THERMAL PROPERTIES OF MATTER
Heat is a form of energy that is transferred from a
hotter to a colder region by one or more of the
following methods conduction, convection and
radiation.
When heat is given to a body one or more of the
following may happen
(i)
Increase in temperature
(ii)
Increase in length ( area, volume)
(iii)
Change in state of the body
Solid
Liquid
Liquid
Gas
(iv)
Change in chemical composition
(v)
Change in electrical properties
(vi)
Change in colour
TEMPERATURE.
Temperature is a measure of the Heat energy that a
body contains or a measure of the Kinetic energy of the
molecules of the material.
By heating a material the kinetic energy of the
molecules of the material also increase.
The temperature also controls the direction of flow of
Heat Energy.
Heat Energy always flows from a higher temperature to
a lower temperature.
S.I. Unit of temperature is Kelvin.
The Kelvin scale and the kelvin are named after the
Scottish physicist and engineer William Thomson, 1st
Baron Kelvin (1824–1907). Unlike the degree
Fahrenheit and degree Celsius, the kelvin is not referred
to as a "degree" it is written K. The Kelvin and the
degree Celsius are often used together, as they have the
same interval, and 0 Kelvin is -273.15 degrees Celsius.
Thermometer.
A thermometer is an instrument for measuring
temperature.
To calibrate a thermometer two temperature called
FIXED POINTS are chosen and the interval in between
them is divided into a number of equal spaces called
DEGREES. In this way we obtain a temperature scale.
The lower fixed point called the ICE POINT is the
temperature of pure melting ice a standard pressure i.e.
0 0C
The upper fixed point called the MELTING POINT is
the temperature of steam from pure boiling water at
standard pressure i.e. 100 oC
On the Celsius or Centigrade scale
Lower point = 0 oC
And
o
Upper point = 100 C
The interval divided into 100 degrees.
We may also use the Kelvin scale Where
Kelvin = degree celsius + 273
K
= oC + 273
We know that to convert any temperature in degrees
Celsius to Kelvin we must add 273
In many applications it is a change in temperature that
is more important than the actual initial and final
temperatures.
Where the change in temperature is delta T and each
temperature T1 and T0 are given in degrees Celsius
Therefore
T = T1 – T0
In degrees Celsius
T (oC) = T1 (oC) – T0 (oC) = T1 – T0
In Kelvin
T(K) = T1(K) – T0(K) = (T1(oC)+273) – (T0(oC)+273)
Since the 273 values cancel each other out
T(K) = T1(oC) – T0(oC) = T (oC)
So for changes in temperature the value of the change
in temperature in degrees Celsius is the same as the
value of the change in temperature in Kelvin.
CHANGE IN TEMPERATURE
The change in temperature which a material experiences depends on three factors
(i)
The type of material
(ii)
The mass of the material
(iii)
The amount of heat energy given to the material
And
When there are more than two factors which determine
a value to compare only two of them we must make the
third factor a common standard value. To compare how
different materials react to the same amount of heat i.e
how two of the above factors depend on each other we
must make the third factor common i.e. all materials
must have equal mass and we take unit mass or 1 kg as
the standard value.
SPECIFIC HEAT CAPACITY
The specific heat capacity, c, of a material is defined as the heat energy required to change the temperature
of unit mass (1 kg) of the material by one degree.
c
Heat Energy added to the body
mass of the body x rise in tempera ture
or
c
Heat Energy lost by the body
mass of the body x decrease in tempera ture
S.I. Units
Joules
 J kg 1 oC 1
kg x degree Celsuis
or
Joules
 J kg 1 K 1
kg x Kelvin
The symbol used for heat energy is Q Therefore using
Q
c

