CHAPTER ELEVEN
THERMAL PROPERTIES OF MATTER
11.1 INTRODUCTION
We all have common sense notions of heat and temperature.
Temperature is a measure of ‘hotness’ of a body. A kettle
with boiling water is hotter than a box containing ice. In
physics, we need to define the notion of heat, temperature,
etc., more carefully. In this chapter, you will learn what heat
is and how it is measured, and study the various proceses by
which heat flows from one body to another. Along the way,
you will find out why blacksmiths heat the iron ring before
fitting on the rim of a wooden wheel of a horse cart and why
the wind at the beach often reverses direction after the sun
goes down. You will also learn what happens when water boils
or freezes, and its temperature does not change during these
processes even though a great deal of heat is flowing into or
out of it.
11.2 TEMPERATURE AND HEAT
We can begin studying thermal properties of matter with
definitions of temperature and heat. Temperature is a relative
measure, or indication of hotness or coldness. A hot utensil
is said to have a high temperature, and ice cube to have a
low temperature. An object that has a higher temperature
than another object is said to be hotter. Note that hot and
cold are relative terms, like tall and short. We can perceive
temperature by touch. However, this temperature sense is
somewhat unreliable and its range is too limited to be useful
for scientific purposes.
We know from experience that a glass of ice-cold water left
on a table on a hot summer day eventually warms up whereas
a cup of hot tea on the same table cools down. It means that
when the temperature of body, ice-cold water or hot tea in
this case, and its surrounding medium are different, heat
transfer takes place between the system and the surrounding
medium, until the body and the surrounding medium are at
the same temperature. We also know that in the case of glass
tumbler of ice-cold water, heat flows from the environment to
11.1 Introduction
11.2 Temperature and heat
11.3 Measurement of
temperature
11.4 Ideal-gas equation and
absolute temperature
11.5 Thermal expansion
11.6 Specific heat capacity
11.7 Calorimetry
11.8 Change of state
11.9 Heat transfer
11.10 Newton’s law of cooling
Summary
Points to ponder
Exercises
Additional Exercises
2020-21
the glass tumbler, whereas in the case of hot
tea, it flows from the cup of hot tea to the
environment. So, we can say that heat is the
form of energy transferred between two (or
more) systems or a system and its
surroundings by virtue of temperature
difference. The SI unit of heat energy
transferred is expressed in joule (J) while SI unit
of temperature is Kelvin (K), and degree Celsius
(
o
C) is a commonly used unit of temperature.
When an object is heated, many changes may
take place. Its temperature may rise, it may
expand or change state. We will study the effect
of heat on different bodies in later sections.
11.3 MEASUREMENT OF TEMPERATURE
A measure of temperature is obtained using a
thermometer. Many physical properties of
materials change sufficiently with temperature.
Some such properties are used as the basis for
constructing thermometers. The commonly used
property is variation of the volume of a liquid
with temperature. For example, in common
liquid–in–glass thermometers, mercury, alcohol
etc., are used whose volume varies linearly with
temperature over a wide range.
Thermometers are calibrated so that a
numerical value may be assigned to a given
temperature in an appropriate scale. For the
definition of any standard scale, two fixed
reference points are needed. Since all
substances change dimensions with
temperature, an absolute reference for
expansion is not available. However, the
necessary fixed points may be correlated to the
physical phenomena that always occur at the
same temperature. The ice point and the steam
point of water are two convenient fixed points
and are known as the freezing and boiling
points, respectively. These two points are the
temperatures at which pure water freezes and
boils under standard pressure. The two familiar
temperature scales are the Fahrenheit
temperature scale and the Celsius temperature
scale. The ice and steam point have values
32 °F and 212 °F, respectively, on the Fahrenheit
scale and 0 °C and 100 °C on the Celsius scale.
On the Fahrenheit scale, there are 180 equal
intervals between two reference points, and on
the Celsius scale, there are 100.
Fig. 11.1 A plot of Fahrenheit temperature (t
F
) versus
Celsius temperature (t
c
).
A relationship for converting between the two
scales may be obtained from a graph of
Fahrenheit temperature (t
F
) versus celsius
temperature (t
C
) in a straight line (Fig. 11.1),
whose equation is
t
t
F
C
– 32
180 100
=
(11.1)
11.4 IDEAL-GAS EQUATION AND
ABSOLUTE TEMPERATURE
Liquid-in-glass thermometers show different
readings for temperatures other than the fixed
points because of differing expansion properties.
