2.3 How to directly measure the specific heat capacity of a solid substance

The experiment apparatus and set-up for a block of solid material

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.

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 29.5^oC, final temperature 38.5^oC, temperature rise ∆T = 9.0^oC

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 SHC_Al (J/kg^oC) x ∆T

4036.5 = 0.5 x SHC_Al x 9.0 = SHC_Al x 4.5

therefore on rearranging SHC_Al = 4036.5 / 4.5 = 897

so, the specific heat capacity of aluminium = 897 J/kg^oC

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 SHC_Al (J/kg^oC) x ∆T

rearranging gives: SHC_Al = ∆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 0.5^oC like a typical 0-100^oC school laboratory thermometer (a more accurate thermometer, mercury or digital reading to 0.1^oC 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

You then plot a graph of temperature versus energy transferred from eg 29.5^oC to 38.5^oC. By assuming the temperature reading is at best to the nearest 0.5^oC, 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 = SHC_Cu

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

energy transferred = mass of Cu x SHC_Cu x temperature change

Rearranging ∆E = m x c x ∆θ 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 = SHC_Cu = 1 / (m x gradient)

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

therefore specific heat capacity of copper = SHC_Cu = 1 / (1.10 x 0.002424) = 1 / 0.002666 = 376 J/kg^oC

For an individual measurement you can use the Key Equation:

c = E / mΔθ,  where:

• c = specific heat capacity (J/kg°C) of the solid
• E = energy transferred by heater (J)
• m = mass of the block (kg)
• Δθ = temperature change (°C)

 

Sources of error

However well insulated, the system will always be losing a small amount of its thermal energy store as it is being heated up. The system should be well insulated e.g. cotton wool or bubble wrap sheeting.

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 a poorer conductor like 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.