SPECIFIC HEAT CAPACITY

How to determine the specific heat capacity of a material

Applications of data - thermal energy stores and calculations

Doc Brown's Physics Revision Notes

Suitable for GCSE/IGCSE Physics/Science courses or their equivalent

Whenever any material is heated to a higher temperature you increase the thermal energy store.

A measure of how much energy is needed to raise the temperature of a given amount of material is called the heat capacity.


1. The specific heat capacity of materials

The energy transferred to given material acting as a thermal energy store can vary quite widely e.g. you need over four times more heat energy to raise a given mass of water to specified temperature than that for the same mass of central heating oil or aluminium.

Different substances store different amounts of energy per kilogram for each °C temperature rise – this is called the specific heat capacity and varies from material to material, whether it be a gas liquid or a solid.

Materials with a high heat capacity will release lots of heat energy when cooling down from a higher to a lower temperature.

The specific heat capacity (SHC) of a substance is the amount of energy required to change the temperature of one kilogram of the substance by one degree Celsius.

The formula for expressing the amount of heat transferred in a given situation is ...

∆E = m x c x ∆θ

E = energy transferred in Joules (change in thermal energy), m = mass of material in kilograms kg

c = SHC = specific heat capacity J/kgoC, θ = ∆T = temperature change in Celsius oC

The specific heat capacity of water is 4180 J/kgoC (Joules per kilogram per degree),

this means it takes 4180 J of heat energy to raise the temperature of 1 kg of water by 1oC.

The amount of energy stored in or released from a system as its temperature changes can be calculated using this equation.

Other specific heat capacity values (J/kgoc): aluminium 902, copper 385,


2. How to measure the specific heat capacity of a substance

The experiment apparatus and set-up

You need a block of material of known mass eg 0.5 to 1.5 kg.

The block must be surrounded by a good layer of insulation to minimise heat losses. Polystyrene would be a good insulator and because it is mainly pockets of CO2 gas of low density with a low heat capacity (low thermal energy store), but watch you don't 'overheat' and soften the polystyrene. Layers of cotton or newspaper might do.

The block must have two holes drilled in it - one for a thermometer and another for the heating element.

The heating element is connected in series with an ammeter (to measure the current I in amperes) and a d.c. power supply e.g. 5-15 volts. The voltmeter must connected in parallel across the heating element connections.

You also need a stop clock or stopwatch.

In the experiment electrically energy is transferred and converted to heat energy which is absorbed by the block, increasing its temperature and increasing its thermal energy store.

Procedure and measurements

Method (i) one set of measurements using a 0.50 kg block of aluminium

Switch on the heater setting the voltage at eg 12V (but use the accurate digital voltmeter reading for calculations).

When the block seems to be heating up steadily, start the clock/stopwatch and record the temperature.

Record the p.d. voltage and the current in amps with an accurate digital ammeter, both readings of which should be constant throughout the experiment.

After eg 15 minutes, record the final temperature and check the voltage and current readings and still the same and turn of the power.

Method (ii) multiple measurements using a 1.1 kg block of copper

Another approach is to take the temperature reading every minute for eg 15 minutes once the copper block seems to be steadily heats up. The voltage and current readings should be constant.

This produces more data AND more reliable results than method (i)  and sorts out inconsistencies in the temperature readings.

The procedure is the same as method (i) BUT taking more temperature readings between the initial and final thermometer readings.

I have assumed the same current and voltage, however, there is a lot more work in the calculations!

How to calculate the specific heat capacity of the solid

Results data and calculation for method (i)

Mass of an eg aluminium block 500g = 0.50 kg

Initial temperature 29.5oC, final temperature 38.5oC, temperature rise = 9.0oC

Current 0.39A, p.d. 11.5V, time 15 mins = 15 x 60 = 900 s

Power P = current x p.d. =  I x V = 0.39 x 11.5 = 4.485 W = 4.485 J/s

therefore total electrical energy = heat energy transferred = P x time = 4.485 x 900 = 4036.5 J

energy transferred = E (J) = m x c x ∆θ = mass of Al (kg) x SHCAl (J/kgoC) x ∆T

4036.5 = 0.5 x SHCAl x 9.0 = SHCAl x 4.5

therefore on rearranging SHCAl = 4036.5 / 4.5 = 897

so, the specific heat capacity of aluminium = 897 J/kgoC

Note that this method relies on only two temperature readings.