A thermometer that uses a gas, however, gives
the same readings regardless of which gas is
used. Experiments show that all gases at low
densities exhibit same expansion behaviour. The
variables that describe the behaviour of a given
quantity (mass) of gas are pressure, volume, and
temperature (P, V, and T)(where T = t + 273.15;
t is the temperature in °C). When temperature
is held constant, the pressure and volume of a
quantity of gas are related as PV = constant.
This relationship is known as Boyle’s law, after
Robert Boyle (1627–1691), the English Chemist
who discovered it. When the pressure is held
constant, the volume of a quantity of the gas is
related to the temperature as V/T = constant.
This relationship is known as Charles’ law,
after French scientist Jacques Charles (1747–
1823). Low-density gases obey these
laws, which may be combined into a single
THERMAL PROPERTIES OF MATTER 279
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280 PHYSICS
relationship. Notice that since PV = constant
and V/T = constant for a given quantity of gas,
then PV/T should also be a constant. This
relationship is known as ideal gas law. It can be
written in a more general form that applies not
just to a given quantity of a single gas but to any
quantity of any low-density gas and is known as
ideal-gas equation:
PV
R
T
µ
=
or PV =
µ
RT (11.2)
where,
µ
is the number of moles in the sample
of gas and R is called universal gas constant:
R = 8.31 J mol
–1
K
–1
In Eq. 11.2, we have learnt that the pressure
and volume are directly proportional to
temperature : PV ∝ T. This relationship allows a
gas to be used to measure temperature in a
constant volume gas thermometer. Holding the
volume of a gas constant, it gives P ∝T. Thus,
with a constant-volume gas thermometer,
temperature is read in terms of pressure. A plot
of pressure versus temperature gives a straight
line in this case, as shown in Fig. 11.2.
However, measurements on real gases deviate
from the values predicted by the ideal gas law
at low temperature. But the relationship is linear
over a large temperature range, and it looks as
though the pressure might reach zero with
decreasing temperature if the gas continued to
be a gas. The absolute minimum temperature
for an ideal gas, therefore, inferred by
extrapolating the straight line to the axis, as in
Fig. 11.3. This temperature is found to be
– 273.15 °C and is designated as absolute zero.
Absolute zero is the foundation of the Kelvin
temperature scale or absolute scale temperature
named after the British scientist Lord Kelvin. On
this scale, – 273.15 °C is taken as the zero point,
that is 0 K (Fig. 11.4).
The size of unit in Kelvin and Celsius
temperature scales is the same. So, temperature
on these scales are related by
T = t
C
+ 273.15 (11.3)
11.5 THERMAL EXPANSION
You may have observed that sometimes sealed
bottles with metallic lids are so tightly screwed
that one has to put the lid in hot water for some
time to open it. This would allow the metallic lid
to expand, thereby loosening it to unscrew
easily. In case of liquids, you may have observed
that mercury in a thermometer rises, when the
thermometer is put in slightly warm water. If
we take out the thermometer from the warm
Fig. 11.2 Pressure versus temperature of a low
density gas kept at constant volume.
Fig. 11.3 A plot of pressure versus temperature and
extrapolation of lines for low density gases
indicates the same absolute zero
temperature.
Fig. 11.4 Comparision of the Kelvin, Celsius and
Fahrenheit temperature scales.
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THERMAL PROPERTIES OF MATTER 281
water the level of mercury falls again. Similarly,
in case of gases, a balloon partially inflated in a
cool room may expand to full size when placed
in warm water. On the other hand, a fully
inflated balloon when immersed in cold water
would start shrinking due to contraction of the
air inside.
It is our common experience that most
substances expand on heating and contract on
cooling. A change in the temperature of a body
causes change in its dimensions. The increase
in the dimensions of a body due to the increase
in its temperature is called thermal expansion.
The expansion in length is called linear
expansion. The expansion in area is called area
expansion. The expansion in volume is called
volume expansion (Fig. 11.5).
Fig. 11.5 Thermal Expansion.
