Measure specific heat capacity with electrical heating

Key idea: Heat a known mass electrically and measure its temperature rise to find c. Compare heating and cooling slopes at the same temperature to correct heat loss, then subtract the apparatus heat capacity.

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GCE A-Level H2 Physics 2027

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The simplest way to measure a specific heat capacity is to heat a block with an electrical heater and see how fast it warms. The catch is that not all the heater’s energy stays in the block: some warms the heater and container, and some leaks away to the room. Comparing how fast the block warms with how fast it cools separates these.

Measuring specific heat capacity electrically

An electrical heater of power P = VI, switched on for a time t, supplies energy Pt. If all of it went into a sample of mass m, warming it by Δθ,

Pt = mcΔθ ⇒ c = Pt/mΔθ

In practice this gives a value of c that is too large, because part of Pt warms the apparatus (heater, container, thermometer) and part is lost to the surroundings. Insulating the sample reduces the losses, but does not remove them.

Correcting for the losses

Write Cₜₒₜ = mc + Cₐₚₚ for the total heat capacity of the sample and the apparatus that warms with it, and H for the power lost to the surroundings. While the heater is on, the temperature rises at a rate sₕₑₐₜ; after it is switched off, it falls at a rate s_cool, which is negative. The energy balances, per second, are

P-H = Cₜₒₜsₕₑₐₜ -H = Cₜₒₜs_cool

The power lost depends on how much hotter the sample is than the room. Use local gradients at the same sample temperature, with the same surroundings and apparatus, so H is the same in both. For a liquid, stir gently to keep its temperature uniform; for a block, ensure good thermal contact with the heater and thermometer. Short intervals approximate a local gradient; a whole heating run need not have one constant loss power. Subtracting the second balance from the first removes it:

Cₜₒₜ = P/(sₕₑₐₜ-s_cool)

Then subtract the heat capacity of the apparatus and divide by the mass of the sample: c = (Cₜₒₜ-Cₐₚₚ)/m.

Heating rises from 40 to 44 degrees Celsius over twenty seconds; cooling falls from 40 to 39 degrees Celsius over the same interval. — Two calculated trials start at 40 °C. Constant-loss approximation over a short interval: heater power 50.0 W, loss rate 10.0 W and total heat capacity 200 J K⁻¹. These are model traces, not measured data.

Figure data: Heating and cooling near the same temperature
Time since trial start, t (s)	Heater on	Heater off
0	40.0	40.00
1	40.2	39.95
2	40.4	39.90
3	40.6	39.85
4	40.8	39.80
5	41.0	39.75
6	41.2	39.70
7	41.4	39.65
8	41.6	39.60
9	41.8	39.55
10	42.0	39.50
11	42.2	39.45
12	42.4	39.40
13	42.6	39.35
14	42.8	39.30
15	43.0	39.25
16	43.2	39.20
17	43.4	39.15
18	43.6	39.10
19	43.8	39.05
20	44.0	39.00

Figure data: Heating and cooling near the same temperature
Time since trial start, t (s)	Heater on	Heater off
0	40.0	40.00
1	40.2	39.95
2	40.4	39.90
3	40.6	39.85
4	40.8	39.80
5	41.0	39.75
6	41.2	39.70
7	41.4	39.65
8	41.6	39.60
9	41.8	39.55
10	42.0	39.50
11	42.2	39.45
12	42.4	39.40
13	42.6	39.35
14	42.8	39.30
15	43.0	39.25
16	43.2	39.20
17	43.4	39.15
18	43.6	39.10
19	43.8	39.05
20	44.0	39.00

Problem

For the trials in the graph, the heater power is 50.0 W, the sample has a mass of 0.400 kg, and the apparatus has a heat capacity of 40.0 J K⁻¹. Find the specific heat capacity of the sample and the power lost to the surroundings near 40 °C.

Show full solution

Slopes
Method
Read each temperature change over the 20 s shown.
Reason
The plotted short straight sections approximate local gradients near 40 °C. The losses are taken as equal there.
Working
Answer
sₕₑₐₜ = (+4.00)/20.0 = +0.200 K s⁻¹ and s_cool = (-1.00)/20.0 = -0.0500 K s⁻¹
Heat capacities
Reason
Subtracting the slopes removes the losses; the apparatus is then subtracted.
Working
Answer
Cₜₒₜ = 50.0/(0.200-(-0.0500)) = 200 J K⁻¹, so c = (200-40.0)/0.400 = 400 J kg⁻¹ K⁻¹
Power lost
Working
Answer
H = -Cₜₒₜs_cool = 200 × 0.0500 = 10.0 W. Check: 50.0-10.0 = 40.0 W = 200 × 0.200.

Quick check

Using c = Pt/mΔθ alone would give 50.0/(0.400 × 0.200) = 625 J kg⁻¹ K⁻¹, much too large.

Problem

A 0.400 kg sample is heated at 60.0 W. At the same temperature, it warms at + 0.150 K s⁻¹ with the heater on and cools at -0.0500 K s⁻¹ with it off. The heat capacity of the apparatus is negligible. Find the specific heat capacity of the sample.