Data and calculation for method (ii) a lot of work!

From the voltage (V) and current (I) readings you calculate the total energy transferred for all the 15 minutes of readings.

total energy transferred = P x t = I x V x t = current (A) x p.d. (V) x time in seconds

So you then have 15 total transferred energy numbers, steadily increasing from 1 to 15 minutes

Let us assume the current, voltage as method (i)

I'm assuming the thermometer can be read to the nearest 0.5oC like a typical 0-100oC school laboratory thermometer (a more accurate thermometer, mercury or digital reading to 0.1oC is most desirable!)

Therefore P = IV = 0.39A x 11.5V = 4.485J/s, energy transferred per second.

So after 1 minute energy transfer = 4.485 x 1 x 60 = ~269 J, this finally rises to 4.485 x 15 x 60 = ~4037 J

Time / mins 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Energy transferred / J 0 269 538 807 1076 1346 1614 1884 2153 2422 2691 2960 3229 3498 3767 4037
Temperature / oC 29.0 29.5 30.0 31.0 31.5 32.0 32.5 33.0 33.5 34.5 35.0 35.5 36.5 37.0 38.0 38.5

You then plot a graph of temperature versus energy transferred from eg 29.5oC to 38.5oC. By assuming the temperature reading is at best to the nearest 0.5oC, it makes the 'calculated' data more realistic AND justifying the multiple reading method (ii).

Mass of copper = 1.10 kg

The specific heat capacity equation is: E = m x c x ∆θ

energy transferred = mass of Cu x SHCCu x temperature change

Rearranging gives: ∆θ = E / (m x c) and ∆θ / ∆E = 1 / (m x c)

This means the gradient of the graph = 1 / (m x c)

so, c = SHCCu = 1 / (m x gradient)

From the graph the gradient = (38 - 30) / (3800 - 500) = 8 / 3300 = 0.002424

therefore specific heat capacity of copper = SHCCu = 1 / (1.10 x 0.002424) = 1 / 0.002666 = 376 J/kgoC

Sources of error

The heat energy has to conduct throughout the block and be evenly distributed, I doubt if that's the case, so the measured temperature reading might be different than the average temperature of the whole block.

The better the heat conduction of the solid, the better the results, so an aluminium or copper block should be ok.

The results would not be as good with eg concrete?

Its difficult to eliminate heat losses so the temperature rise might be a bit less than that expected for perfect insulation.

Experiment extension

You can repeat for any suitable material in solid block form. You could also put other materials in a polystyrene container eg sand, soil  etc.

You can swap the block and insulation for an insulating polystyrene cup filled with a know mass of liquid. It would need a lid with two holes in it. The procedure and calculation would be the same to determine the specific heat capacity of a liquid.


3. Applications of heat capacity data - examples of thermal energy storage systems

The greater the heat capacity of a material, the more heat energy it can hold for a given mass of material.

This means that high heat capacity materials can store lots of energy when heated and can then release a lot if cooled down. In other words, materials with a high specific heat capacity are good for storing heat energy - a good material for a thermal energy store.

Materials used in heaters/heating systems, usually have a high specific heat capacity eg water (SHC H2O = 4180 J/kgoC, very high) is used in central heating systems and is easily pumped around house to distribute lots of heat where needed, an excellent 'mobile' thermal energy store.

Concrete (SHC 750-960 J/kgoC, quite high) is used in night storage heaters (using cheap night-time electricity), the more dense the concrete the greater the capacity to store heat.

Oil-filled heaters are used for a small scale heat storage (SHC oil = 900 J/kgoC, not as good as water) but will convect in the oil radiator and steadily release heat.

More SHCs J/kgoC: Aluminium 902, copper 385


  • Check out your practical work you did or teacher demonstrations you observed in Unit P1.1, all of this is part of good revision for your module examination context questions and helps with 'how science works'.

    • Passing white light through a prism and detecting the infrared radiation with a thermometer.

    • Demonstration using balls in a tray to show the behaviour of particles in substances in different states i.e. gas, liquid and solid.

    • Measuring the cooling effect produced by evaporation by putting wet cotton wool over the bulb of a thermometer or temperature probe.

    • Plan and carry out an investigation into factors that affect the rate of cooling of a can of water, eg shape, volume, and colour of can using Leslie’s cube to demonstrate the effect on radiation of altering the nature of the surface.

    • Investigating thermal conduction using rods of different materials.


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