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
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.
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.
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. 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
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.
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 29.5oC, 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
(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.
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
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.
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.
You can repeat for any suitable material in solid block form.
You can swap the block and insulation for an insulating polystyrene cup filled with a know mass of 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.
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
Heat Transfer and explaining physical changes and physical properties using a particle model
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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