Six original questions to test recall, calculations, graph analysis and evaluation of the required practical. Attempt each question before revealing the answer and marking guidance.
Question 1. Define specific heat capacity. (2 marks)
Reveal answer and marking guidance
Mark value: 2 marks.
AO focus: AO1 – knowledge of a scientific definition.
Bullet-point answer points:
- Energy needed to increase temperature by 1°C (1).
- Of 1 kg of a substance (1).
Full model answer: Specific heat capacity is the energy needed to raise the temperature of 1 kg of a substance by 1°C.
Common pitfalls and how to avoid them: Do not confuse it with the energy needed to change state, or forget “1 kg”.
A 0.20 kg sample of water has a specific heat capacity of 4,200 J/kg°C and warms by 10°C.
Question 2. Calculate the increase in thermal energy stored by the water. (3 marks)
Reveal answer and marking guidance
Mark value: 3 marks.
AO focus: AO2 – applying the energy equation.
Bullet-point answer points:
- Uses ΔE = m × c × Δθ (1).
- Correct substitution: 0.20 × 4,200 × 10 (1).
- Correct answer: 8,400 J (1).
Full model answer: ΔE = m × c × Δθ = 0.20 × 4,200 × 10 = 8,400 J.
Common pitfalls and how to avoid them: Do not substitute the final temperature in place of Δθ or omit J.
Two metal blocks have equal masses and receive equal amounts of energy. A graph plots temperature rise on the y-axis against energy supplied on the x-axis. The line for metal A is steeper than the line for metal B.
Question 3. Explain which metal has the smaller specific heat capacity. (4 marks)
Reveal answer and marking guidance
Mark value: 4 marks.
AO focus: AO2/AO3 – applying concepts to graphical evidence.
Bullet-point answer points:
- Identifies metal A (1).
- A has the greater temperature rise per joule (1).
- Therefore A needs less energy per kg per degree Celsius (1).
- Links to the relation gradient = 1 ÷ (m × c), with equal mass (1).
Full model answer: Metal A has the smaller specific heat capacity. Its steeper gradient means that each joule causes a larger temperature rise. For equal masses, needing less energy for the same rise means smaller c, because gradient = 1 ÷ (m × c).
Common pitfalls and how to avoid them: A steeper temperature-rise-versus-energy line does not mean larger c. Check which quantity is on each axis.
An electric heater operates at 12 V and 2.5 A for 120 s. It warms a 0.50 kg metal block by 7.4°C.
Question 4. Calculate the experimental specific heat capacity of the block. Give your answer to three significant figures. (4 marks)
Reveal answer and marking guidance
Mark value: 4 marks.
AO focus: AO2 – multi-step calculation.
Bullet-point answer points:
- Calculates power: P = 12 × 2.5 = 30 W (1).
- Calculates energy: ΔE = 30 × 120 = 3,600 J (1).
- Uses c = 3,600 ÷ (0.50 × 7.4) (1).
- Correct result: 973 J/kg°C to 3 significant figures (1).
Full model answer: P = VI = 12 × 2.5 = 30 W. ΔE = Pt = 30 × 120 = 3,600 J. c = ΔE ÷ (mΔθ) = 3,600 ÷ (0.50 × 7.4) = 973 J/kg°C (3 s.f.).
Common pitfalls and how to avoid them: Time must be in seconds; use the full 3,600 J, not 30 W, in the heat capacity equation.
A student is given a metal block with holes, an electric heater, a thermometer, a balance, insulation, an ammeter, a voltmeter and a stopwatch.
Question 5. Describe how the student could determine the specific heat capacity of the metal using this equipment. Include the measurements and how the results are processed. (6 marks)
Reveal answer and marking guidance
Mark value: 6 marks.
AO focus: AO1/AO2 – planning and explaining a required practical.
Bullet-point answer points:
- Measure block mass with a balance in kg (1).
- Insert heater and thermometer, and insulate block (1).
- Connect ammeter in series and voltmeter across the heater (1).
- Record starting temperature, then switch on heater and timer together (1).
- Record temperature and V/I over measured time, calculating Δθ and ΔE = VIt (1).
- Calculate c = ΔE ÷ (m × Δθ), with sensible safety/consistent conditions (1).
Full model answer: Measure the metal block’s mass in kilograms. Place the heater and thermometer in its holes and wrap insulation around the block. Connect the ammeter in series and voltmeter across the heater. Record the starting temperature, then switch on the low-voltage heater and start a stopwatch. Record temperature, voltage and current after a measured time. Calculate the temperature rise and electrical energy supplied using ΔE = VIt. Finally calculate c = ΔE ÷ (m × Δθ). Switch off and let the hot block cool before handling it.
Common pitfalls and how to avoid them: Do not describe only “heat it and measure temperature”. Include how energy is measured or calculated and how mass is used.
An insulated 0.50 kg aluminium block starts at 20.0°C. A heater supplies 7,200 J and its measured temperature becomes 34.1°C. A reference value for aluminium is approximately 900 J/kg°C.
Question 6. Calculate the experimental specific heat capacity, compare it with the reference value and suggest why it differs and how the investigation could be improved. (5 marks)
Reveal answer and marking guidance
Mark value: 5 marks.
AO focus: AO2/AO3 – calculation, interpretation and evaluation.
Bullet-point answer points:
- Temperature rise = 14.1°C (1).
- c = 7,200 ÷ (0.50 × 14.1) ≈ 1,020 J/kg°C (1).
- Identifies measured value as higher than 900 J/kg°C (1).
- Explains that some electrical energy is lost to surroundings/apparatus, giving a smaller Δθ and larger calculated c (1).
- Suggests a linked improvement, such as thicker insulation or better thermal contact (1).
Full model answer: The temperature rise is 34.1 − 20.0 = 14.1°C. c = 7,200 ÷ (0.50 × 14.1) ≈ 1,020 J/kg°C. This is higher than the approximate 900 J/kg°C reference value. Some input energy heated the surroundings or apparatus, so the block’s temperature rose less than if all 7,200 J had entered it. Thicker insulation would reduce this energy loss.
Common pitfalls and how to avoid them: Do not claim heat loss makes the calculated c too low. A smaller measured temperature rise appears in the denominator and makes calculated c larger.