symbols the equation becomes
m x T
Rearranging gives
Q = m c T Joules
Is the quantity of heat energy required to raise the
temperature of m kg of a substance, of specific heat
capacity c, by T degrees
Typical Values
Substance
specific heat capacity J / kg K
Water
4,200
Ice
2,100
Copper
400
Aluminium
900
Iron
460
Brass
380
Note
1.water has a very large specific heat capacity value in
comparison to most other materials
2. Ice and water do not have the same specific heat
capacity value.
Heat Flow between bodies at different temperatures.
When a body at a higher temperature is in direct
contact with a body at a lower temperature heat energy
flows between the two bodies.
The heat energy flows from the hotter body to the
cooler body until they are both at the same temperature.
If we assume that there is no heat loss to the
surroundings as the heat flow occurs then
Heat lost
by hotter body
=
Heat gained
by cooler body
Therefore when two or more bodies are in contact they
have or will achieve the same temperature.
When doing problems where substances at different
temperatures are being put together(i.e mixing
substances at different temperatures) always
1. Assume that there is no heat loss to the surroundings
unless otherwise told
And
2. always use the equation
Heat lost
=
by hotter body/bodies
Heat gained
by cooler body/bodies
Experiment to measure Specific Heat Capacity of a
Metal by Method of Mixtures
In this experiment, a piece of copper is heated to 100oC
and then transferred to some water whose temperature
change is noted.
A simple heat balance equation enables the specific
heat capacity for the metal to be calculated if that for
the water is known.
1. Weigh a piece of metal using the analytical balance, attach it to a thread and leave it in
boiling water for about 5 - 10 minutes, as shown
thread
boiling water
solid metal
2. Place some water into a polystyrene or plastic cup
and weigh the mass of water using the analytical
balance m w . Also place a thermometer into the water
and note the temperature T i.
thermometer
polystyrene
beaker with lagging
cup
m
e
t
a
l
3. Rapidly transfer the metal to the plastic cup
contained in a beaker lagged with cotton wool. Use the
thermometer to stir the mixture and note the highest
temperature reached.
4. Assuming that the polystyrene cup has zero specific
heat capacity and given that the specific heat capacity
of water = 4,200 J kg-1 K-1
Heat lost by metal = Heat gained by water
i.e.
mm cm ( 100 - Tf )
= mw cw ( Tf - Ti )
where Tf is the final temperature of the mixture and Ti
is the initial temperature of the water (the subscripts m
and w stand for metal and water respectively).
Calculate the value for the specific heat capacity of the
metal.
Problem Sheet Specific Heat Capacity
Question 1. If 101 J of heat is added to a 257 g piece
of aluminium at 16.0. 0C calculate the final temperature
of the aluminium?
Question 2. A 235 g lead ball at a temperature of
84.4oC is placed in a light calorimeter of negligible
specific heat capacity containing 178 g of water at
20.5oC. Calculate the equilibrium temperature of the
system.
Question 3. To determine the specific heat of an
object, a student heats it to 100oC in boiling water. She
then places the 82.8g object in a 150 g copper
calorimeter containing 100 g of water. The copper and
water are initially at a temperature of 19.4C, and are
thermally insulated from their surroundings. If the final
temperature is 30.3C calculate the specific heat of the
object?
Question 4. How much heat energy is required to raise
the temperature of a 125g iron cylinder by 58oC
Question 5. A copper block of mass 1 kg is being
heated by an electric immersion heater. If the
temperature of the block changes from 38oC to 68oC in
12 minutes calculate the power rating of the immersion
heater.
Question 6. A 1 kW electric heater takes 6 mins 6
seconds to heat 2.5 kg of water from 10oC to 40oC.
Calculate (a) The % heat loss to the surroundings and
(b) if the % heat loss remains constant how long will it
take to bring the temperature of the water up to 100oC.
Question 7. A ceramic coffee cup of mass 116 g and
specific heat capacity
1090Jkg-1K-1, is initially at room temperature (22.00C).
If 215g of 80.3oC coffee and 12.2 g of 6.00oC cream
are added to the cup, what is the equilibrium
temperature of the system? Assume that no heat is
exchanged with the surroundings, and that the specific
heat of coffee and cream are the same as the specific
heat of water.
EXPANSION.
Materials expand when heated and contract when
cooled. This is true for all three states of matter
however gases expand more than liquids which expand
more than solids.
Expansion of Solids.
Different solids e.g metals expand by different amounts
when heated through the same temperature range.
Solids expand in
(i)
Length
(ii)
Area
(iii)
Volume
And
For each we define a coefficient of expansion
Coefficient of expansion
Linear ( linear)
Area
( superficial )
Volume ( cubical )


Symbol



Linear Expansion.
Experiments show that the increase in length or
expansion of a solid depends of three factors
(i)
The change in temperature
(ii)
The material of which the solid is made
(iii)
the original length of the material
And
Bimetallic Strip.
A bi-metallic strip is used to convert a temperature
change into mechanical displacement. The strip
consists of two strips of different metals which expand
at different rates as they are heated, usually steel and
copper.
The strips are joined together throughout their length.
The different expansions force the flat strip to bend one
way if heated, and in the opposite direction if cooled
below its normal temperature. The metal with the
higher expansion is on the outer side of the curve when
the strip is heated and on the inner side when cooled.
The coefficient of linear expansion is defined as the
length by which 1 m of a material expands for every
degree rise in temperature.