If the substance is in the form of a long rod,
then for small change in temperature, ∆T, the
fractional change in length, ∆l/l, is directly
proportional to ∆T.
∆
∆
l
l
T=
α
1
(11.4)
where
α
1
is known as the coefficient of linear
expansion (or linear expansivity) and is
characteristic of the material of the rod. In Table
11.1, typical average values of the coefficient of
linear expansion for some material in the
temperature range 0 °C to 100
°C are given. From
this Table, compare the value of
α
l
for glass and
copper. We find that copper expands about five
times more than glass for the same rise in
temperature. Normally, metals expand more and
have relatively high values of
α
l
.
Table 11.1 Values of coefficient of linear
expansion for some material
Material
αα
αα
α
l
(10
–5
K
–1
)
Aluminium 2.5
Brass 1.8
Iron 1.2
Copper 1.7
Silver 1.9
Gold 1.4
Glass (pyrex) 0.32
Lead 0.29
Similarly, we consider the fractional change
in volume,
∆V
V
, of a substance for temperature
change ∆T and define the coefficient of volume
expansion (or volume expansivity),
α
V
as
α
V
=
∆
∆
V
V T
1
(11.5)
Here
α
V
is also a characteristic of the
substance but is not strictly a constant. It
depends in general on temperature (Fig 11.6). It
is seen that
α
V
becomes constant only at a high
temperature.
Fig. 11.6 Coefficient of volume expansion of copper
as a function of temperature.
Table 11.2 gives the values of coefficient of
volume expansion of some common substances
in the temperature range 0–100 °C. You can see
that thermal expansion of these substances
(solids and liquids) is rather small, with material,
l
l
a T
l
∆
= ∆
l
2
A
a T
A
∆
= ∆
l
3
V
a T
V
∆
= ∆
(a) Linear expansion (b) Area expansion (c) Volume expansion
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282 PHYSICS
like pyrex glass and invar (a special iron-nickel
alloy) having particularly low values of
α
V
. From
this Table we find that the value of
α
v
for
alcohol (ethanol) is more than mercury and
expands more than mercury for the same rise
in temperature.
Table 11.2 Values of coefficient of volume
expansion for some substances
Material
αα
αα
α
v
(
K
–1
)
Aluminium 7 × 10
–5
Brass 6 × 10
–5
Iron 3.55 × 10
–5
Paraffin 58.8 × 10
–5
Glass (ordinary) 2.5 × 10
–5
Glass (pyrex) 1 × 10
–5
Hard rubber 2.4 × 10
–4
Invar 2 × 10
–6
Mercury 18.2 × 10
–5
Water 20.7 × 10
–5
Alcohol (ethanol) 110 × 10
–5
Water exhibits an anomalous behaviour; it
contracts on heating between 0 °C and 4 °C.
The volume of a given amount of water decreases
as it is cooled from room temperature, until its
temperature reaches 4 °C, [Fig. 11.7(a)]. Below
4 °C, the volume increases, and therefore, the
density decreases [Fig. 11.7(b)].
This means that water has the maximum
density at 4
°C. This property has an important
environmental effect: bodies of water, such as
lakes and ponds, freeze at the top first. As a lake
cools toward 4 °C, water near the surface loses
energy to the atmosphere, becomes denser, and
sinks; the warmer, less dense water near the
bottom rises. However, once the colder water on
top reaches temperature below 4 °C, it becomes
less dense and remains at the surface, where it
freezes. If water did not have this property, lakes
and ponds would freeze from the bottom up,
which would destroy much of their animal and
plant life.
Gases, at ordinary temperature, expand more
than solids and liquids. For liquids, the
coefficient of volume expansion is relatively
independent of the temperature. However, for
gases it is dependent on temperature. For an
ideal gas, the coefficient of volume expansion at
constant pressure can be found from the ideal
gas equation:
PV =
µ
RT
At constant pressure
P∆V =
µ
R ∆T
∆ ∆V
V
T
T
=
i.e.,
α
v
T
=
1
for ideal gas (11.6)
At 0 °C,
α
v
= 3.7 × 10
–3
K
–1
, which is much
larger than that for solids and liquids.
Equation (11.6) shows the temperature
dependence of
α
v
; it decreases with increasing
temperature. For a gas at room temperature and
constant pressure,
α
v
is about 3300 × 10
–6
K
–1
, as
Temperature (°C) Temperature (°C)
(a) (b)
Fig. 11.7 Thermal expansion of water.
