Interactive practical
Specific heat capacity of a solid
Weigh an aluminium block, heat it electrically for ten minutes while reading the thermometer and meters, then use a graph of energy supplied against temperature rise to find its specific heat capacity.
Open interactive simulationA-Level Physics
Specific heat capacity of a solid
WJEC 1420QS — matched to your course.
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At a glance
Practical focus
- Determine the specific heat capacity of a solid from measured electrical energy and temperature rise.
Practical method
Aim
Determine the specific heat capacity of aluminium from the gradient of a temperature–time graph.
Variables
Independent variable: time of heating
Dependent variable: temperature
Control variables: heater power; mass of the block; insulation.
Apparatus
- 1.00 kg aluminium block with holes for heater and thermometer
- 12 V heater
- Ammeter and voltmeter (or joulemeter)
- Thermometer with a drop of oil
- Insulation
- Stopwatch
Method
- Insulate the block, insert the heater and the thermometer (with a drop of oil for thermal contact).
- Switch on, record V and I, and record the temperature every 120 s for 10 minutes.
- Plot θ against t. The gradient of the straight section is \(\dfrac{\Delta \theta }{\Delta t}\).
- c = \(\dfrac{P}{m \times \text{gradient}}\), where P = VI.
Sample results
m = 1.00 kg, V = 12.0 V, I = 4.17 A, P = 50.0 W.
| t / s | θ / °C |
|---|---|
| 0 | 20.0 |
| 120 | 26.7 |
| 240 | 33.3 |
| 360 | 40.0 |
| 480 | 46.7 |
| 600 | 53.3 |
Analysis
The graph is a straight line with gradient 0.0556 °C s⁻¹. c = \(\dfrac{50.0}{1.00 \times 0.0556}\) = 900 J kg⁻¹ K⁻¹. In practice the curve starts slowly (the heater warming up) and flattens at higher temperatures as losses grow; use the straight section.
Uncertainty
Thermometer ±0.5 °C over a 33 °C rise is 1.5%; meters add about 1% each. Heat losses make the measured c too high, a systematic error.
Errors and improvements
| Problem | Effect | Improvement |
|---|---|---|
| Heat lost to the surroundings | c too high | Insulate; start below room temperature and finish the same amount above |
| Thermometer lag at the start | Initial curve | Use the gradient of the straight section |
| Poor thermal contact | Readings lag | Oil in the thermometer hole |
Risk assessment
| Hazard | Risk | Control measure |
|---|---|---|
| Hot heater and block | Burns | Allow to cool; do not touch the heater |
| Electrical supply | Shock | Low-voltage heater |
Common mistakes
- Using the first and last temperature instead of the gradient.
- Forgetting P = VI.
- Using total time including the warm-up lag.
Worked analysis
Gradient = \(\dfrac{53.3 - 20.0}{600}\) = 0.0556 °C s⁻¹. c = \(\dfrac{P}{m \times \text{gradient}}\) = \(\dfrac{50.0}{1.00 \times 0.0556}\) = 900 J kg⁻¹ K⁻¹.
Common misconception
Specific heat capacity is per kilogram per kelvin; a temperature change of 1 °C equals a change of 1 K.
Assessment coverage
Practice this practical
2 questions available, worth 4 marks in total.
Retrieval cards
Specific heat capacity?
Energy to raise 1 kg by 1 K.
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How this practical fits your course
The label below uses the wording from your course specification.
- WJEC 1420QS — Specified practical work: Specified practical work – Measurement of the specific heat capacity for a solid. The specification requires determination of the specific heat capacity of a solid. Checked 18 September 2026. View official course source