**
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 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. 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^{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 0-100^{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.

INDEX for my physics notes on specific
heat capacity

**
Keywords, phrases and learning objectives for measuring directly the specific heat capacity
of a solid material**

Understand how to directly measure the specific heat capacity of a
solid material, the apparatus you need , procedure and method of
calculating the specific heat capacity of the solid under
investigation

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