AQA GCSE Physics Physics Paper 1 (Foundation), June 2023: Question 9

7 marks · Standard Demand difficulty · Short Answer

Investigate the specific heat capacity of an iron block using experimental data and a temperature-time graph, and evaluate the effect of insulation.

Practise this question

Question

The question shows the apparatus for Required Practical 1: a power supply connected to a joulemeter, which connects to an immersion heater inside an iron block alongside a thermometer, and a stopclock showing 0:00. Part 09.1 asks why the thermometer was left in the iron block for a few minutes before reading the initial temperature. Part 09.2 presents Figure 13, a line graph plotting Temperature in degrees Celsius from 0 to 80 against Time in minutes from 0 to 15. The temperature starts at 20 degrees Celsius, remains constant until about 3 minutes, then rises steadily, reaching 28 degrees Celsius at 5 minutes and 54 degrees Celsius at 10 minutes, and 80 degrees Celsius at 15 minutes. Candidates must calculate specific heat capacity using 26,000 Joules of energy transferred between 5 and 10 minutes and a mass of 2.0 kilograms. Part 09.3 asks students to select two correct effects of wrapping insulation around the block.
Question text

09 Figure 12 shows the equipment a student used to determine the

specific heat capacity of iron.

The iron block the student used has two holes, one for the heater and one for

the thermometer.

Figure 12

09.1 Before the power supply was switched on, the thermometer was used to measure the

temperature of the iron block.

The student left the thermometer in the iron block for a few minutes before recording

the initial temperature.

Suggest why.

[1 mark]

09.2 Figure 13 shows how the temperature changed after the power supply was

switched on.

Figure 13

The energy transferred to the iron block between 5 and 10 minutes was 26 000 J.

The mass of the iron block was 2.0 kg.

Calculate the specific heat capacity of iron.

Use information from Figure 13 and the Physics Equations Sheet.

[4 marks]

Specific heat capacity = J/kg °C

09.3 The student repeated the investigation but wrapped insulation around the iron block.

*35* What effect will adding insulation have had on the investigation?

[2 marks]

Tick ( ) two boxes.

The calculated specific heat capacity will be more accurate.

The iron block will transfer thermal energy to the surroundings at a lower rate.

The power output of the heater will be lower than expected.

The temperature of the iron block will increase more slowly than expected.

The uncertainty in the temperature measurement will be greater.

Mark scheme

Show the mark scheme The mark scheme shows: for 09.1, 1 mark for 'so the thermometer temperature was the same as the temperature of the iron block'. For 09.2, 4 marks: temperature change = 54 - 28 = 26 degrees Celsius (1 mark); substitution into 26000 = 2.0 * c * 26 (1 mark); rearrangement c = 26000 / (2.0 * 26) (1 mark); final answer c = 500 J/kg degrees Celsius (1 mark). For 09.3, 2 marks for selecting 'The calculated specific heat capacity will be more accurate' and 'The iron block will transfer thermal energy to the surroundings at a lower rate'. Total = 7 marks.

Question 9

AO /

Question Answers Extra information Mark

Spec. Ref.

09.1 so the thermometer temperature 1 AO3

was the same as the 4.1.1.3

temperature of the iron block RPA1

AO /

Spec. Ref.

09.2 Δθ = (54 – 28) = 26 (°C) 1 AO2

4.1.1.3

26 000 = 2.0 × c × 26 allow a correct substitution using 1 RPA1

an incorrect value of Δθ

obtained from the graph

26 000 allow a correct rearrangement 1

c = using an incorrect value of Δθ

2.0 × 26

obtained from the graph

c = 500 (J/kg °C) allow an answer consistent with

their value of Δθ obtained from

the graph

AO /

Spec. Ref.

09.3 the calculated specific heat 1 AO3

capacity will be more accurate 4.1.1.3

RPA1

the iron block will transfer 1

thermal energy to the

surroundings at a lower rate

Total Question 9 7

How to answer it

Determining Specific Heat Capacity of Iron

📋 What This Question Tests

This question assesses practical and mathematical skills from AQA Required Practical 1 (Energy: Specific Heat Capacity):

  • Thermal Equilibrium: Understanding why instruments need time to reach the same temperature as the object measured.
  • Graph Reading & Analysis: Determining a specific temperature change (Δθ) over a stated time interval.
  • Quantitative Physics: Applying and rearranging the equation ΔE = m × c × Δθ to calculate specific heat capacity ( c ).
  • Experimental Evaluation: Identifying the effect of thermal insulation on reducing heat dissipation and improving experimental accuracy.
Part 09.1 • 1 Mark

Thermometer Thermal Equilibrium

Why leave the thermometer in the iron block before turning on the power?