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THERMAL PROPERTIES OF MATTER 283
t
t
much as order(s) of magnitude larger than the
coefficient of volume expansion of typical liquids.
There is a simple relation between the
coefficient of volume expansion (
α
v
) and
coefficient of linear expansion (
α
l
). Imagine a
cube of length, l, that expands equally in all
directions, when its temperature increases by
∆T. We have
∆l =
α
l
l ∆T
so, ∆V = (l+∆l)
3
– l
3
3l
2
∆l (11.7)
In Equation (11.7), terms in (∆l)
2
and (∆l)
3
have
been neglected since ∆l is small compared to l.
So
∆
∆
∆V
V l
l
V T
l
= =
3
3 α
(11.8)
which gives
α
v
= 3
α
l
(11.9)
What happens by preventing the thermal
expansion of a rod by fixing its ends rigidly?
Clearly, the rod acquires a compressive strain
due to the external forces provided by the rigid
support at the ends. The corresponding stress
set up in the rod is called thermal stress. For
example, consider a steel rail of length 5 m and
area of cross-section 40 cm
2
that is prevented
from expanding while the temperature rises by
10 °C. The coefficient of linear expansion of steel
is
α
l(steel)
= 1.2 × 10
–5
K
–1
. Thus, the compressive
strain is
∆l
l
=
α
l(steel)
∆T = 1.2 × 10
–5
× 10=1.2 × 10
–4
.
Youngs modulus of steel is Y
(steel)
= 2 × 10
11
N m
–2
.
Therefore, the thermal stress developed is
∆ ∆F
A
Y
l
l
steel
=
=
2.4 × 10
7
N m
–2
, which
corresponds to an external force of
∆F = AY
steel
∆l
l
= 2.4 × 10
7
× 40 × 10
–4
j 10
5
N.
If two such steel rails, fixed at their outer ends,
are in contact at their inner ends, a force of this
magnitude can easily bend the rails.
Example 11.1 Show that the coefficient
of area expansion, (∆A/A)/∆T, of a
rectangular sheet of the solid is twice its
linear expansivity,
α
l
.
Answer
Fig. 11.8
Consider a rectangular sheet of the solid
material of length a and breadth b (Fig. 11.8 ).
When the temperature increases by ∆T, a
increases by ∆a =
α
l
a∆T and b increases by ∆b
=
α
l
b ∆T. From Fig. 11.8, the increase in area
∆A = ∆A
1
+∆A
2
+ ∆A
3
∆A = a ∆b + b ∆a + (∆a) (∆b)
= a
α
l
b ∆T + b
α
l
a ∆T + (
α
l
)
2
ab (∆T)
2
=
α
l
ab ∆T (2 +
α
l
∆T) =
α
l
A ∆T (2 +
α
l
∆T)
Since
α
l
10
–5
K
–1
, from Table 11.1, the
product α
l
∆T for fractional temperature is small
in comparision with 2 and may be neglected.
Hence,
t
Example 11.2 A blacksmith fixes iron ring
on the rim of the wooden wheel of a horse
cart. The diameter of the rim and the iron
ring are 5.243 m and 5.231 m, respectively
at 27 °C. To what temperature should the
ring be heated so as to fit the rim of the
wheel?
Answer
Given, T
1
= 27 °C
L
T1
= 5.231 m
L
T2
= 5.243 m
So,
L
T2
=L
T1
[1+α
l
(T
2
–T
1
)]
5.243 m = 5.231 m [1 + 1.20×10
–5
K
–1
(T
2
–27 °C)]
or T
2
= 218 °C. t
∆A
3
= (∆a) (∆b)
∆A
l
= a (∆b)
∆A
2
= b (∆a)
a
b
∆b
∆a
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284 PHYSICS
11.6 SPECIFIC HEAT CAPACITY
Take some water in a vessel and start heating it
on a burner. Soon you will notice that bubbles
begin to move upward. As the temperature is
raised the motion of water particles increases
till it becomes turbulent as water starts boiling.
