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SQA National 5 Physics

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4.2.2 Calculating Specific Latent Heat

4.2.2 Calculating Specific Latent Heat: Use Eh = ml to find the energy for melting or boiling, and combine it with Eh = cmΔT for multi-stage problems.

4.2.2 Calculating Specific Latent Heat

The energy for a change of state depends only on the mass and the latent heat.

Relationship
  • Eh = ml, where Eh is the heat energy in joules, m is mass in kilograms and l is specific latent heat in J/kg.
Rearranged
  • l = Eh ÷ m finds the specific latent heat, and m = Eh ÷ l finds the mass changed.
No temperature term
  • There is no ΔT in this equation because the temperature does not change.
Choosing the value
  • Use the fusion value for melting and freezing and the vaporisation value for boiling and condensing.
Process Latent heat used Value for water (J/kg)
Melting ice Fusion 3.34 × 10⁵
Boiling water Vaporisation 22.6 × 10⁵

Calculations are straightforward once the correct stage is identified.

Examples
Melting ice
  • For 0.50 kg of ice at 0 °C: Eh = 0.50 × 3.34 × 10⁵ = 1.67 × 10⁵ J.
Boiling water
  • For 0.20 kg of water at 100 °C: Eh = 0.20 × 22.6 × 10⁵ = 4.52 × 10⁵ J.
Finding l
  • A 60 W heater running for 5.0 minutes supplies 18 000 J; if this melts 0.054 kg, l is about 3.3 × 10⁵ J/kg.
Heat losses
  • In experiments, some energy is lost, so the measured value of l is usually larger than the true value.

Break the process into separate heating and state-change steps.

1

Identify each stage, for example melting at 0 °C and then warming the liquid.
2

Use Eh = ml for each change of state, with the correct latent heat value.
3

Use Eh = cmΔT for each stage where the temperature changes.
4

Add the energies from all stages to find the total.
5

State the final answer with its unit and a sensible number of significant figures.

Question: Calculate the energy needed to change 0.20 kg of ice at 0 °C into water at 30 °C. (l fusion = 3.34 × 10⁵ J/kg, c water = 4180 J/kg °C.)

  1. Melting: Eh = ml = 0.20 × 3.34 × 10⁵ = 66 800 J.
  2. Warming the water: Eh = cmΔT = 4180 × 0.20 × 30 = 25 080 J.
  3. Total = 66 800 + 25 080 = 91 880 J.

Answer: About 9.2 × 10⁴ J is needed.

4.2.2 – Key Points for the Exam

  • Eh = ml: Heat energy equals mass multiplied by specific latent heat.
  • Latent heat of fusion: Energy to melt or freeze a substance without a temperature change.
  • Latent heat of vaporisation: Energy to boil or condense a substance without a temperature change.
  • Multi-stage: A process with more than one stage, each needing its own calculation.

Common Mistakes

  • Using Eh = cmΔT while a substance is melting or boiling.
  • Choosing the vaporisation value when the question describes melting.
  • Forgetting to add the energy for warming the liquid in a multi-stage question.
  • Leaving the mass in grams.
YOUR STUDY PATH

4.2.2 Calculating Specific Latent Heat

SQA National 5 Physics / 4. Properties of Matter

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