AQA GCSE Physics Physics Paper 1 (Higher), June 2025: Question 6

📐 Step-by-Step Calculation

1. Select the equation:
   ΔE = m × c × Δθ
2. Substitute known values:
   8000 = 0.20 × 1600 × Δθ [1 mark]
3. Rearrange to make Δθ the subject:
   Δθ = 8000 / (0.20 × 1600)
   Δθ = 8000 / 320 [1 mark]
4. Calculate the final value:
   Δθ = 25 °C [1 mark]

✅ Mark Scheme Breakdown

• Mark 1 (AO2): Correct substitution into specific heat formula:
  8000 = 0.20 × 1600 × Δθ
• Mark 2 (AO2): Correct rearrangement:
  Δθ = 8000 / (0.20 × 1600)
• Mark 3 (AO2): Correct answer:
  25 (°C)

🧠 Exam Technique

Always substitute the values directly into the formula before rearranging. This guarantees Mark 1 even if you make an algebra error when rearranging later!

❌ Common Errors

• Dividing 8000 / 0.20 and then multiplying by 1600 due to missing brackets on the calculator. Put brackets around the denominator: 8000 / (0.20 × 1600) .
• Forgetting that mass must be in kilograms (here it was already 0.20 kg, so no conversion needed).

📐 Step-by-Step Calculation

1. Convert energy to standard units (J):
   180 kJ = 180 × 1000 = 180 000 J [1 mark]
2. Substitute into the energy equation:
   E = Q × V
   180 000 = Q × 5.0 [1 mark]
3. Rearrange to solve for Q:
   Q = 180 000 / 5.0 [1 mark]
4. Calculate the final answer:
   Q = 36 000 C [1 mark]

✅ Mark Scheme Breakdown

• Mark 1 (AO2): Unit conversion of 180 kJ = 180 000 J
• Mark 2 (AO2): Correct substitution:
  180 000 = Q × 5.0
• Mark 3 (AO2): Correct rearrangement:
  Q = 180 000 / 5.0
• Mark 4 (AO2): Final answer:
  36 000 (C)

💡 Key Knowledge: The Prefix "kilo-"

Always watch out for prefixes on the physics exam:

• 1 kJ = 1 000 J (multiply by 10³)
• The formula sheet lists: energy transferred = charge flow × potential difference (E = Q V)

❌ Common Errors

• Leaving energy as 180, leading to an answer of 36 C (losing the 1st mark).
• Multiplying voltage by energy instead of dividing ( 180 000 × 5 ).

✅ Model Answer (3 Marks)

• Mark 1: Closing both switches puts the components in parallel, which decreases the total resistance of the circuit.
• Mark 2: Because total resistance decreases, the total current increases. (Dependent on Mark 1)
• Mark 3: Since the potential difference (V) remains constant and P = I × V , the total power output increases.

💡 Key Knowledge: Adding Resistors in Parallel

• Adding a branch in parallel provides an additional pathway for current to flow.
• Total resistance of the circuit decreases whenever an additional parallel pathway is opened.
• Because I = V / R , a smaller total resistance draws a larger total current from the supply.

🧠 Exam Technique: Chain of Reasoning

To get all 3 marks on circuit explanation questions, build a clear cause-and-effect chain:

Resistance ↓  →  Current ↑  →  Power Output ↑

Explicitly mention the formula P = I V and state that the potential difference across the battery remains unchanged (5.0 V).

❌ Common Errors & Examiner Warnings

• Confusing series with parallel: Stating that adding another component increases total resistance. (That only applies to series circuits!).
• Formula confusion: Quoting P = I²R without explaining both changing factors. Because both I and R change, using P = IV (where V is fixed at 5.0 V) is much clearer and safer.
• Failing to mention current before concluding about power (Mark 2 depends on Mark 1).