What are the factors on which the quantity of
heat required to raise the temperature of a
substance depend? In order to answer this
question in the first step, heat a given quantity
of water to raise its temperature by, say 20 °C
and note the time taken. Again take the same
amount of water and raise its temperature by
40 °C using the same source of heat. Note the
time taken by using a stopwatch. You will find
it takes about twice the time and therefore,
double the quantity of heat required raising twice
the temperature of same amount of water.
In the second step, now suppose you take
double the amount of water and heat it, using
the same heating arrangement, to raise the
temperatur
e by 20 °C, you will find the time
taken is again twice that required in the first
step.
In the third step, in place of water, now heat
the same quantity of some oil, say mustard oil,
and raise the temperature again by 20 °C. Now
note the time by the same stopwatch. You will
find the time taken will be shorter and therefore,
the quantity of heat required would be less than
that required by the same amount of water for
the same rise in temperature.
The above observations show that the quantity
of heat required to warm a given substance
depends on its mass, m, the change in
temperature, ∆T and the nature of substance.
The change in temperature of a substance, when
a given quantity of heat is absorbed or rejected
by it, is characterised by a quantity called the
heat capacity of that substance. We define heat
capacity, S of a substance as
Q
S
T
∆
=
∆
(11.10)
where ∆Q is the amount of heat supplied to
the substance to change its temperature from T
to T + ∆T.
You have observed that if equal amount of
heat is added to equal masses of different
substances, the resulting temperature changes
will not be the same. It implies that every
substance has a unique value for the amount of
heat absorbed or given off to change the
temperature of unit mass of it by one unit. This
quantity is referred to as the specific heat
capacity of the substance.
If ∆Q stands for the amount of heat absorbed
or given off by a substance of mass m when it
undergoes a temperature change ∆T, then the
specific heat capacity, of that substance is given
by
1
S Q
s
m m T
∆
= =
∆
(11.11)
The specific heat capacity is the property of
the substance which determines the change in
the temperature of the substance (undergoing
no phase change) when a given quantity of heat
is absorbed (or given off) by it. It is defined as the
amount of heat per unit mass absorbed or given
off by the substance to change its temperature
by one unit. It depends on the nature of the
substance and its temperature. The SI unit of
specific heat capacity is J kg
–1
K
–1
.
If the amount of substance is specified in
terms of moles
µ
, instead of mass m in kg, we
can define heat capacity per mole of the
substance by
C
S Q
T
= =
µ µ
1 ∆
∆
(11.12)
where C is known as molar specific heat
capacity of the substance. Like S, C also
depends on the nature of the substance and its
temperature. The SI unit of molar specific heat
capacity is J mol
–1
K
–1
.
However, in connection with specific heat
capacity of gases, additional conditions may be
needed to define C. In this case, heat transfer
can be achieved by keeping either pressure or
volume constant. If the gas is held under
constant pressure during the heat transfer, then
it is called the molar specific heat capacity at
constant pressure and is denoted by C
p
. On
the other hand, if the volume of the gas is
maintained during the heat transfer, then the
corresponding molar specific heat capacity is
called molar specific heat capacity at constant
volume and is denoted by C
v
. For details see
Chapter 12. Table 11.3 lists measured specific
heat capacity of some substances at atmospheric
pressure and ordinary temperature while Table
11.4 lists molar specific heat capacities of some
gases. From Table 11.3 you can note that water
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THERMAL PROPERTIES OF MATTER 285
t
has the highest specific heat capacity compared
to other substances. For this reason water is also
used as a coolant in automobile radiators, as
well as, a heater in hot water bags. Owing to its
high specific heat capacity, water warms up
more slowly than land during summer, and
consequently wind from the sea has a cooling
effect. Now, you can tell why in desert areas,
the earth surface warms up quickly during the
day and cools quickly at night.
Table 11.4 Molar specific heat capacities of
some gases
Gas C
p
(J mol
–1
K
–1
) C
v
(J mol
–1
K
–1
)
He 20.8 12.5
H
2
28.8 20.4
N
2
29.1 20.8
O
2
29.4 21.1
CO
2
37.0 28.5
11.7 CALORIMETRY
A system is said to be isolated if no exchange or
transfer of heat occurs between the system and
its surroundings. When different parts of an
isolated system are at different temperature, a
quantity of heat transfers from the part at higher
temperature to the part at lower temperature.
