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Latent Heat & Specific Latent Heat

Key idea: Learn latent heat and use Q=ml for melting and boiling calculations, including the power-time method and common O Level exam pitfalls.

  • SEC G3 Physics 2027
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Learning objectives

  • Compare physical properties of solids, liquids and gases
  • Explain state properties using particle arrangement, motion, forces and separation
  • Infer random molecular motion from a Brownian-motion experiment
  • Relate temperature rise to increased average kinetic energy of particles
  • Explain gas pressure using particle collisions with container walls
  • Explain heating from higher to lower temperature until thermal equilibrium
  • Describe conduction in solids through particle vibration and mobile electrons
  • Describe convection in fluids through density changes and bulk motion
  • Explain that energy transfer by electromagnetic radiation needs no material medium
  • Explain how surface colour, texture, temperature and area affect radiation transfer rate
  • Apply conduction, convection and radiation in everyday systems
  • describe internal energy as an energy store that is made up of the total kinetic energy associated with the random motion of the particles and the total potential energy between the particles in the system
  • Define heat capacity and specific heat capacity
  • Apply energy transferred = mass × specific heat capacity × temperature change
  • describe melting/solidification and boiling/condensation as processes of energy transfer without a change in temperature
  • Explain the difference between boiling and evaporation
  • Define latent heat and specific latent heat
  • Apply energy transferred for a change of state = mass × specific latent heat
  • Explain latent heat using particle behaviour
  • Sketch and interpret a cooling curve

1. Definition

A. Latent heat

The latent heat, L, of a sample is the energy absorbed or released when that sample changes state without a change in temperature. Its unit is the joule (J), so it depends on how much substance changes state.

B. Specific latent heat

Specific latent heat, l, is the energy needed to change the state of 1 kg of a substance without a change in temperature.

The two quantities are linked by L = ml. If Q represents the energy transferred during the change of state, then Q = L = ml for the sample.

2. Key Ideas

  • Energy transfer during a change of state: Q = ml
    • Q = energy transferred, equal to the sample’s latent heat L (J)
    • m = mass (kg)
    • l = specific latent heat (J kg⁻¹)
  • Fusion: solid ↔ liquid, specific latent heat of fusion l_f.
  • Vaporisation: liquid ↔ gas, specific latent heat of vaporisation lᵥ.
  • During melting/boiling, energy transfer increases potential energy between particles (separating them), so temperature stays constant.
  • During freezing/condensation, energy transfer decreases potential energy (particles come closer), so latent heat is released.

3. Detailed Explanations

A. Why is temperature constant during a change of state?

Temperature depends on the average kinetic energy of the particles. During a change of state, energy transfer changes the particles’ arrangement and potential energy as attractive forces are overcome or re-form. The average kinetic energy therefore stays constant, so the temperature does not change.

B. Specific latent heat of fusion (l_f)

The specific latent heat of fusion, l_f, is the energy needed to change 1 kg of a substance from solid → liquid without a temperature change.

Q = ml_f

C. Specific latent heat of vaporisation (lᵥ)

The specific latent heat of vaporisation, lᵥ, is the energy needed to change 1 kg of a substance from liquid → gas without a temperature change.

Q = mlᵥ

D. Particle explanation (what the energy is used for)

  • Melting: energy is transferred to overcome enough attraction for particles to leave fixed positions and move past one another.
  • Boiling: energy is transferred to separate particles into the gas state throughout the liquid.
Heating curve for a pure substanceA temperature-time graph rises through solid, liquid, and gas regions, with constant-temperature plateaus during melting and boiling.TimeTemperaturesolid warmsmelting: solid + liquidliquid warmsboiling: liquid + gasgas warmsenergy transferred to substance →
A plateau identifies energy transfer without temperature change. Its duration cannot determine specific latent heat unless mass, heater power, and energy losses are accounted for.

E. Practical note (how l can be found from data)

If a heater supplies energy E to a substance during a change of state and a mass m changes state, then:

l = E/m

If the heater power is P and it runs for time t, then E = Pt.

In a real experiment, not all of Pt changes the sample’s state: some energy warms the apparatus or escapes to the surroundings. Insulation, a control measurement, or an energy-loss correction improves the estimate.

4. Common Mistakes

  • Using Q = mcΔ θ during a change of state (use Q = ml when temperature is constant).
  • Forgetting to convert grams to kilograms before using Q = ml.
  • Mixing up l_f (fusion) and lᵥ (vaporisation).
  • Assuming all electrical energy Pt reaches the substance without checking the question’s loss assumptions.

5. Exam Tips

  • Always write the formula first: Q = ml (then substitute).
  • State the unit of l: J kg⁻¹.
  • When asked “what happens to particles?”, mention intermolecular forces and potential energy.
  • A flat section on a heating/cooling curve means latent heat (temperature constant).

6. Worked Examples

Modelled example 1

Energy to melt ice

Core

Problem

Ice at 0°C melts to water at 0°C. Its mass is 0.50 kg and l_f = 3.34 × 10⁵ J kg⁻¹. Find the energy transferred.
Study the worked solution

  1. Select the state-change equation

    Method

    Use Q = ml_f.

