Knetic Theory of Matter, Specific Heat Capacity, Specific Latent Heat and The First Law of Thermodynamics

1. THERMAL PROPERTIES OF MATERIALS
A2 Physics Topic 4
By Allen Tobve

2. KINETIC MODEL
Gases, liquids and solids are all made up of atoms,
molecules, and/or ions, but the behaviour of these
particles differ in the three phases.
Particles in gases are well separated with no regular
arrangement.
liquid are close together with no regular arrangement.
solid are tightly packed, usually in a regular pattern.
Particles in gases vibrate and move freely at high speeds.
liquid vibrate, move about, and slide past each other.
solid vibrate but generally do not move from place to
place

3. CHANGES OF STATE
Particles in a body have a mixture of kinetic
energy and potential energy. The molecules
have kinetic energy because they are constantly
in motion and potential energy because they are
held by attractive intermolecular forces and
energy is needed to separate them.
Kinetic energy determines the temperature of
the body and potential energy determines the
state of the body.
Matter can be in four states like solid, liquid,
gas and plasma.

5. CHANGE OF STATE
In order to 'change state' energy is either 'taken in' or 'given out'.
The temperature is constant during a change of state, e.g. during
melting or boiling, even though energy is absorbed. This energy is
called the latent (or hidden) heat because it does not produce a change
of temperature.
The average kinetic energy of the molecules stays constant, but the
latent heat supplied increases the potential energy of the molecules, as
they break free from the attraction of their neighbouring particles.
The latent heat of vaporisation is higher than the latent heat of fusion
for the same substance. There is a large change in volume when a
liquid changes to a vapour, so the molecular separation in the gaseous
phase is several times that in the liquid phase. The energy used to
increase the potential energy of the molecules is greater during
vaporisation. The latent heat also supplies energy for the vapour to do
external work in pushing back the atmosphere when the liquid changes
to a vapour.

6. EVAPORATION AND BOILING
Evaporation takes place over a wide range of temperatures, whereas
boiling takes place takes place at a single temperature. Unlike boiling,
evaporation is affected by surface area and wind. Both are affected by
pressure.
Evaporation occurs because some of the molecules in the liquid have
more energy than other molecules. These fast moving molecules are
able to escape from the attraction of their neighbouring molecules and
leave the surface to form a vapour outside. The average kinetic of the
molecules left behind decreases, so the temperature of the liquid falls.
Hence evaporation has a cooling effect.
The boiling point of a liquid is defined as the temperature at which its
saturated vapour pressure (SVP) becomes equal to the external
atmospheric pressure. The space above a liquid is saturated with its
vapour and the pressure it exerts is called the saturated vapour
pressure. When a liquid is heated the SVP increases, and when it
becomes equal to the external atmospheric pressure boiling occurs and
bubbles of vapour form in the liquid.

7. SPECIFIC HEAT CAPACITY AND SPECIFIC LATENT
HEAT
Specific Heat Capacity
The temperature rise of an object when it is heated depends on:
the amount of energy supplied (ΔQ)
its mass (m)
what the substance is made of, e.g. water or copper
where c is the constant of proportionality, called the specific heat
capacity of the material.
The specific heat capacity (c) of a substance is the heat energy required
to increase the temperature of 1 kg of the substance by 1 °C or 1 K. The
unit of specific heat capacity (c) is J kg-1 K-1 or J kg-1 C-1.
The heat capacity (C) of an object is the heat energy required to
increase the object’s temperature by 1 °C or 1 K. The unit of heat
capacity (C) is J K-1 or J °C-1.

8. MEASURING SPECIFIC HEAT CAPACITY

9. The principle is to supply a known amount of energy to a known
mass of material and measure the rise in temperature. The metal
block of mass 1 kg is heated using an electrical heater, while
keeping the current and p.d. constant.
Measurements of the temperature are taken with time, as shown.
If the power of the heater is 50 W and the rise in temperature is
16.4 °C in 400 s.
Heat losses to the surroundings have been neglected, so the
value obtained is too high. The block is insulated to reduce heat
loss.
Another way is to cool the block below room temperature before
heating it. Then turn the heater off when the block is at an
equivalent temperature above room temperature.
The block will gain energy from the surroundings when it is below
room temperature and lose an equal amount when it is above
room temperature. The block should be heated at a low rate so
the heat spreads throughout the block.

10. THE CONTINUOUS-FLOW CALORIMETER
This was first developed by Callender and
Barnes in 1902 for the measurement of the
specific heat capacity of a liquid, and is shown in
Figure 2.
Its main advantage is that the thermal capacity of
the apparatus itself need not be known.

11.  Liquid flows in from a constant-head apparatus at a constant rate past a thermometer (θ1).
It then flows around the heater coil and out past a second thermometer where the outlet
temperature (θ2) may be measured.
 When steady-state conditions have been reached (a temperature difference between inlet
and outlet points of 5oC is reasonable) the temperatures and the flow rate of the liquid are
measured. A vacuum jacket round the heater coil reduces heat losses.
 The electrical energy supplied to the heater coil (E = V I t) may be found with an ammeter
and voltmeter, or with a joulemeter.
 Two sets of measurements are carried out. For a first experiment we have:
 Electrical energy supplied
E1 = V1I1t1 = m1 c (θ2 – θ1) + H
where c is the specific heat capacity of the liquid and the heat loss to the surroundings and
to the apparatus.
 The flow rate and rate of energy input are now altered to give a second set of results.
However, if the inlet and outlet temperatures are the same as in the first experiment the
heat loss will also be the same. Therefore:
Electrical energy supplied
E2 = V2I2t2 = m2 c (θ2 – θ1) + H
Eliminating the heat loss (H) gives:
Specific heat capacity of the liquid
c = (E2 – E1) / (m2 – m1) (θ2 – θ1)

12. FIRST LAW OF THERMODYNAMICS
1. Thermodynamics is the study of heat and its transformation
to mechanical energy in a system.
2. A system is a well defined group of objects, e.g. atoms or
molecules whose energy is being measured. An example of
a system is the petrol air mixture in the cylinder of a car
engine.
Adding heat to the mixture or igniting it
increases its internal energy, and the mixture
may do work on a piston and cause it to
move.

