Measuring Specific Heat Capacity

Start with the energy model

You already know that Q = mcΔθ. An electric heater supplies energy at a measured rate, but some energy may warm the apparatus or reach the surroundings. The experiment tests how well we can estimate the part that warms the block.

If the heater’s power is P, its energy input in time t is E = Pt. If electrical measurements are used, P = VI, where V is the potential difference across the heater and I is the current through it. A watt is one joule per second.

Under the ideal assumption that all input energy raises the block’s temperature,

Use this model only when the block does not change state and its specific heat capacity is approximately constant over the temperature interval.

Collect the measurements

1. Measure the block’s mass m with a balance, in kilograms.
2. Insert the heater and temperature probe into their holes. Follow the apparatus instructions for improving thermal contact; trapped air slows energy transfer.
3. Insulate the block. Keep the probe clear of the heater so that it measures the block, rather than a local hot spot.
4. Record the initial temperature. Connect the ammeter in series and the voltmeter across the heater, as shown.
5. Switch on the heater and start the timer together. Record V, I and temperature at regular times; check that the power remains approximately constant.
6. Use the temperature rise and heating time to estimate c. Repeat under comparable conditions to check consistency.

Using a temperature–time graph

A sequence of readings shows the trend more clearly than one start–finish pair. For a nearly straight section, find the gradient from two well-separated points on a suitable best-fit line, rather than from two neighbouring readings.

Finding specific heat capacity from a temperature–time gradient

A straight temperature–time line with six points rises from 20 °C at 0 s to 40 °C at 100 s. Temperature rises by equal amounts over equal time intervals.

Scroll across the figure to read all labels.

View figure data

Values for Finding specific heat capacity from a temperature–time gradient

Time (s)	Illustrative block temperature
0	20
20	24
40	28
60	32
80	36
100	40

Values for Finding specific heat capacity from a temperature–time gradient

Time (s)	Illustrative block temperature
0	20
20	24
40	28
60	32
80	36
100	40

The example uses a 0.50 kg block and a constant 50 W heater. Use the graph and its table to find the gradient, then estimate the specific heat capacity using

Do not automatically choose the earliest interval. Avoid an initial response lag, and inspect whether later readings curve as energy transfer to the surroundings grows. A straight section makes the gradient easier to estimate, but it does not prove that all heater energy reaches the block.

Evaluate the estimate

If some input energy warms the heater or probe, or leaves for the surroundings, the block’s temperature rise is smaller than the ideal model predicts. Treating all Pt as energy absorbed by the block then tends to overestimate c.

• Insulation reduces energy transfer out of the block; it does not eliminate it.
• Good thermal contact and a suitable heating rate reduce temperature differences within the apparatus.
• A clear temperature rise reduces the relative effect of temperature-reading uncertainty. Heating much farther above room temperature can also increase unwanted transfer, so a larger rise is not automatically better.
• Repeated readings and runs reveal scatter and inconsistent results. Repetition alone does not remove a systematic energy error.

The error direction depends on the energy path. If an initially colder block also receives energy from its warmer surroundings, assuming heater energy is the only input can instead underestimate c.

Worked calculation

Practise the reasoning

An estimate of c is larger than an accepted value. Explain why “the heater was on for too long” is not enough to diagnose the error.