P t to link energy and time for an electrical heater.

Worked example: a kettle left on

A kettle holds 0.40 kg of water at 20 °C. It boils, and 0.10 kg of water boils away before it switches off. c = 4200 J/kg °C, LV = 2.26 × 10⁶ J/kg.
heating to 100 °C: ΔE = 0.40 × 4200 × 80 = 134 400 J
boiling 0.10 kg: E = 0.10 × 2 260 000 = 226 000 J
total = 360 400 J. For a 2400 W kettle: t = E ÷ P = 360 400 ÷ 2400 = 150 s

Answer: about 3.6 × 10⁵ J, taking about 150 s

Measuring the specific latent heat of fusion of ice

  • Control run: with the heater off, collect the water that drips into the beaker in a set time (e.g. 8 minutes) and find its mass, m₁. This is the ice melted by the warm room.
  • Heater run: empty the beaker, switch the heater on and collect water for exactly the same time. Record the joulemeter readings at the start and end, and the mass collected, m₂.
  • Mass melted by the heater: m = m₂ − m₁. Energy E = difference in joulemeter readings. LF = E ÷ (m₂ − m₁).
  • Instead of a joulemeter you can use a voltmeter, ammeter and stopclock: E = V I t.

Worked example: results

m₁ = 0.018 kg, m₂ = 0.062 kg, joulemeter: 0 J to 14 700 J
m = 0.062 − 0.018 = 0.044 kg; LF = 14 700 ÷ 0.044 = 334 000 J/kg (3 s.f.)

Answer: 3.34 × 10⁵ J/kg

Safety: the heater gets hot – do not touch it, and switch it off before handling. Wear eye protection.

Measuring the specific latent heat of vaporisation of water

Heat the water until it boils. Then record the balance reading and the joulemeter reading, keep the water boiling for a measured time (e.g. 5 minutes), and record both again. The fall in mass is the mass of water boiled away. LV = E ÷ m. (This is usually a teacher demonstration, behind a safety screen, because of the steam and the hot heater.)

Worked example: results

balance: 248.0 g → 240.0 g, so m = 8.0 g = 0.0080 kg. Energy supplied = 19 200 J.
LV = 19 200 ÷ 0.0080 = 2 400 000 J/kg

Answer: 2.4 × 10⁶ J/kg (higher than the accepted 2.26 × 10⁶ J/kg)

Exam tip: Results for L are usually too high: some of the energy from the heater is transferred to the surroundings (or the beaker), so not all of it changes the state of the water. Improve by insulating and using a lid with a small hole.

Common mistakes

  • Using E = m c Δθ for a change of state. While the state changes, Δθ = 0 – use E = m L.
  • Forgetting to convert grams to kilograms and kJ or MJ to joules before using E = m L.
  • In the ice experiment, forgetting to subtract the mass melted in the control run.
  • Saying no energy is transferred when a vapour condenses. Latent heat is released to the surroundings.

Revision questions

Try these, then check your answers below each one.

Question 1. What is the specific latent heat of fusion of a substance? (Choose from: The energy needed to melt 1 kg of it with no change in temperature, The energy needed to raise the temperature of 1 kg of it by 1 °C, The energy needed to boil 1 kg of it with no change in temperature, The temperature at which 1 kg of it melts.)

Answer: The energy needed to melt 1 kg of it with no change in temperature

Working: Specific latent heat of fusion: the energy needed to change 1 kg of a substance from solid to liquid at its melting point, without changing its temperature.

Question 2. Calculate the energy needed to boil away 0.40 kg of water at 100 °C. Specific latent heat of vaporisation of water = 2.26 × 10⁶ J/kg.

Answer: 904000 J

Working: E = 0.40 × 2 260 000 = 904 000 J (9.04 × 10⁵ J).

Question 3. 50 100 J of energy melts some ice at 0 °C. Specific latent heat of fusion of ice = 334 000 J/kg. What mass of ice melts?

Answer: 0.15 kg

Working: m = E ÷ L = 50 100 ÷ 334 000 = 0.15 kg.

Related lessons

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