Find the specific heat capacity

Show the steps Optional working

Step 1: Heating slope minus signed cooling slope

Unit: K s⁻¹

Step 2: Heat capacity of the sample

Unit: J K⁻¹

Specific heat capacity

Unit: J kg⁻¹ K⁻¹

Hints

Hint 1

the sign

sₕₑₐₜ-s_cool = 0.150-(-0.0500).

Show solution step by step

Heat capacity
Reason
The losses cancel because the slopes are at the same temperature.
Working
Answer
C = 60.0/(0.150 + 0.0500) = 300 J K⁻¹
Specific heat capacity
Working
Answer
c = 300/0.400 = 750 J kg⁻¹ K⁻¹

Problem

With a heater power of 72.0 W, a heating slope of + 0.250 K s⁻¹ and a cooling slope of -0.0500 K s⁻¹ at the same temperature, a learner calculates Cₜₒₜ = 72.0/(0.250-0.0500) = 360 J K⁻¹. Find the correct value.

Find the heat capacity

Show the steps Optional working

Step 1: Heating slope minus cooling slope

Unit: K s⁻¹

Total heat capacity

Unit: J K⁻¹

Show solution step by step

Correct denominator
Reason
The cooling slope is -0.0500, not + 0.0500.
Working
Answer
0.250-(-0.0500) = 0.300 K s⁻¹
Heat capacity
Working
Answer
Cₜₒₜ = 72.0/0.300 = 240 J K⁻¹

Common mistake

The heater has to supply the energy that warms the sample and the energy that leaks away, so the correction must make the heat capacity smaller than P/sₕₑₐₜ = 288 J K⁻¹, not larger.

Problem

A 0.500 kg sample is heated at 90.0 W; the apparatus has a heat capacity of 150 J K⁻¹. Three temperature–time slopes were measured:

heater on, at 40 °C: + 0.0800 K s⁻¹;
heater off, at 40 °C: -0.0200 K s⁻¹;
heater off, at 70 °C: -0.0600 K s⁻¹.

Find the specific heat capacity of the sample and the power lost at 40 °C.

Find both values

Cooling slope to pair with heating at 40 °C

−0.0200 K s⁻¹ at 40 °C−0.0600 K s⁻¹ at 70 °C

Specific heat capacity

Unit: J kg⁻¹ K⁻¹

Show the steps Optional working

Step 1: Specific heat capacity

Unit: J kg⁻¹ K⁻¹

Power lost at 40 °C

Unit: W

Hints

Hint 1

matching temperatures

The losses cancel only if both slopes are at the same temperature, so ignore the 70 °C reading.

Show solution step by step

Total heat capacity
Reason
Both slopes at 40 °C have the same losses.
Working
Answer
Cₜₒₜ = 90.0/(0.0800 + 0.0200) = 900 J K⁻¹
Sample and losses
Working
Answer
c = (900-150)/0.500 = 1500 J kg⁻¹ K⁻¹ and H = 900 × 0.0200 = 18.0 W. The 70 °C cooling slope is larger because the sample loses energy faster when it is hotter; using it would give a wrong answer, about 986 J kg⁻¹ K⁻¹.

Check your understanding 1 Explain the correction

A student insulates a heater and sample, uses c = P/(msₕₑₐₜ), and says insulation makes any correction unnecessary. Explain why this can overestimate c. Describe how to use a cooling curve and the apparatus heat capacity to improve the result.

Energy account and corrected method

Show answer

Some heater power still leaves to the room, and some warms the apparatus. At the same temperature and surroundings, use the local signed heating and cooling gradients:

P-H = Cₜₒₜsₕₑₐₜ, -H = Cₜₒₜs_cool

Subtract to obtain Cₜₒₜ = P/(sₕₑₐₜ-s_cool), then c = (Cₜₒₜ-Cₐₚₚ)/m. Check that your method matches temperatures, preserves the negative cooling sign, and removes apparatus capacity before dividing by sample mass. Insulation reduces losses; it does not remove the apparatus contribution.

Common mistakes

Using c = Pt/mΔθ without allowing for losses and the apparatus, which makes c too large.
Comparing heating and cooling slopes at different temperatures.
Treating the cooling slope as positive when subtracting it.
Dividing the total heat capacity by the sample’s mass without first subtracting the apparatus.

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Syllabus and review details

GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30

Last reviewed: 2026-08-28

Measure specific heat capacity with electrical heating

Guided lesson

Guided steps

Reading settings

Measuring specific heat capacity electrically

Correcting for the losses

Specific heat capacity from the two slopes

Slopes

Heat capacities

Power lost

Heating and cooling at the same temperature

Heat capacity

Specific heat capacity

The sign of the cooling slope

Correct denominator

Heat capacity

Choosing the right cooling data

Total heat capacity

Sample and losses

Check your understanding 1 Explain the correction

Common mistakes