Allowing for heat loss in a calorimeter

Part (a) marking points

Final answer: 4.74 × 10³ J kg⁻¹ K⁻¹. Numerical checks assess the final value only; they do not check working or the number of digits typed.

1. Electrical energy is VIt = 7200 J. (1 method mark)
2. Use VIt = (mc + 80.0)ΔT. (1 method mark)
3. Obtain c = 4.74 × 10³ J kg⁻¹ K⁻¹. (1 accuracy mark)

No earlier numerical result is required. Credit a valid alternative method. Accuracy credit requires a physically valid method.

Part (b) marking points

Final answer: 3.00 × 10³ J kg⁻¹ K⁻¹. Numerical checks assess the final value only; they do not check working or the number of digits typed.

1. Let total heat capacity be H = mc + 80.0; estimated loss power is H × 0.0120. (1 method mark)
2. Use 7200 = H × 7.0 + H × 0.0120 × 300. (1 method mark)
3. Solve H = 679 J K⁻¹ and c = 3.00 × 10³ J kg⁻¹ K⁻¹. (1 accuracy mark)

No earlier numerical result is required. Credit a valid alternative method. Accuracy credit requires a physically valid method.

Part (c) marking points

1. Heat loss usually increases as the water becomes warmer relative to the room. (1 independent mark)
2. Using the final loss rate throughout overestimates the lost energy and so tends to underestimate c. (1 independent mark)
3. The cooling rate also depends on the heat capacity being determined; treating it as a fixed loss power without H is inconsistent. (1 independent mark)

Credit equivalent correct explanations. Award each distinct point once; a contradiction invalidates the affected point.

Acceptable alternatives: Discuss changes in thermal gradients after switch-off that make the measured cooling rate unrepresentative; support the criticism physically.

Part (d) marking points

Assess the response as a whole. Indicative content:

• Measure water mass, heater voltage/current and temperature at regular times while stirring; avoid the heater touching the vessel.
• Measure or independently calibrate the heat capacity of the vessel and immersed apparatus, and retain its contribution.
• Reduce losses using insulation and a lid; keep the temperature rise modest while large enough to resolve.
• Record a cooling curve over the same temperature range and use loss rates at matched temperatures in an iterative energy balance, or use matched heating/cooling slopes.
• For matched slopes at the same temperature, H = VI/(heating slope + magnitude of cooling slope), assuming equal heat loss at that temperature; then c = (H − 80.0)/m.
• Repeat and assess sensor resolution/lag, voltage/current variation, heat-capacity calibration and unequal cooling/heating conditions. Explain why repetition alone cannot fix a shared bias.

Level 1 (1–2): Relevant measurements or improvements are stated, with limited energy-balance reasoning.

Level 2 (3–4): A workable procedure includes vessel heat capacity and a plausible heat-loss treatment, but analysis or limitations are incomplete.

Level 3 (5–6): A coherent method connects measured electrical energy, total heat capacity and a justified heat-loss correction, with meaningful controls and evaluation of the remaining systematic limitations.

Judge the response as a whole against the levels. The indicative content is not a checklist of six marks. Choose the lower mark in a band for a borderline response and the upper mark for a response that consistently meets the descriptor. No relevant correct physics: 0 marks.