    Reason

    The substance changes from solid to liquid at constant temperature.

    Working

    Q = (0.50)(3.34 × 10⁵)

  2. Calculate the energy

    Method

    Multiply mass by specific latent heat of fusion.

    Reason

    The mass is already in kilograms, matching J kg⁻¹.

    Working

    Q = 1.67 × 10⁵ J

Guided practice 2

Mass that changes state

About 4 min

Problem

6.68 × 10⁵ J melts ice at 0°C, where l_f = 3.34 × 10⁵ J kg⁻¹. Find the mass melted.

Make mass the subject

Unit: kg

Hints

Hint 1: start from the fusion equation
Use Q = ml_f.
Hint 2: divide by latent heat
m = Q/l_f.
View solution step by step

  1. Rearrange for mass

    Method

    Divide energy by specific latent heat.

    Reason

    Mass multiplies l_f in Q = ml_f.

    Working

    m = Q/l_f

  2. Calculate mass

    Method

    Divide the two standard-form values.

    Reason

    The joule units cancel to leave kilograms.

    Working

    m = (6.68 × 10⁵)/(3.34 × 10⁵) = 2.0 kg

Common misconception 3

Finding lᵥ from data

Find and correct the mistake

Learner response

9.04 × 10⁵ J changes 0.40 kg of water at 100°C into steam at the same temperature. A student multiplies Q by m to find lᵥ. Locate the error and calculate lᵥ.

Use energy per kilogram

Unit: J kg^-1

View solution step by step

  1. Rearrange for specific latent heat

    Method

    Divide energy by mass.

    Reason

    l is the energy required per kilogram, and Q = ml.

    Working

    lᵥ = Q/m

  2. Calculate the vaporisation value

    Method

    Use lᵥ because the change is liquid to gas.

    Reason

    The temperature remains constant at the boiling point.

    Working

    lᵥ = (9.04 × 10⁵)/0.40 = 2.26 × 10⁶ J kg⁻¹

Examiner practice 4

Heater power method

5 marks

Examination question

A 500 W heater melts ice at 0°C for 8.0 min. Assume all heater energy melts ice and use l_f = 3.34 × 10⁵ J kg⁻¹. Find the mass melted. [5 marks]

Convert time, find energy, then mass

View solution step by step

  1. Convert time and calculate energy

    2 marks

    Method

    Convert minutes to seconds, then use E = Pt.

    Reason

    Watts are joules per second.

    Working

    t = 8.0 × 60 = 480 s, E = 500(480) = 2.40 × 10⁵ J

  2. Calculate mass melted

    3 marks

    Method

    Divide supplied energy by l_f.

    Reason

    The stated assumption allows all Pt to equal the fusion energy ml_f.

    Working

    m = (2.40 × 10⁵)/(3.34 × 10⁵) ≈ 0.72 kg

Challenge 5

Time taken to boil (power method)

Minimal support

Changed-state transfer

A 1.0 kW heater changes 0.30 kg of water at 100°C into steam at the same temperature. Assuming no unwanted energy transfer and using lᵥ = 2.26 × 10⁶ J kg⁻¹, find the time.

Link vaporisation energy to heater time

Hints

Hint 1: find phase-change energy
Use Q = mlᵥ.
Hint 2: match power units
Convert 1.0 kW to 1000 W, then use t = Q/P.
View solution step by step

  1. Calculate vaporisation energy

    Method

    Multiply mass by lᵥ.

    Reason

    The change is liquid to gas at constant temperature.

    Working

    Q = (0.30)(2.26 × 10⁶) = 6.78 × 10⁵ J

  2. Calculate heater time

    Method

    Convert power and divide energy by power.

    Reason

    P = Q/t, so t = Q/P.

    Working

    t = (6.78 × 10⁵)/1000 = 678 s ≈ 11.3 min

7. Mind Stretchers

Mind stretcher 1: Heating vs boilingExtension

Which needs more energy for 1.0 kg of water: heating it from 0°C to 100°C, or boiling it at 100°C? Use c = 4200 J kg⁻¹°C⁻¹ and lᵥ = 2.26 × 10⁶ J kg⁻¹.

Show Answer

Heating: Q = mcΔ θ = (1.0)(4200)(100) = 4.2 × 10⁵ J.
Boiling: Q = mlᵥ = (1.0)(2.26 × 10⁶) = 2.26 × 10⁶ J.
Boiling needs much more energy.

Mind stretcher 2: Interpreting a plateauExtension

On a heating curve, the temperature stays constant while energy is supplied. What two statements must be true about the substance during this time?

Show Answer
  • The substance is undergoing a change of state (two states present, e.g. liquid + gas).
  • The supplied energy is going into latent heat (changing potential energy between particles), not increasing temperature.

8. Practice and next step

Practice specific latent heat

Match each plateau to Q = ml in the Thermal Physics Explorer, then practise:

Thermal Physics Quiz  Structured Thermal Practice

Continue to cooling curves, where plateaux provide graph evidence for constant-temperature changes of state.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027