SPECIFIC HEAT CAPACITY  explanation and its applications
How to determine the specific heat capacity of a material
Applications of data  thermal energy stores and calculations
Doc Brown's
school physics revision notes: GCSE physics, IGCSE physics, O level
physics, ~US grades 8, 9 and 10 school science courses or equivalent for ~1416 year old
students of physics
Whenever any material is heated to a higher temperature
you increase the thermal energy store
of the material. A measure of how much energy is needed to raise the
temperature of a given amount of material to a specific temperature is called the heat capacity
of the material. The
specific heat capacity of
a substance can be defined as the amount of energy required to change the temperature of
one kilogram of the substance by one degree Celsius. From the specific heat capacity of a
material, the amount of material and the temperature change the material
experiences, you can calculate the increase or decrease of that material's
thermal energy store.
Subindex for this page
1.
Explaining and defining
the specific heat
capacity of materials
2.
Examples of questions
involving specific heat
3a.
How to measure the
specific heat capacity of a substance  solid
3b.
Measuring
the specific heat capacity of a liquid like water
3c.
Two ways of measuring the heat capacity of a solid using an indirect method
4.
Applications of heat
capacity data  examples of thermal energy storage systems  thermal energy
transfers
1.
Explaining and defining the specific heat
capacity of materials
It's a good idea to read
Examples of energy store conversions in systems first.
Specific latent
heat is dealt with on a separate page.
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.
Don't confuse heat and temperature!
When some object is heated, the
thermal energy ('heat') transferred increases the thermal energy
store of the object.
The temperature increases, but
the temperature only indicates how hot or cold the object is.
When you heat a material, thermal energy is
absorbed and its
internal energy is increased
due to an increase in its
thermal energy and potential energy stores.
At a particle level
this is due:
(i) An increase in the kinetic energy
store caused by increased vibration of solid particles or increased kinetic energy of the free movement
of liquid and gas particles from one place to another.
From kinetic particle theory, a
temperature value is a measure of the average kinetic energy of the
particles  much of the average internal energy of the material.
(ii) An increase in the potential
energy caused by the increase in kinetic energy opposing the
interparticle forces of attraction  the particles on average a bit
further apart with increase in temperature.
The internal energy store is the sum of
the kinetic energy store plus the potential energy store  the latter can
often be ignored in the situations described here concerning heat capacity.
The energy transferred to a given material
acting as a thermal energy store to raise its temperature by a specific amount 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 (they have
different specific heat capacities  but more on this later).
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 thermal energy.
This water may then be used to
heat buildings or provide domestic hot water.
Water is the usual conveyer of
thermal energy in central heating systems.
Water is a very good thermal
energy store in a hot bottle for cold winter nights in bed.
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
or just c) 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
(transferred to) 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):
ice 2100, aluminium 902, concrete 800,
glass 670, steel 450, brass 380, copper 385, lead 130
Because each material has a different
heat capacity, although you can heat the same mass of substance from one
temperature to another, you cannot assume they store the same amount of
thermal heat energy per kilogram.
The materials with the highest heat
capacity will store the most thermal energy per kilogram for the same
increase in temperature  they are effectively a more concentrated
thermal energy store.
Conversely, when allowing materials
to cool, the materials with the highest specific heat capacity will
release more thermal energy per kilogram for the same decrease in
temperature.
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2. Example 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
Calculate the thermal energy in kJ required to heat 2 kg of water (SHC = 4180
J/kg^{o}C) from 20^{o}C to 100^{o}C.
∆E = m x c x ∆T = 2 x 4180 x
(10020) = 6.68 x 10^{5} J
Thermal energy required =
668 kJ
Q4 A storage tank of water contains 500 kg of
hot water at 80^{o}C. (SHC water = 4180 J/kg^{o}C)
How many MJ of thermal energy can be
transferred to a heating system if the temperature of the water falls to 25^{o}C?
∆E = m x c x ∆T = 500 x 4180
x (8025) = 1.15 x 10^{8} J
Thermal energy transferred = 1.15 x 10^{8}
/ 10^{6} = 115 MJ
(1.15 x 10^{2} MJ, 3 sf)
Q5 A more complicated problem to solve  so
'double' think carefully!
Suppose 2 kg of steel at a temperature of
80^{o}C is placed in 10 kg (~10 litres) of water at 15^{o}C.
Calculate the final temperature of the
materials.
SHC steel = 450 J/kg^{o}C,
SHC water = 4180 J/kg^{o}C
The loss in J from the steel energy
store = gain in J of the thermal energy store of the water.
At 'thermal equilibrium' they will be
both at the same temperature, lets call this T.
