Interactive practical

Specific heat capacity

Weigh aluminium and copper blocks, heat each one in insulation while reading a joulemeter and thermometer every two minutes, and find each specific heat capacity from a graph.

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GCSE Physics

Determining specific heat capacity

WJEC 3420QS — matched to your course.

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Practical focuses1

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Available question marks5

Practical focus

Measure energy transferred, mass and temperature rise to determine a material's specific heat capacity.

Practical method

Aim

Determine the specific heat capacity of a metal block using an electrical heater, c = \(\dfrac{\Delta E}{m\Delta \theta }\).

Variables

Independent variable: energy supplied by the heater (or heating time)
Dependent variable: temperature rise of the block
Control variables: mass and material of the block; insulation around the block; heater power.

Apparatus

1.00 kg metal block with holes for heater and thermometer
12 V immersion heater
Ammeter and voltmeter (or joulemeter)
Thermometer (0.1 °C) or temperature probe
Stopwatch
Insulating jacket
A drop of oil or water in the thermometer hole

Method

Measure the mass of the block.
Wrap the block in insulation and put the heater and thermometer in their holes, with a drop of water or oil for good thermal contact.
Record the starting temperature.
Switch on, record the p.d. and current, and start the stopwatch.
Switch off after 10 minutes and record the highest temperature reached.
Calculate energy supplied E = VIt, the temperature rise Δθ and c = \(\dfrac{E}{m\Delta \theta }\). Alternatively plot temperature against energy every minute; the gradient is \(\dfrac{1}{mc}\).

Sample results

1.00 kg aluminium block, heater at 12.0 V and 4.0 A for 600 s.

Quantity	Value
Energy supplied E = VIt	12.0 × 4.0 × 600 = 28 800 J
Start temperature	21.0 °C
Highest temperature	51.0 °C
Temperature rise Δθ	30.0 °C
c = \(\dfrac{E}{m\Delta \theta }\)	960 J/kg °C

Analysis

c = 28 800 ÷ (1.00 × 30.0) = 960 J/kg °C. The accepted value for aluminium is about 900 J/kg °C. The experimental value is too high because some energy heated the surroundings, the heater and the thermometer rather than the block, so the measured temperature rise was smaller than it should have been.

Errors and improvements

Problem	Effect	Improvement
Energy lost to the surroundings	Temperature rise too small, c too large	Use better insulation and a lid
Poor thermal contact with the thermometer	Temperature readings lag behind	Add a drop of water or oil to the thermometer hole
Block keeps warming after switch-off	Final temperature recorded too early	Record the highest temperature reached after switching off

Risk assessment

Hazard	Risk	Control measure
Hot heater and block	Burns	Do not touch until cooled; use a heat-proof mat
Water near electrical connections	Electric shock or short circuit	Use a low-voltage supply and keep water away from connections

Common mistakes

Using the final temperature instead of the temperature rise.
Using minutes instead of seconds in E = VIt.
Saying energy is 'lost' without saying where it goes.

Worked analysis

A 2.0 kg block receives 36 000 J and warms from 20.0 °C to 40.0 °C. c = \(\dfrac{E}{m\Delta \theta }\) = 36 000 ÷ (2.0 × 20.0) = 900 J/kg °C.

Common misconception

A higher specific heat capacity does not mean a material gets hotter. It means more energy is needed to raise the temperature of 1 kg by 1 °C.

Assessment coverage

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