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/kg^{o}C) x change in temperature (^{o}C)
∆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/kg^{o}C,
∆θ =
∆T = temperature change in Celsius ^{o}C
The specific heat capacity of
water is 4180 J/kg^{o}C (Joules per kilogram per degree),
this means it takes 4180 J of heat energy
to raise the temperature of 1 kg of water by 1^{o}C.
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/kg^{o}C):
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
8.0^{o}C.
Calculate the specific heat capacity of
copper.
∆E = m x c x ∆T
SHC_{copper} = ∆E / (m x ∆T)
SHC_{copper} = ∆E / (m x ∆T)
= 1520 / (0.5 x 8.0) =
380 J/kg^{o}C
Q2
A block bronze has a specific heat capacity of 400 J/kg^{o}C.
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 SHC_{bronze}) = 3000 / (1.5 x 400) =
5.0^{o}C
Q3
?
2.
How to measure the specific heat capacity of a substance
The
experiment apparatus and setup
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 CO_{2} 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. 515 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^{o}C, final
temperature 38.5^{o}C, temperature rise ∆T
= 9.0^{o}C
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^{o}C) 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^{o}C
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^{o}C) 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^{o}C like a typical 0100^{o}C school laboratory
thermometer (a more accurate thermometer, mercury or digital reading to 0.1^{o}C 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 / ^{o}C 
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 29.5^{o}C
to 38.5^{o}C. By assuming the temperature reading is at best to the
nearest 0.5^{o}C, 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 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^{o}C
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 setup 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
SHC_{H2O }(J/kg^{o}C) x ∆T
rearranging gives: SHC_{H2O
}= ∆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 H_{2}O = 4180 J/kg^{o}C, 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 750960 J/kg^{o}C,
quite high) is used in night storage heaters (using cheap nighttime
electricity), the more dense the concrete the greater its capacity to store heat.
Oilfilled
heaters are used for a small scale heat storage (SHC oil = 900 J/kg^{o}C, not
as good as water) but will convect in the oil radiator and steadily release
heat.
More SHCs J/kg^{o}C: Aluminium 902,
copper 385
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