Change in length
l

Original length x change in tempera ture l0 xT
S.I. Units
m
m oC

m
 o C 1  K 1
mK
The coefficient of linear expansion is a constant for a
given material.
lo is the original length of the material
T0 is the original temperature of the material
l1 is the final length of the material
T1 is the final temperature of the material
Then
l
l1  l0


l0 xT l0 xT
Cross multiply gives
 l0 xT   l1  l0
 l0 xT   l0  l1
l0 1   T   l1
l1  l0 1   T 
It can also be shown that



Typical values
Material
Aluminium
coefficient of linear
expansion (x 10-5 K-1)
2.2
Brass
1.9
Copper
1.8
Glass
0.9
Iron
1.2
Steel
1.1
Zinc
2.6
Superficial expansion
Change in area
A


 2
Original area x change in tempera ture A0 xT
Cubical expansion
Change in Volume
V


 3
Original Volume x change in tempera ture V0 xT
PROBLEM SHEET
Question 1.(a) A bar of steel is 150 mm long at 20oC
calculate its length at 175oC
(b)Repeat part (a) for and original length of 25 m (c)
Repeat part (a) for a zinc bar
Question 2.A concrete highway is built of slabs 12 m
long (150C). How wide should the expansion cracks
between the slabs be (at 150C).to prevent buckling if
the range of temperature is - 250C to +500C?
Question 3. An aluminum sphere is 18 cm in diameter.
What will be its change in volume if it is heated from
100C to 1000C?
Question 4. A glass and an iron bar are the same length
at 10oC. At 120oC they differ in length by 1.2 mm,
calculate their original lengths.
Question 5 A brass plug is to be placed in a ring made
of iron. At 200C, the diameter of the plug is 26.80 cm
and that of the inside of the ring is 26.699 cm.
(a)Calculate the common temperature in they must both
be brought to in order to fit? (b) Repeat part (a)if the
plug were iron and the ring brass?
Laboratory Experiment to Measure the coefficient
of the linear expansivity of copper
In this experiment you will measure the linear
expansion of a copper bar at various temperatures
ranging from room temperature to 100oC. A graph of
expansion versus temperature change will enable you to
find a value for the coefficient of linear expansion of
copper.
One slight difficulty is that you will not be able to
measure the expansion directly, because the scale
which registers changes in length is not calibrated
directly. Thus, prior to commencing the experiment
you will need to produce a calibration graph for this
scale.
Experimental Method
Fig.1
scale
thermometer
from steam
generator
Ratchet and pawl
micrometer screw
thread
steam out
a) Calibration of the scale
(
1. Turn the micrometer screw thread (see Fig.1) until
the pointer just begins to move over the scale.
The pitch of the screw thread is 0.5mm. Thus one
complete turn of the screw will cause the bar to move
against the ratchet and pawl device by 0.5 mm. Notice
that the screw has a scale showing quarter turns
2.
Continue turning the screw until the scale reads 50
divisions. Note the scale reading corresponding to
every quarter turn of the screw.
3. Plot a calibration graph of Scale readings versus
expansion, and refer to this graph when performing the
main experiment.
(b) Measurement of the expansivity
1. Measure the length of the bar L at room temperature.
Adjust the micrometer screw until the scale reading
corresponds to zero expansion on your calibration
curve.
2. Pass steam through the apparatus for a few minutes
until the temperature is steady at about 100oC.
3. Turn off the steam and note the scale reading versus
temperature as the bar cools.
4. Use your calibration graph to draw up a data table of
expansion and temperature change.
5. Plot an graph of expansion versus temperature
change
Given that
L =
 L0 (T)
obtain a value for the coefficient of linear expansion of
copper from its slope.