✅ Correct Answer

So the thermometer reaches the same temperature as the iron block (they reach thermal equilibrium).

1 Mark: Explicitly linking the thermometer temperature to the block temperature.

💡 Key Knowledge

Glass and the liquid inside a thermometer take time to conduct heat to or from the metal block. Waiting ensures the initial reading accurately reflects the starting temperature of the iron, not the surrounding room air.

❌ Common Errors

  • Vague statements like "to let it settle" or "to make it accurate" without stating that temperatures must equalize.
  • Claiming it is to let the heater warm up (the power supply has not been turned on yet!).
Part 09.2 • 4 Marks

Calculating Specific Heat Capacity from Graph Data

Using Figure 13 to calculate specific heat capacity (c)

📐 Step-by-Step Calculation

  1. Read temperatures from graph (Figure 13):
    At time = 5 min, Temperature = 28 °C
    At time = 10 min, Temperature = 54 °C
    Temperature change: Δθ = 54 − 28 = 26 °C [1 mark]
  2. State values from the question:
    Energy transferred, ΔE = 26 000 J
    Mass, m = 2.0 kg
  3. Substitute into equation ΔE = m × c × Δθ :
    26 000 = 2.0 × c × 26 [1 mark]
  4. Rearrange for c:
    c = 26 000 / (2.0 × 26)
    c = 26 000 / 52 [1 mark]
  5. Calculate final answer:
    c = 500 J/kg °C [1 mark]
Total: 4 Marks (1 for Δθ from graph, 1 for substitution, 1 for rearrangement, 1 for final value).

🧠 Exam Technique & Graph Skills

  • Check axes carefully: The question specifically asked for energy transferred between 5 and 10 minutes. Do not use the initial temperature at 0 minutes (20 °C) or the final temperature at 15 minutes!
  • Scale check: On the y-axis, 10 small squares = 10 °C, so each 1 small square = 1 °C. Reading 28 °C and 54 °C is precise.
  • Error carried forward (ecf): If you misread the graph, you can still gain 3 out of 4 marks if your substitution, rearrangement, and final calculation are carried out correctly.

❌ Calculation Traps to Avoid

  • Using the wrong time interval: Calculating Δθ = 80 − 20 = 60 °C uses the entire 15 minutes, but the 26 000 J given was strictly for the 5 to 10 minute window.
  • Converting minutes to seconds: Because the question already provided the energy in Joules ( 26 000 J ), time does NOT need to be converted to seconds. You do not use P = E / t here.
Part 09.3 • 2 Marks

Effect of Adding Thermal Insulation

Identify the two correct statements

✅ Correct Options (Tick 2 Boxes)

  • ☑ The calculated specific heat capacity will be more accurate. [1 mark]
  • ☑ The iron block will transfer thermal energy to the surroundings at a lower rate. [1 mark]
2 Marks: 1 mark for each correct box ticked.

💡 Why These Are Correct

Insulation reduces unwanted heat loss via conduction and convection to the surrounding air. When less thermal energy escapes:

  • Almost all electrical energy measured by the joulemeter stays inside the block, meaning temperature rises faster per joule supplied.
  • Because c = ΔE / (m × Δθ) , when energy lost is minimized, the calculated value is closer to the true value (more accurate).

❌ Incorrect Distractors Explained

  • "The power output of the heater will be lower than expected" – The heater's power output depends solely on the voltage and current supplied, not external insulation.
  • "The temperature of the iron block will increase more slowly than expected" – It will actually increase faster because less heat is lost to the room.
  • "The uncertainty in the temperature measurement will be greater" – Insulation does not change the resolution or precision of the thermometer.

Topics

Physics · Required Practicals · P1: Energy · Required Practicals

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Foundation), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.