The heat lost by the part at higher temperature
is equal to the heat gained by the part at lower
temperature.
Calorimetry means measurement of heat.
When a body at higher temperature is brought
in contact with another body at lower
temperature, the heat lost by the hot body is
equal to the heat gained by the colder body,
provided no heat is allowed to escape to the
surroundings. A device in which heat
measurement can be done is called a
calorimeter. It consists of a metallic vessel and
stirrer of the same material, like copper or
aluminium. The vessel is kept inside a wooden
jacket, which contains heat insulating material,
like glass wool etc. The outer jacket acts as a
heat shield and reduces the heat loss from the
inner vessel. There is an opening in the outer
jacket through which a mercury thermometer
can be inserted into the calorimeter (Fig. 11.20).
The following example provides a method by
which the specific heat capacity of a given solid
can be determinated by using the principle, heat
gained is equal to the heat lost.
Example 11.3 A sphere of 0.047 kg
aluminium is placed for sufficient time in a
vessel containing boiling water, so that the
sphere is at 100
°
C. It is then immediately
transfered to 0.14 kg copper calorimeter
containing 0.25 kg water at 20
°
C. The
temperatur
e of water rises and attains a
steady state at 23
°
C. Calculate the specific
heat capacity of aluminium.
Answer In solving this example, we shall use
the fact that at a steady state, heat given by an
aluminium sphere will be equal to the heat
absorbed by the water and calorimeter.
Mass of aluminium sphere (m
1
) = 0.047 kg
Initial temperature of aluminium sphere = 100 °C
Final temperature = 23 °C
Change in temperature (∆T)=(100 °C-23°C)= 77 °C
Let specific heat capacity of aluminium be s
Al
.
Table 11.3 Specific heat capacity of some substances at room temperature and atmospheric
pressure
Substance Specific heat capacity Substance Specific heat capacity
(J kg
–1
K
–1
) (J kg
–1
K
–1
)
Aluminium 900.0 Ice 2060
Carbon 506.5 Glass 840
Copper 386.4 Iron 450
Lead 127.7 Kerosene 2118
Silver 236.1 Edible oil 1965
Tungesten 134.4 Mercury 140
Water 4186.0
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286 PHYSICS
The amount of heat lost by the aluminium
sphere =
∆ = × × °
1
0.047kg 77 C
Al Al
m s T s
Mass of water (m
2
) = 0.25 kg
Mass of calorimeter (m
3
) = 0.14 kg
Initial temperature of water and calorimeter=20 °C
Final temperature of the mixture = 23 °C
Change in temperature (∆T
2
) = 23 °C – 20 °C = 3 °C
Specific heat capacity of water (s
w
)
= 4.18 × 10
3
J kg
–1
K
–1
Specific heat capacity of copper calorimeter
= 0.386 × 10
3
J kg
–1
K
–1
The amount of heat gained by water and
calorimeter = m
2
s
w
∆T
2
+ m
3
s
cu
∆T
2
= (m
2
s
w
+ m
3
s
cu
) (∆T
2
)
= (0.25 kg × 4.18 × 10
3
J kg
–1
K
–1
+ 0.14 kg ×
0.386 × 10
3
J kg
–1
K
–1
) (23 °C – 20 °C)
In the steady state heat lost by the aluminium
sphere = heat gained by water + heat gained by
calorimeter.
So, 0.047 kg × s
Al
× 77 °C
= (0.25 kg × 4.18 × 10
3
J kg
–1
K
–1
+ 0.14 kg ×
0.386 × 10
3
J kg
–1
K
–1
)(3 °C)
s
Al
= 0.911 kJ kg
–1
K
–1
t
11.8 CHANGE OF STATE
Matter normally exists in three states: solid,
liquid and gas. A transition from one of these
states to another is called a change of state. Two
common changes of states are solid to liquid
and liquid to gas (and, vice versa). These changes
can occur when the exchange of heat takes place
between the substance and its surroundings.
To study the change of state on heating or
cooling, let us perform the following activity.
Take some cubes of ice in a beaker. Note the
temperature of ice. Start heating it slowly on a
constant heat source. Note the temperature after
every minute. Continuously stir the mixture of
water and ice. Draw a graph between
temperature and time (Fig. 11.9). You will
observe no change in the temperature as long
as there is ice in the beaker. In the above process,
the temperature of the system does not change
even though heat is being continuously supplied.