The temperature fall of the steel is (80
 T), the temperature rise of the water is (T  15)
The loss from the steel thermal energy
store = ∆E = m x c x ∆T = 2 x 450 x (80  T) = 900 x (80  T) J
The gain of the water thermal energy
store = ∆E = m x c x ∆T = 10 x 4180 x (T  15) = 41800 x
(T  15) J
So, change in steel thermal energy store
= change in water thermal energy store
900 x (80  T) = 41800 x (T 
15)
72000  900T = 41800T  627000
(changing sides, changing signs, we
get)
72000 + 627000 = 900T + 41800T
699000 = 42700T
T = 699000 / 42700 =
16.4^{o}C
(3 sf)
Q6 A night storage heater contains 75 kg of
concrete (SHC = 800 J/kg^{o}C)
At night when electricity is cheaper and
the house cooler, how many kJ of thermal energy is needed to raise the
temperature of the concrete from 8^{o}C to 35^{o}C.
∆E = m x c x ∆T = 75 x
800 x (35  8) = 1.62 x 10^{6} J
1.62 x 10^{6} / 1000 =
1620 J
Q7
This question is based on a physics experiment I did in school back in circa
1961.
Its a good little experiment to do.
Accurately weigh 100.0 g (~100 ml) of
water into a beaker at room temperature.
A 50.0 g brass weight was held by a thin
wire string or with a pair tongs in the centre of a roaring Bunsen flame.
After a few minutes the brass weight was
plunged into a beaker of 100 g cold water whose initial temperature was 19.5^{o}C.
After gently stirring carefully with a
thermometer the temperature of the water rose to 55.5^{o}C.
SHC of brass = 380 J/kg^{o}C, SHC
water = 4180 J/kg^{o}C
(a) If the flame temperature is T,
calculate the two thermal energy store transfers.
Thermal energy transfer from brass
weight = ∆E = m x c x ∆T = (50/1000) x 380 x (T  19.5) = 19.0 (T 
19.5) J
Thermal energy transfer from brass
weight = ∆E = m x c x ∆T = (100/1000) x 4180 x (55.5  19.5) = 15048 J
(b) Calculate the flame temperature T
Thermal energy transfer from brass
weight (J) = thermal energy transfer to water (J)
19.0 (T  19.5) = 15048
19.0T  370.5 = 15048
19.0T = 15418.5
T = 15418.5 / 19.0 =
812^{o}C
(3 sf)
(b) Suggest some sources of error in the
experiment
(i) the weight loses some thermal
energy to air on transfer  possibly quite a bit, since the experimental
answer you get is often well below the real answer for the flame
temperature of a bunsen flame ~10001200^{o}C.
(ii) the glass beaker absorbs some
heat, therefore its heat capacity has been ignored
(iii) the lower part of the wire will
transfer a little heat  but if you use brass wire, you could weigh that
too and make sure everything is immersed in the water!
Q8
A car of mass 1000 kg is moving at 20.0 m/s is brought to a halt by sharp
braking.
Assume all the kinetic energy is
converted to thermal energy by four steel brake drums.
If the total mass of the steel drums of
the brakes is 25.0 kg, assuming no heat losses, calculate the maximum rise
in temperature the brake drums will attain.
(a) First calculate the kinetic energy
(KE) of the car.
KE = ˝ mv^{2}
KE = ˝ x 1000 x 20^{2} = 2.0
x 10^{5} J
(b) Calculate the temperature rise of the
braking system (SHC of steel = 450 J/kg^{o}C)
∆E = m x c x ∆T
∆T = ∆E / (m x c) = 2.0 x 10^{5}
/ (25 x 450) =
17.8^{o}C
(3sf)
The next two questions are more tricky
and involve using electricity formulae:
Electricity power calculations, P = IV
The specific heat capacity of water is
4180 J/kg^{o}C. Be careful with all the units.
Q9
A small electric kettle is connected to the 240 V AC mains supply and uses a
current of 8.0 A.
It contains 1.5 kg of water (~1.5 litres,
1500 ml, 1500 cm^{3}) at 20^{o}C.
(a) Calculate the power of the kettle and
the rate of energy transfer.
P (W) = I (a) x V (p.d. in volts) = 8
x 240 =
1920 W =
1920
J/s
(b) If the kettle is switched on for 2.0
minutes, how much energy is transferred to the thermal energy store of the
water?
P = 1920 J/s, energy transferred =
power x time = 1920 x 2 x 60 = 230400 J =
2.3
x 10^{5}
J
(c) What temperature will the hot water
rise too?
∆E = mass of water x SHC_{H2O} x temperature rise
∆E (J) = m (kg) x c (J/kgoC) x ∆T (^{o}C)
Rearranging: ∆T = ∆E / (m x SHC_{H2O})
∆T = 230400 / (1.5 x 4180) = 36.7 ^{o}C
Therefore temperature after 2 minutes
of heating is 20 + 36.7 =
56.7^{o}C
(d) How much energy is needed to raise
the water temperature from 20^{o}C to 100^{o}C?
