AQA GCSE Physics Physics Paper 1 (Higher), November 2021: Question 11

✅ Model Answer (2/2 Marks)

• Mark 1: The gradient of the line for ice is steeper than the gradient for water (liquid). (or: the temperature of the ice increases faster than the temperature of the water)
• Mark 2: This means less thermal energy is required to raise the temperature of ice by a given amount.

💡 Key Knowledge

On a heating curve supplied at a constant rate:

• Gradient = Rate of temperature rise.
• A lower specific heat capacity ( c ) means a material needs less energy per 1 °C rise.
• Therefore, materials with lower heat capacity heat up faster (steeper slope).

🧠 Exam Technique

Always structure "Explain how the graph shows this" into two linked steps:

1. State the graph feature (e.g., steeper gradient / temperature rises faster).
2. State the physical meaning (e.g., needs less energy to increase temperature).

❌ Common Errors

• Confusing the sloping regions with the flat regions (the flat regions show latent heat, not specific heat capacity).
• Only stating that the gradient is steeper without linking it to the energy needed.

✅ Model Answer (2/2 Marks)

• Mark 1: The water took more time to vaporise/boil than the ice took to melt. (or: the horizontal line at 100 °C is longer than at 0 °C)
• Mark 2: Which means less energy was needed to change state from solid to liquid than from liquid to gas.

💡 Key Knowledge

• Horizontal lines (plateaus): Temperature remains constant during state changes because energy is used to break intermolecular bonds.
• Since power is constant ( E = P × t ), time = energy transferred.
• Melting took ~75 s, while boiling took ~550 s. Longer time = more energy needed.

❌ Common Errors

• Saying "boiling happens at a higher temperature" – temperature relates to kinetic energy, not latent heat quantity!
• Failing to mention the time duration or the length of the horizontal sections.

Examiner Tip: Check the horizontal axis carefully. Fusion took from ~25 s to 100 s (75 s), whereas vaporisation took from 200 s to 750 s (550 s). This visual difference is key.

✅ Model Answer (Any 2 for 2 Marks)

• Gradients would be less steep / It would take more time to increase in temperature.
• Horizontal lines would be longer / It would take more time to change state (melt and boil).
• The change in temperature with time would become non-linear (curved rather than straight).

🧠 Exam Technique

When energy is "lost" to the surroundings, less useful energy enters the water per second. Think about how that alters both parts of the curve:

• Heating stages: Rate of heating slows down → shallower slope.
• State changes: Takes longer to supply the needed bond-breaking energy → longer plateau.

Step 1: Convert units to standard SI units [Mark 1]
Energy E = 69 kJ = 69 × 1000 = 69 000 J

Step 2: State the formula and substitute [Mark 2]
E = m × L
69 000 = 0.030 × L

Step 3: Rearrange for L [Mark 3]
L = 69 000 / 0.030

Step 4: Calculate numerical answer [Mark 4]
L = 2 300 000 (or in standard form: 2.3 × 10⁶ )

Step 5: Provide the correct unit [Mark 5]
Unit: J/kg (Note: if energy was kept as 69 kJ, the value is 2300 and the unit must be kJ/kg)

❌ Common Calculation Traps

• Forgetting to convert kJ to J: Calculating 69 / 0.030 = 2300 and giving the unit as J/kg loses marks. Either write 2 300 000 J/kg or 2300 kJ/kg .
• Incorrect units: Confusing specific latent heat ( J/kg ) with specific heat capacity ( J/kg °C ).

Mark Allocation Summary:
• 1 mark: Energy converted to 69 000 J
• 1 mark: Correct substitution into equation
• 1 mark: Correct rearrangement
• 1 mark: Correct numerical value (2 300 000 or 2.3 × 10⁶)
• 1 mark: Correct unit consistent with answer (J/kg)