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 and calculations

It's a good idea to read Examples of energy store conversions in systems first.

Whenever you get an increase in temperature of a system, energy must be transferred from one energy store to another.

However, for the same quantity of heat energy transferred, the temperature rise will vary.

The temperature rise will depends on the amount of material heated and its structure.

When you heat a material, energy is absorbed and its internal energy is increased - an increase in its thermal energy store - due to an increase in the kinetic energy of vibration (solids) or kinetic energy of the free movement liquid and gas particles.

From kinetic particle theory, a temperature value is a measure of the average kinetic energy of the particles - the average internal energy of the material.

The internal energy store is the sum of the kinetic energy store plus the potential energy store - the latter can usually be ignored in the situations described here.

The energy transferred to a given material acting as a thermal energy store to raise its temperature 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.

Application: Solar panels may contain water that is heated by radiation from the Sun.

Water has a high heat capacity and can store a lot of heat energy.

This water may then be used to heat buildings or provide domestic hot water.

Different substances store different amounts of energy per kilogram for each °C temperature rise.

To put it another way, different materials require different amounts of heat energy to raise a given amount of material by the same increase in temperature.

This is called the specific heat capacity and varies from material to material, whether it be a gas, liquid or a solid - its all to do with the nature and arrangement of the particles - atoms, ions or molecules.

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. This is a way of quantifying an increase or decrease in a material's thermal energy store.

The formula for expressing the amount of heat transferred between energy stores is given by the equation

change in thermal energy store (J) = mass (kg) x specific heat capacity (J/kgoC) x change in temperature (oC)

∆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 the above equation.

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


Questions on specific heat

Q1 A 0.5 kg block of copper absorbed 1520 J of energy and its temperature rose by 8.0oC.

Calculate the specific heat capacity of copper.

∆E = m x c x ∆T

SHCcopper = ∆E / (m x ∆T)

SHCcopper = ∆E / (m x ∆T) = 1520 / (0.5 x 8.0) = 380 J/kgoC

 

Q2 A block bronze has a specific heat capacity of 400 J/kgoC.

If a 1500 g block of bronze absorbs 3000 J of energy, what would be the rise in its temperature?

∆E = m x c x ∆T,  1500 g = 1.500 kg

so rearranging: ∆T = ∆E / (m x SHCbronze) = 3000 / (1.5 x 400) = 5.0oC

 

Q3 ?


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.

So you need a mass balance.

The block must be surrounded by a good layer of insulation to minimise heat losses to the surroundings. 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.

Do diagram with joulemeter

Its mass should be accurately measured.

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.

The electrical current in the circuit does work on the heater and so transferring electrical energy from the power supply to the heaters thermal energy store which in turn is transferred to the metal block's energy store and therefore its temperature rises.

 

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

When the block has cooled down, you can repeat experiment.

 

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 heating 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 over a longer period of time.

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

The calculations assume that all the electrical energy does end up increasing the thermal energy store of the metal block.

In reality, you can't avoid a small loss of heat through the insulation.

 

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 ∆T = 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

(Note: You can do the experiment with a joulemeter, initially set at zero, so no need for the above calculations!)

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.

In SHC experiments you can include in the power supply circuit a joule meter to measure the energy transferred, which makes the calculation a lot easier. By using a joulemeter you don't need the voltmeter or ammeter.

energy transferred = mass of water x specific heat capacity of water x temperature rise

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

rearranging gives: SHCAl = ∆E / (mass of Al x ∆T)

Let the temperature rise a good 10 degrees and repeat the experiment at least twice to get an average - for the most accurate result.

 

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).

Graph note: The block may not heat up steadily at first and you may get a curve upwards at the start, but eventually the plot should become linear AND that is where you measure the gradient.

Calculation

Mass of copper = 1.10 kg, let c = SHCCu

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

You always need to repeat experiments to be more sure of your data, but you should always be aware of sources of error and how to minimise them.

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 faster the heat spreads, so better the results, so an aluminium or copper block should be ok.

The results would not be as good with e.g. concrete?

Its difficult to eliminate heat losses so the temperature rise might be a bit less than that expected for perfect insulation, but you should always use insulation around ALL of the surface of the block for this specific heat capacity experiment.

 

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 for the heating element and accurate thermometer.

The procedures and calculations would be the same to determine the specific heat capacity of a liquid.

 

Measuring the specific heat capacity of a liquid

You can use a similar set-up to that described above for measuring the SHC of a solid block.

Instead of the block you can use a polystyrene cup (good insulation) with a lid.

Measure a mass of liquid into the polystyrene cup = mass of cup + liquid - mass of empty cup (measured on a mass balance).

You can use water for convenience.

Place the cup in an insulated box or beaker.

The double thermal insulation is essential to minimise the loss of heat energy to the surroundings.

Do diagram with joulemeter

The procedure is identical to that described for a solid.

In SHC experiments you can include in the power supply circuit a joule meter to measure the energy transferred, which makes the calculation a lot easier. By using a joulemeter you don't need the voltmeter or ammeter.

energy transferred = mass of water x specific heat capacity of water x temperature rise

energy transferred = E (J) = m x c x ∆θ = mass of water (kg) x SHCH2O (J/kgoC) x ∆T

rearranging gives: SHCH2O = ∆E / (mass of water x ∆T)

Let the temperature rise a good 10 degrees and repeat the experiment at least twice to get an average - for the most accurate result.


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 its 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.


Heat Transfer and explaining physical changes and physical properties using a particle model

Introduction to heat transfer - conduction (and thermal conductivity), convection and radiation gcse physics notes

Specific heat capacity: How to determine it, use of data, calculations and thermal energy stores physics notes

More on methods of reducing heat transfer eg in a house and investigating insulating properties of materials

The density of materials and the particle model of matter gcse physics revision notes

Energy transfer and efficiency - calculations and Sankey diagrams gcse physics revision notes

Particle theory models, internal energy, heat transfer in state changes and latent heat and particle motion in gases (written more from a 'physics' point of view)


OCR GCSE 9-1 Gateway Science Physics: P1.2d Be able to define the term specific heat capacity and distinguish between it and the term specific latent heat. Investigation of the specific heat capacity of different metals or water using electrical heaters and a joulemeter. P1.2e Be able to apply the relationship between change in internal energy of a material and its mass, specific heat capacity and temperature change to calculate the energy change involved.


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