The heat supplied is being utilised in changing
the state from solid (ice) to liquid (water).
Fig. 11.9 A plot of temperature versus time showing
the changes in the state of ice on heating
(not to scale).
The change of state from solid to liquid is
called melting and from liquid to solid is called
fusion. It is observed that the temperature
remains constant until the entire amount of the
solid substance melts. That is, both the solid
and the liquid states of the substance coexist
in thermal equilibrium during the change of
states from solid to liquid. The temperature
at which the solid and the liquid states of the
substance is in thermal equilibrium with each
other is called its melting point. It is
characteristic of the substance. It also depends
on pressure. The melting point of a substance
at standard atomspheric pressure is called its
normal melting point. Let us do the following
activity to understand the process of melting
of ice.
Take a slab of ice. Take a metallic wire and
fix two blocks, say 5 kg each, at its ends. Put
the wire over the slab as shown in Fig. 11.10.
You will observe that the wire passes through
the ice slab. This happens due to the fact that
just below the wire, ice melts at lower
temperature due to increase in pressure. When
the wire has passed, water above the wire freezes
again. Thus, the wire passes through the slab
and the slab does not split. This phenomenon
of refreezing is called regelation. Skating is
possible on snow due to the formation of water
under the skates. Water is formed due to the
increase of pressure and it acts as a
lubricant.
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THERMAL PROPERTIES OF MATTER 287
Fig. 11.10
After the whole of ice gets converted into water
and as we continue further heating, we shall see
that temperature begins to rise (Fig.11.9). The
temperature keeps on rising till it reaches nearly
100 °C when it again becomes steady. The heat
supplied is now being utilised to change water
from liquid state to vapour or gaseous state.
The change of state from liquid to vapour (or
gas) is called vaporisation. It is observed that
the temperature remains constant until the
entire amount of the liquid is converted into
vapour. That is, both the liquid and vapour states
of the substance coexist in thermal equilibrium,
during the change of state from liquid to vapour.
The temperature at which the liquid and the
vapour states of the substance coexist is called
its boiling point. Let us do the following activity
to understand the process of boiling of water.
Take a round-bottom flask, more than half
filled with water. Keep it over a burner and fix a
Triple Point
The temperature of a substance remains constant during its change of state (phase change).
A graph between the temperature T and the Pressure P of the substance is called a phase
diagram or P – T diagram. The following figure shows the phase diagram of water and CO
2
.
Such a phase diagram divides the P – T plane into a solid-region, the vapour-region and the
liquid-region. The regions are separated by the curves such as sublimation curve (BO), fusion
curve (AO) and vaporisation curve (CO). The points on sublimation curve represent states
in which solid and vapour phases coexist. The point on the sublimation curve BO represent
states in which the solid and vapour phases co-exist. Points on the fusion curve AO represent
states in which solid and liquid phase coexist. Points on the vapourisation curve CO represent
states in which the liquid and vapour phases coexist. The temperature and pressure at which
the fusion curve, the vaporisation curve and the sublimation curve meet and all the three
phases of a substance coexist is called the triple point of the substance. For example the
triple point of water is represented by the temperature 273.16 K and pressure 6.11×10
–3
Pa.
(a) (b)
Fig. 11.11: Pressure-temperature phase diagrams for (a) water and (b) CO
2
(not to the scale).
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288 PHYSICS
thermometer and steam outlet through the cork
of the flask (Fig. 11.11). As water gets heated in
the flask, note first that the air, which was
dissolved in the water, will come out as small
bubbles. Later, bubbles of steam will form at
the bottom but as they rise to the cooler water
near the top, they condense and disappear.
Finally, as the temperature of the entire mass
of the water reaches 100 °C, bubbles of steam
reach the surface and boiling is said to occur.
The steam in the flask may not be visible but as
it comes out of the flask, it condenses as tiny
droplets of water, giving a foggy appearance.
Fig. 11.11 Boiling process.
If now the steam outlet is closed for a few
seconds to increase the pressure in the flask,
you will notice that boiling stops. More heat
would be required to raise the temperature
(depending on the increase in pressure) before
boiling begins again. Thus boiling point increases
with increase in pressure.