∆E = mass of water x SHC_{H2O} x temperature rise
∆E = 1.5 x 4180 x (100  20) = 5.016 x 10^{5} J =
5.01 x 10^{5}
J (3 sf)
(e) Starting with the cold water at 20^{o}C,
how long would it take to boil the kettle' in minutes and seconds?
What assumptions have you made for
this calculation?
The kettle's power s 1920 W, that is
1920 J/s, and energy needed = 5.016 x 10^{5} J
P = E / t, so t = E / P = 5.016 x 10^{5}
/ 1920 = 261.25 s.
Time to boil =
4 mins and 21 seconds
(to the nearest second).
This calculation assumes all the
electrical energy converted to thermal energy, actually increased the
thermal energy store of the water.
However, you would always get a small
amount of wasted thermal energy transferring to the thermal energy store
of the kettle casing by conduction, and convection and radiation
transfer to the thermal energy store of he surrounding air.
Q10
A hot water cylinder (immersion heater) is quoted as having a
capacity of '120 litres of water' (120 kg of water).
The tank is fitted with a 3 kW heating
element working off 240 V mains electricity.
(a) How much energy is needed to heat
fresh cold water at 12^{o}C to a temperature of 70^{o}C?
∆E = mass of water x SHC_{H2O} x temperature rise
Energy needed = 120 x 4180 x (70 
12) = 29 092 800 J =
2.91 x 10^{7}
J (3 sf)
(b) How long will it take to heat the
water up? (to the nearest minute)
P (J/s) = E (J) / t (s), 3000 = 29
092 800 / t
time = E / P = 29 092 800 / 3000 =
9697.6 seconds
9697.6 / 60 = 161.6. time
required =
162 minutes (sf)
(c) What current is flowing through the
heating element?
P = I x V, I = P / V = 3000 /
240 = 12.5 A
(d) What is the resistance of the heating
element?
V = IR, R = V / I = 240 / 12.5
= 19.2 Ω
(e) If the immersion was fitted with a 6
kW heating element, how would this affect the time to heat water up to a
certain temperature?
This is twice the power of the 3 kW
heating element.
P = E / t, t = E / P, for the
same quantity of water and the same temperature rise, it would
take half the time
compared to the 3 kW heater,
Q11
?
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3a.
How to measure the specific heat capacity of a substance  solid
The
experiment apparatus and setup 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 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 extra diagram with
a 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 ∆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^{o}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.
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3b. Measuring
the specific heat capacity of a liquid like water
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 extra diagram with
a joulemeter?
The procedure is identical to that
described for a solid.
In SHC experiments you can include in the
power supply circuit a joulemeter to measure the energy transferred, which
makes the calculation a lot easier.
By using a joulemeter you don't need
the voltmeter and a ammeter, plus the extra calculation.
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.
If you have no joulemeter, then, as in
the diagram, take measurements from a voltmeter and ammeter.
The use the equation: total energy transferred
(J) = P x t = I x V x t
= current (A) x p.d. (V) x time in seconds
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3c. Measuring the heat capacity of a solid using an indirect method (A)
(Procedures (iv)(A) and (iv)(B) relate to calculation
Q5 in the set of specific heat capacity questions above)
This method relies on the transfer
between a brass weight's thermal energy store and the thermal energy store
of water.
Illustration of procedure A
Method (procedure A)
A brass weight is accurately weighed and
placed in beaker of icecold water (tap water plus lumps of ice)
This is left for some time and
occasionally stirred, until the brass weight is at the same temperature as
the water.
250 g of water is weighed into a beaker
and heated until its about 80^{o}C (no need to boil, increases
hazard).
After a gentle stir, the temperature of
the warm water is taken (T2), immediately before transfer.
The temperature (T1) of the icedwater is
also taken just before transferring the brass weight from the icedwater to
the warm water using tongs or thin wire and loop.