13. FIRST LAW OF THERMODYNAMICS
3. The internal energy of a system depends on the state of
the system and is the sum of the random distribution of
kinetic and potential energies associated with the
molecules of a system.
4. The internal energy of a gas can be increased by either
(i) heating the body. The molecules gain kinetic energy
when they collide with the hot container walls and bounce
off faster (or if a fuel is burnt within it), or
(ii) doing work on the body. Gas molecules are struck by a
piston moving downwards and bounce off faster, gaining
kinetic energy in the same way as a ball gains kinetic
energy when struck by a moving bat. This explains why a
gas becomes hotter when compressed.

14. FIRST LAW OF THERMODYNAMICS
5. The First Law of
thermodynamics in
equation form applied to a
gas is:
Increase in internal energy
(ΔU) = Heat energy
supplied (ΔQ) + Work
done on the gas (ΔW)
(ii) doing
work
(i) heating

15. FIRST LAW OF THERMODYNAMICS
6. The first law of thermodynamics is the law of conservation of energy, i.e.
energy cannot be created or destroyed but it can be transferred from
one form to another.
7. The first law of thermodynamics states:
The internal energy of a system depends
(i) only on its state; and
(ii) the increase in internal energy of a system (ΔU) equals the sum of the
energy supplied to the system by heating (ΔQ) and the work done on the
system (ΔW).
8. The work done by an expanding gas is given by
If the volume of the gas is decreased by ΔV, then p ΔV is the work done on
the gas.

17. WORKED EXAMPLES
Example 1
If 100 J of energy is added to a system that does no external work, by
how much does the internal energy of the system increase.
Answer
ΔW = 0
ΔU = ΔQ + 0 = 100 J
∆U= 100 J
Example 2.
If 100 J of energy is added to a system that does 40 J of external work,
what is the increase in internal energy of the system?

18. WORKED EXAMPLES
Example 3.
A 24 W filament bulb has been switched on for some
time. Apply the first law of thermodynamics during a
period of 2 s of the lamps operation, and when first
switched on.

19. Example 4
Find the increase in potential energy of a molecule of water when changing from
liquid at 100 °C to vapour at 100 °C.
( lv = 2.26 x 10 6 J kg-1, steam at 100 °C and normal atmospheric pressure has a
density of
0.59 kg m-3. Normal atmospheric pressure
= 1.01 x 10 5 Pa. Density of water
= 1000 kg m-3.)

20. EXAMPLE 5
(A) SUPPOSE THE GAS IN A CYLINDER EXPANDS FROM P TO Q, IN SUCH A WAY THAT
THE TEMPERATURE REMAINS CONSTANT. THE GAS IS THEN COMPRESSED FROM Q
TO R AT CONSTANT PRESSURE AND THEN HEATED AT CONSTANT VOLUME BACK TO P.
WHAT IS THE NET WORK DONE BY THE GAS?
(B) IN THE GRAPH BELOW, WHAT IS
(I) AN ISOTHERMAL CHANGE,
(II) AN ISOBARIC CHANGE AND
(III) AN ISOVOLUMETRIC CHANGE?
STATE WHAT CHANGES TAKE PLACE TO ΔU, ΔQ AND ΔW DURING THESE CHANGES?
Answer
(a) Pressure-volume graphs,
called indicator diagrams, can
be used to show the cycle of
changes taking place in the
cylinder of an engine. The work
done on or by a gas can be
found from the area under the
graph,

21. Since, ΔW = p ΔV = area under the graph.
From P to Q,
 the work done by the gas = area PQST
From Q to R,
 the work done on the gas = area RQST
Along RP,
 no work is done since there is no change in volume.
Net work done by the gas = shaded area enclosed PQR
(b) (i) Isothermal change
 This is a change from P to Q. The gas is expanding at constant temperature (or
contracting at constant temperature from Q to P.) Isothermals are constant
temperature curves. Boyle’s law applies,
Using, ,
There is no change in temperature, ΔU = 0.
Work is done by the gas as it expands, ΔW is negative. The gas takes in heat
from the surroundings to keep the temperature constant, so ΔQ is positive,
(equal to the work done.)

22. (ii) Isobaric change.
This is a change from Q to R when the gas is cooled at constant pressure, (or
from R to Q.) The pressure remains constant. Charles’ law applies,
(at constant pressure).
From
The temperature is decreasing, ΔU is negative. The gas is contracting,
work is done on it, and so ΔW is positive. Heat is removed from the gas to
keep the pressure constant, so ΔQ is negative.
(iii) Iso-volumetric change.
This is a change from R to P. The gas is warmed at constant volume so its
pressure increases. The pressure law applies
From ,
There is no change in volume, so no work is done, ΔW is zero.
Heat goes into the system to raise the pressure, so ΔQ is positive.(= +),
The internal energy increases, ΔU is positive, (and equal to the heat going
in.)

23. Adiabatic change
This is an expansion or contraction in which no heat
enters or leaves the gas.
From ,
ΔQ is zero as no heat enters the system,
Work is done by the gas as it expands, so ΔW is
negative.
ΔU is negative, the internal energy is decreasing.
Work is done by the gas at the expense of the
internal energy, so the gas cools.

24. THE
END