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4.2.5Solids, Liquids and Gases: Part 1

Core Practical: Investigating Specific Heat Capacity

5 min read 5 parts Lesson 9 of 9 Unit quiz

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Section 1 of 5The Apparatus

4.2.5 Core Practical: Investigating Specific Heat Capacity
The Apparatus
Metal block
  • A block of known mass, such as aluminium, with one hole for a heater and one for a thermometer.
Heater
  • A low-voltage electric heater supplies energy, and the energy equals power × time, or voltage × current × time.
Insulation
  • Wrapping the block in cotton wool or a lagging layer reduces the energy lost to the surroundings.
Thermometer
  • A drop of oil in the thermometer hole improves thermal contact with the metal.

Section 2 of 5Method

1
Measure the mass of the block on a balance and record the starting temperature.
2
Insert the heater and thermometer and wrap the block in insulating material.
3
Switch on the heater and the stopwatch together, and record the voltage and current or use a joulemeter.
4
Record the temperature at regular intervals, then switch off and note the highest temperature reached.
5
Calculate the energy supplied and use c = ΔQ ÷ (m × Δθ) to find the specific heat capacity.

Section 3 of 5Analysis and Errors

Energy supplied
  • E = V × I × t, with voltage in volts, current in amperes and time in seconds.
Graph method
  • Plot temperature against energy supplied; the gradient is 1 ÷ (mc), so c = 1 ÷ (gradient × m).
Heat loss
  • Some energy warms the air and apparatus, so the block gains less than the supply gives and the calculated c is too high.
Improvements
  • Better insulation, using a lid, and a larger mass of metal reduce the percentage error from losses.
Safety
  • The block and heater become hot, so use a heatproof mat, allow cooling before touching, and keep electrical connections dry.
MeasurementInstrumentUnit
MassBalancekg
TemperatureThermometer°C
TimeStopwatchs
Voltage and currentVoltmeter and ammeterV and A

Section 4 of 5Worked Example

Question

  • A 1.0 kg aluminium block is heated by a 60 W heater for 5.0 minutes.
  • The temperature rises by 20 °C.
  • Calculate the specific heat capacity of aluminium from these results.

  1. Time = 5.0 × 60 = 300 s.
  2. Energy supplied = power × time = 60 × 300 = 18 000 J.
  3. c = ΔQ ÷ (m × Δθ) = 18 000 ÷ (1.0 × 20).
  4. c = 900 J/kg °C.

Answer 900 J/kg °C.

Section 5 of 5Exam Checklist & Tips

4.2.5 – Key Points for the Exam

  • Specific heat capacity The energy needed to raise 1 kg of a substance by 1 °C.
  • Joulemeter An instrument that measures the energy transferred by an electrical component.
  • Insulation Material that reduces thermal energy loss.
  • Gradient The slope of a graph, found by dividing the change in y by the change in x.
  • Thermal contact Close contact between two objects that lets energy transfer easily.

Common Mistakes

  • Forgetting to convert minutes into seconds when calculating the energy.
  • Using the highest temperature rather than the temperature rise.
  • Stating that heat loss makes the calculated value too low; it makes it too high.
  • Placing the thermometer so that it does not touch the metal block.
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