AQA A-Level Physics Paper 2, June 2023: Question 8

📐 Step-by-Step Calculation

1. Convert time to seconds:
   t = 7 hours = 7 × 3600 s = 25,200 s
2. Calculate useful power:
   P_useful = 0.75 × 1000 W = 750 W (or J s⁻¹)
3. Find useful thermal energy supplied (Q):
   Q = P_useful × t
   Q = 750 × 25,200 = 1.89 × 10⁷ J
4. Calculate temperature change (Δθ):
   Δθ = 85 °C - 10 °C = 75 K (or °C)
5. Rearrange Q = mcΔθ to find mass (m):
   m = Q / (c × Δθ)
   m = 1.89 × 10⁷ / (4200 × 75)
   m = 1.89 × 10⁷ / 315,000 = 60 kg

💡 Key Knowledge

• Specific Heat Capacity ( c ): The energy required to raise the temperature of 1 kg of a substance by 1 K without a change of state.
• Efficiency:
  Efficiency = Useful Energy Out / Total Energy In
  Always use the useful power/energy when working out temperature change.
• SI Units: Power in Watts (J s⁻¹) requires time in seconds (s). Kelvin and Celsius changes are numerically identical: Δθ (in °C) = ΔT (in K) .

❌ Common Errors & Distractor Traps

• Forgetting to convert hours to seconds: Using t = 7 or t = 7 × 60 = 420 gives values around 0.017 kg or 1.0 kg (Distractor A).
• Inverting efficiency or dividing instead of multiplying: Using 1000 / 0.75 instead of 1000 × 0.75 gives roughly 106.7 kg ≈ 110 kg (Distractor D).
• Omitting efficiency entirely: Calculating with 100% efficiency gives m = 80 kg .
• Using absolute temperature instead of Δθ: Dividing by (273 + 85) or similar instead of the difference (85 - 10) = 75 .

🧠 Exam Technique for Speed

• Combine into a single algebra expression:
  η × P × t = m × c × Δθ
  m = (η × P × t) / (c × Δθ)
• Substitute everything at once in your calculator:
  m = (0.75 × 1000 × 7 × 3600) / (4200 × 75)
• Spot quick mental cancellations: Notice that 0.75 × 1000 = 750 and Δθ = 75 , so 750 / 75 = 10 . Then (10 × 7 × 3600) / 4200 = 252,000 / 4200 = 60 kg . This saves huge time under exam pressure!