AQA GCSE Physics Physics Paper 1 (Foundation), June 2025: Question 9
10 marks · Standard Demand difficulty · Extended Answer
Describe an experimental method to determine the specific heat capacity of water, identify hazards and improvements, and calculate percentage difference.
Practise this questionQuestion
Mark scheme
Show the mark scheme
How to answer it
Required Practical 1: Specific Heat Capacity of Water
This question assesses knowledge of AQA GCSE Physics Required Practical 1 (Specific Heat Capacity) and essential practical evaluation skills:
- Identifying lab hazards and risk factors during thermal experiments.
- Writing a clear, logical, step-by-step experimental method to determine specific heat capacity (SHC) using a joulemeter.
- Suggesting practical apparatus modifications to minimize thermal energy loss to surroundings.
- Calculating percentage difference between an experimental value and an accepted actual value.
Lab Hazard Identification
1 Mark • AO3
✅ Acceptable Answers (Choose ONE)
- Hot water or hot beaker
- Heater (it gets very hot)
- Spilt water (slip hazard / electric shock risk)
- Broken glass / beaker
❌ Common Errors & Examiner Tips
- Hazard vs Risk: A hazard is the thing causing harm (e.g. the hot heater). A risk is the action or consequence (e.g. burning your skin). The mark scheme states: ignore descriptions of risk. Simply naming "burns" will not gain credit.
Extended Response: Method for Determining Specific Heat Capacity
6 Marks • AO1 (Levels of Response)
💡 Core Governing Formula
The entire method must provide data to satisfy the equation:
ΔE = m × c × Δθ ➔ c = ΔE / (m × Δθ)
Where: ΔE = energy supplied (J), m = mass of water (kg), c = specific heat capacity (J/kg °C), Δθ = temperature increase (°C).
✅ Full-Mark Model Answer (Level 3, 6 Marks)
- Measure Mass: Place the empty beaker on the digital balance and record its mass. Add water, record the total mass, and subtract the empty beaker's mass to calculate the mass of water (or tare/zero the balance first).
- Initial Temperature: Place the thermometer and heater into the water. Measure and record the initial temperature of the water.
- Energy Input: Turn on the power supply to heat the water. Measure the energy transferred (ΔE) directly using the joulemeter.
- Final Temperature: Switch off the heater and record the highest final temperature reached by the water to find temperature change (Δθ = final temp - initial temp).
- Calculation: Calculate specific heat capacity using the rearranged equation: c = ΔE / (m × Δθ) .
🧠 Level 3 Examiner Criterion
Crucial Requirement: The mark scheme explicitly states:
"A Level 3 answer (5–6 marks) must have a clear method of how the mass of water is determined."
Many students lose marks because they say "weigh the water" without explaining how they account for the beaker. You must state difference in mass or zeroing the balance.
❌ Common Slip-ups
- Using a stopclock: The diagram has a joulemeter, NOT an ammeter and voltmeter. Do not describe calculating energy from E = I × V × t ; just read energy directly from the joulemeter.
- Missing the final step: Forgetting to state how the measurements are used to calculate SHC using E = mcΔθ .
Reducing Systematic Experimental Error
1 Mark • AO3
✅ Acceptable Answers (Choose ONE)
- Insulate the beaker (e.g. wrap with bubble wrap, foam, or cotton wool).
- Add a lid to the beaker (reduces energy loss by convection and evaporation).
- Use an insulating cup (e.g. polystyrene cup or plastic cup instead of glass).
🧠 Why does this matter?
Energy lost to the surroundings means the joulemeter measures more energy than what actually heated the water. This causes the calculated value of c to be too high compared to the true value.
Percentage Difference Calculation
2 Marks • AO2
📐 Step-by-Step Calculation
Given Data:
- Student's experimental value = 4410 J/kg °C
- Actual accepted value = 4200 J/kg °C
Step 1: Calculate the absolute difference [1 mark]
Difference = 4410 − 4200 = 210 J/kg °C
Step 2: Divide by the actual value and multiply by 100 [1 mark]
Percentage difference = (210 / 4200) × 100 = 5%
❌ Common Trap (Dividing by the Wrong Number)
Percentage difference is always calculated relative to the true/actual accepted value:
Incorrect: (210 / 4410) × 100 = 4.76%
The mark scheme explicitly says: do not accept 4.76 (%) .
🧠 Final Answer Box
Write your working clearly so that even if an arithmetic error occurs, you still secure 1 method mark for the difference.
Percentage difference = 5 %
Topics
Physics · Required Practicals · P1: Energy · P3: Particle Model of Matter · Required Practicals
Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Foundation), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.