A-Level Physics May/June 2025 Q3(b)(v): Use the first law of thermodynamics to determine, to three significant figures, a value…

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  The first law of thermodynamics states that the increase in internal energy $\Delta U$ of a system is equal to the thermal energy supplied to the system $Q$ plus the work done on the system $W$.

  $\Delta U = Q + W$

  Reasoning: The question provides the increase in internal energy, $\Delta U = 4.38 \times 10^6$ J, and the work done by the system during expansion, which is $11.1$ J. Therefore, the work done on the system is $W = -11.1$ J.

  To find the total thermal energy $Q$ supplied to the aluminium, we rearrange the first law:

  $Q = \Delta U - W$ $Q = (4.38 \times 10^6) - (-11.1)$ $Q = 4.38 \times 10^6 + 11.1 \text{ J}$

  The work done (11.1 J) is negligible compared to the large increase in internal energy. Therefore, we can approximate that the thermal energy supplied is used almost entirely to increase the internal energy:

  $Q \approx \Delta U = 4.38 \times 10^6 \text{ J}$

  This thermal energy supplied causes a temperature rise, given by the formula for specific heat capacity, $c$:

  $Q = mc\Delta\theta$

  Rearranging for $c$ and substituting the values:

  $c = \frac{Q}{m\Delta\theta}$ $c = \frac{4.38 \times 10^6 \text{ J}}{(9.75 \text{ kg}) \times (500 \text{ °C})}$ $c = 898.4615... \text{ J kg}^{-1} \text{°C}^{-1}$

  Rounding to three significant figures:

  $c = 898 \text{ J kg}^{-1} \text{°C}^{-1}$

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  How the marks are awarded

  • B1 — Correctly applying the first law of thermodynamics to find the total thermal energy supplied, Q. This is shown by stating that Q is the sum of the increase in internal energy and the work done by the system: (4.38 × 10⁶) + 11.1.
  • C1 — Correctly substituting values into the specific heat capacity formula. This involves using the approximated energy (equal to the change in internal energy) and the given mass and temperature change, as in the expression: (4.38 × 10⁶) / (9.75 × 500).
  • A1 — Calculating the final correct answer of 898, ensuring it is rounded to three significant figures and stated with the correct units, J kg⁻¹ °C⁻¹.

  Common mistakes

  • Making a sign error when applying the first law of thermodynamics, for example subtracting the work done (Q = 4.38 × 10⁶ - 11.1) instead of adding it.
  • Incorrectly using the value for work done (11.1 J) as the thermal energy Q in the specific heat capacity equation, leading to a very small and incorrect answer.
  • Failing to round the final answer to the required three significant figures, for example writing 898.5 or 900.
  • Omitting the units for specific heat capacity or providing incorrect units, such as J kg °C or J g⁻¹ °C⁻¹.

  Examiner tip: Pay close attention to the relative magnitudes of quantities to identify where valid simplifying approximations can be made, as the work done here is negligible compared to the change in internal energy.

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