A-Level · Physics · AQA · Mark scheme decoded

AQA A-Level Physics: Internal Energy and Heat Transfer — mark scheme explained

Machine-verifiedchecked against the AQA A-Level Physics specificationlast verified 2 July 2026

The short answer

Understanding internal energy and heat transfer is crucial in the study of thermodynamics, which forms a significant part of A-Level Physics. Internal energy is the sum of the randomly distributed kinetic energies and potential energies of the particles within a system.

The question

A 2 kg block of ice at -10°C is heated until it completely melts. The specific heat capacity of ice is 2100 J/kg°C, and the specific latent heat of fusion for ice is 334,000 J/kg. Calculate the total energy required. [Paraphrased for study — not reproduced from any exam paper.]

Mark scheme, decoded

What each mark is really for — in plain English — and the wording trap that loses it.

  • S1

    Calculate the energy required to raise the temperature of the ice from -10°C to 0°C using Q = mcΔθ.

  • S2

    Q 1 = 2 kg × 2100 J/kg°C × (0°C - (-10°C))

  • S3

    Q 1 = 2 kg × 2100 J/kg°C × 10°C

  • S4

    Q 1 = 42,000 J

  • S5

    Calculate the energy required to melt the ice at 0°C using Q = ml.

  • S6

    Q 2 = 2 kg × 334,000 J/kg

  • S7

    Q 2 = 668,000 J

  • S8

    Add the two energies to find the total energy required.

  • S9

    Total Q = Q 1 + Q 2

  • S10

    Total Q = 42,000 J + 668,000 J

  • S11

    Total Q = 710,000 J

Model answer

Worked through, with each step tagged to the mark it earns.

  1. S1

    Calculate the energy required to raise the temperature of the ice from -10°C to 0°C using Q = mcΔθ.

  2. S2

    Q 1 = 2 kg × 2100 J/kg°C × (0°C - (-10°C))

  3. S3

    Q 1 = 2 kg × 2100 J/kg°C × 10°C

  4. S4

    Q 1 = 42,000 J

  5. S5

    Calculate the energy required to melt the ice at 0°C using Q = ml.

  6. S6

    Q 2 = 2 kg × 334,000 J/kg

  7. S7

    Q 2 = 668,000 J

  8. S8

    Add the two energies to find the total energy required.

  9. S9

    Total Q = Q 1 + Q 2

  10. S10

    Total Q = 42,000 J + 668,000 J

  11. S11

    Total Q = 710,000 J

  12. Final answer: 710,000 J

Common mistakes

  • Confusing specific heat capacity with specific latent heat. — Review the definitions: Specific heat capacity is used for temperature changes (Q = mcΔθ), while specific latent heat is used for phase changes (Q = ml).
  • Forgetting to convert units, especially when dealing with mass and time. — Always check that the units for mass (kg), temperature (°C), and time (s) are consistent. Convert if necessary.
  • Using the wrong formula for continuous flow systems. — Remember to use the correct continuous-flow approach: P = (m/t)cΔθ + H, eliminating heat loss by comparing two different flow rates (P₁ − P₂ = ((m₁ − m₂)/t)cΔθ).
  • Neglecting to account for phase changes in energy calculations. — Always check if the problem involves a phase change and use the appropriate formula (Q = ml) for that part of the calculation.
  • Failing to identify and correct systematic errors in experiments. — Calibrate equipment, use more accurate instruments, and correct for known biases. Always check for consistency in measurements.
  • Incorrectly applying the first law of thermodynamics. — Ensure a clear understanding of the first law: The change in internal energy is equal to the heat added minus the work done by the system. Practice applying this concept in various scenarios.

Where the marks go

  • Full worked solution (all marking points)5 marks

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