After transferring the cold brass weight
to the warm water, allow time for the thermal heat transfer into the brass
weight to complete and after a gentle stir read the final temperature (T3)
Results and calculation (A)
Typical results:
Mass of brass weight 200 g (0.200
kg), mass of water 250 g (0.250 kg), SHC water = 4180 J/kg^{o}C
Initial temperature of iced water T1
= 0.5^{o}C
Initial temperature of the warm water
(T2) = 79.5^{o}C
Final temperature of water/brass in
2nd beaker (T3) = 74.0^{o}C (cooler, as brass weight absorbs
heat)
∆E
= Thermal energy transfer to brass weight = Thermal energy transfer from
water (all in J)
∆E
= m x c x ∆T = 0.200 x SHC_{brass} x (T3  T1) = 0.250 x 4180 x
(T2  T3) = 5747.5
∆E
= 0.200 x SHC_{brass} x (74.0  0.5) = 0.250 x 4180 x (79.5 
74.0) = 5747.5
0.20 x SHC_{brass} x 73.5 =
5747.5
14.7 x SHC_{brass} = 5747.5
SHC_{brass} = 5747.5 / 14.7 =
391 J/kg^{o}C
(3 sf)
Improvements and sources of error (A)
(i) Could not be quite sure if the
brass weight has completely cooled down to ~0^{o}C (T1).
(ii) The hot water in the beaker is
losing heat all the time giving a greater cooling effect than just from
the brass weight  might be better to use a lower start temperature in
the 2nd beaker e.g. 4050^{o}C.
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3c. continued:
Alternative procedure, but a similar methodology (B)
Illustration of procedure B
Method (procedure B)
You could do a similar experiment to (A) by
putting the brass weight into hot water for some to heat up to ~8090^{o}C.
You could heat the water directly in a beaker to
~8090^{o}C.
Allow time for the brass weight to heat
up.
After a careful gentle stir, measure the
initial temperature of the brass weight (T1, ~90^{o}C) in the hot
water.
Measure the initial temperature of the
cold water in the 2nd beaker (T2, ~20^{o}C)
Pick up the brass weight with tongs or a
thin wire and loop and
transfer it to a beaker of water at room temperature.
Allow time for the thermal energy store
transfers to take place.
After a careful gentle stir, measure the
final temperature of the brass weight/water (T3) in the 'warmed up' water.
You can then do a similar calculation as
above e.g.
Results and calculation (B)
Typical results:
Mass of brass weight 200 g (0.200 kg),
mass of water 250 g (0.250 kg), SHC water = 4180 J/kg^{o}C
Initial temperature of heated water/brass
weight (T1) = 85.0^{o}C
Initial temperature of the cold water
(T2) = 20.5^{o}C
Final temperature of water/brass in 2nd
beaker(T3) = 24.9^{o}C
∆E
= Thermal energy transfer from hot brass weight = Thermal energy transfer to
water (all in J)
∆E
= m x c x ∆T = 0.200 x SHC_{brass} x (T1  T3) = 0.250 x 4180 x (T3
 T2) = ?
∆E
= 0.200 x SHC_{brass} x (85.0  24.9) = 0.250 x 4180 x (24.9  20.5)
= 4598
0.20 x SHC_{brass} x 60.1 = 4598
12.02 x SHC_{brass} = 4598
SHC_{brass} = 4598 / 12.02 =
383 J/kg^{o}C
(3 sf)
Improvements and sources of error (B)
(i) Could not be quite sure if the
brass weight has completely warmed up to ~8090^{o}C (T1)
(ii) Not absolutely sure the thermal
energy transfer from the hot brass weight to the water is complete  if left
too long the 2nd beaker of water will start to cool give a greater
temperature change than should be measured.
(iii) The beaker absorbs some of the
extra thermal energy transferred to the water's thermal energy store.
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4.
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.
Water is also used as a coolant in
car engines because it can absorb a lot of thermal energy for a given
temperature increase. The thermal energy store of the engine block is
reduced and the thermal energy store of the water increased. The thermal
energy in the water is then transferred to the surrounding air to
increase its thermal energy store via the radiator grill.
The good old fashioned hot water
bottle is a nice convenient thermal energy store to heat up the bed.
Concrete (SHC 750960 J/kg^{o}C,
quite high) is used in night storage heaters (using cheap nighttime
electricity).
The greater the mass of concrete, the greater
its rise in temperature rise (safely!) the greater its capacity to store
thermal energy, to be later released into the house in daytime..
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.
An archaeological note!
Prehistoric man learned thousands of
years ago that hot stone retained a lot of thermal energy.
The heat capacity of natural stone is
usually around 840 J/kg^{o}C.
Large stones were heated in a fire and
dropped into stone line cooking troughs like the one shown below.
The heat from the thermal energy store of
the stone increases the thermal energy store of the cooler water, so boiling
the water and cooking food like meat placed in the water filled trough.
It may seem crude, but brass cooking pots
were something of a luxury item for many prehistoric people!
This stone cooking trough is by the
Bronze age stone circle (shown below) at Drombeg, Co. Cork, Ireland.
Several of them were constructed on this
site and fed and connected by a diverted spring stream.
They can be found all over Ireland and
the UK, and presumably on continental Europe.
Native American Indians also used the
same technique by dropping hot stones into a wooden bowl of food and water.
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