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 ...
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
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.
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.
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
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