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Specific Latent Heat

Key idea: Define and use specific latent heat, interpret constant-temperature phase changes, and solve multi-stage thermal energy balances.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Define and use heat capacity and specific heat capacity in energy balances.
  • Define and use specific latent heat in phase-change energy balances.

1. Definitions (Must Know)

Specific latent heat, l (J kg⁻¹), is the energy required per unit mass to change the state of a substance without changing its temperature: l = Q/m or Q = ml

  • l_f: specific latent heat of fusion, for solid–liquid changes.
  • lᵥ: specific latent heat of vaporisation, for liquid–gas changes.

2. Key Ideas (What Earns Marks)

  • During a phase change of a pure substance at fixed pressure, energy changes microscopic potential energy while the temperature remains constant.
  • Melting and vaporisation require energy transfer to the substance; freezing and condensation transfer energy from it.
  • Use Q = ml only for the phase-change stage.
  • Use Q = mcΔ T before or after the phase change when temperature changes.
  • In a multi-stage process, calculate each energy term separately and then add them.
Heating curve and phase-change energyA 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 →
On a schematic constant-power heating curve, sloping sections represent temperature change and plateaux represent phase change. The time width of a section reflects its energy requirement only when input power and losses are controlled.

3. Detailed Explanations

A. Why temperature stays constant

Temperature is proportional to mean microscopic kinetic energy. During melting or boiling, the transferred energy changes the arrangement and separation of particles, increasing microscopic potential energy rather than mean kinetic energy. The temperature therefore stays constant until the phase change is complete.

B. Energy balance with ice or steam

For ice initially below 0°C that becomes water above 0°C, write separate terms: Q = m c_ice(0-Tᵢ) + ml_f + m c_water(T_f-0)

Do not combine these into one mcΔ T expression because the state and relevant property change.

Check whether all of the sample changes state

An energy balance may end with a mixture of phases. Compare the available energy with ml before assuming all the ice melts or all the vapour condenses.

4. Common Mistakes

  • Saying energy is used to increase kinetic energy during a constant-temperature phase change.
  • Using grams with l in J kg⁻¹.
  • Omitting the warming or cooling that occurs before and after the phase change.
  • Assuming every plateau on a measured curve is perfectly horizontal despite heat loss or changing power.

5. Exam Tips

  • Sketch the sequence of states and label the temperature at each boundary.
  • Write one energy term per stage.
  • State whether each part gains or loses energy before forming the balance.
  • Check units: kg × J kg⁻¹ = J.

6. Worked Examples

Modelled example 1

Melting ice at its melting point

Core

Problem

Find the energy needed to melt 35 g of ice at 0°C if l_f = 3.34 × 10⁵ J kg⁻¹.
Study the worked solution

  1. Identify the single stage

    Method

    The ice is already at its melting point, so only the phase-change term is needed.

    Reason

    There is no temperature change before melting.

    Working

    Q = ml_f

  2. Convert the mass

    Method

    35 g = 0.035 kg.

    Reason

    The latent heat is stated per kilogram.

    Working

    m = 35 × 10⁻³ kg

  3. Calculate

    Method

    The required energy is 1.17 × 10⁴ J.

    Reason

    Energy changes microscopic potential energy while temperature remains at the melting point.

    Working

    Q = (0.035)(3.34 × 10⁵) = 1.17 × 10⁴ J

Common misconception 2

Warm ice, melt it, then warm the water

Find and correct the mistake

Learner claim

0.050 kg of ice at -10°C becomes water at 20°C. A learner uses only Q = mc_water(30). Diagnose the model and find the energy using c_ice = 2.1 × 10³ J kg⁻¹ K⁻¹, l_f = 3.34 × 10⁵ J kg⁻¹ and c_water = 4.2 × 10³ J kg⁻¹ K⁻¹.

Try this before viewing the solution

Unit: J

View solution step by step

  1. Separate the states

    Method

    Use one term to warm ice to 0°C, one to melt it, and one to warm the resulting water.

    Reason

    The material property and energy mechanism change at each boundary.

    Working

    Q = mc_ice(10) + ml_f + mc_water(20)

  2. Evaluate the stages

    Method

    The three contributions are 1050 J, 16700 J and 4200 J.

    Reason

    The latent stage is calculated without a temperature-change factor.

    Working

    Q = (0.050)(2100)(10) + (0.050)(3.34 × 10⁵) + (0.050)(4200)(20)

  3. Add the transfers

    Method

    The total is 2.20 × 10⁴ J to three significant figures.

    Reason

    Every stage requires energy supplied to the sample.

    Working

    Q = 1050 + 16700 + 4200 = 2.195 × 10⁴ J ≈ 2.20 × 10⁴ J

7. Mind Stretchers

Mind stretcher 1: Test for incomplete meltingExtension

5.0 kJ is supplied to 0.020 kg of ice at 0°C. Does all the ice melt? Use l_f = 3.34 × 10⁵ J kg⁻¹.

Show Answer

Melting all the ice requires ml_f = (0.020)(3.34 × 10⁵) = 6.68 kJ Only 5.0 kJ is available, so some ice remains and the equilibrium mixture stays at 0°C in the ideal model.

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Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027