Let us now r
emove the burner. Allow water to
cool to about 80 °C. Remove the thermometer and
steam outlet. Close the flask with the airtight
cork. Keep the flask turned upside down on the
stand. Pour ice-cold water on the flask. Water
vapours in the flask condense reducing the
pressure on the water surface inside the flask.
Water begins to boil again, now at a lower
temperature. Thus boiling point decreases with
decrease in pressure.
This explains why cooking is difficult on hills.
At high altitudes, atmospheric pressure is lower,
reducing the boiling point of water as compared
to that at sea level. On the other hand, boiling
point is increased inside a pressure cooker by
increasing the pressure. Hence cooking is faster.
The boiling point of a substance at standard
atmospheric pressure is called its normal
boiling point.
However, all substances do not pass through
the three states: solid-liquid-gas. There are
certain substances which normally pass from
the solid to the vapour state directly and vice
versa. The change from solid state to vapour
state without passing through the liquid state
is called sublimation, and the substance is said
to sublime. Dry ice (solid CO
2
) sublimes, so also
iodine. During the sublimation process both the
solid and vapour states of a substance coexist
in thermal equilibrium.
11.8.1 Latent Heat
In Section 11.8, we have learnt that certain
amount of heat energy is transferred between a
substance and its surroundings when it
undergoes a change of state. The amount of heat
per unit mass transferred during change of state
of the substance is called latent heat of the
substance for the process. For example, if heat
is added to a given quantity of ice at –10 °C, the
temperature of ice increases until it reaches its
melting point (0 °C). At this temperature, the
addition of more heat does not increase the
temperature but causes the ice to melt, or
changes its state. Once the entire ice melts,
adding more heat will cause the temperature of
the water to rise. A similar situation
occurs during liquid gas change of state at the
boiling point. Adding more heat to boiling water
causes vaporisation, without increase in
temperature.
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THERMAL PROPERTIES OF MATTER 289
t
The heat required during a change of state
depends upon the heat of transformation and
the mass of the substance undergoing a change
of state. Thus, if mass m of a substance
undergoes a change from one state to the other,
then the quantity of heat required is given by
Q = m L
or L = Q/m (11.13)
where L is known as latent heat and is a
characteristic of the substance. Its SI unit is
J kg
–1
. The value of L also depends on the
pressure. Its value is usually quoted at standard
atmospheric pressure. The latent heat for a solid-
liquid state change is called the latent heat of
fusion (L
f
), and that for a liquid-gas state change
is called the latent heat of vaporisation (L
v
).
These are often referred to as the heat of fusion
and the heat of vaporisation. A plot of
temperature versus heat for a quantity of water
is shown in Fig. 11.12. The latent heats of some
substances, their freezing and boiling points, are
given in Table 11.5.
Fig. 11.12 Temperature versus heat for water at
1 atm pressure (not to scale).
Note that when heat is added (or removed)
during a change of state, the temperature
remains constant. Note in Fig. 11.12 that the
slopes of the phase lines are not all the same,
which indicate that specific heats of the various
states are not equal. For water, the latent heat of
fusion and vaporisation are L
f
= 3.33 × 10
5
J kg
–1
and L
v
= 22.6 × 10
5
J kg
–1
,
respectively. That is,
3.33 × 10
5
J of heat is needed to melt 1 kg ice at
0 °C, and 22.6 × 10
5
J of heat is needed to convert
1 kg water into steam at 100
°C. So, steam at
100
°C carries 22.6 × 10
5
J kg
–1
more heat than
water at 100
°C. This is why burns from steam
are usually more serious than those from
boiling water.
Example 11.4 When 0.15 kg of ice at 0 °C
is mixed with 0.30 kg of water at 50
°C in a
container, the resulting temperature is
6.7
°C. Calculate the heat of fusion of ice.
(s
water
= 4186 J kg
–1
K
–1
)
Answer
Heat lost by water = ms
w
(
θ
f
–
θ
i
)
w
= (0.30 kg) (4186 J kg
–1
K
–1
) (50.0 °C – 6.7 °C)
= 54376.14 J
Heat required to melt ice = m
2
L
f
= (0.15 kg) L
f
Heat required to raise temperature of ice
water to final temperature = m
I
s
w
(
θ
f
–
θ
i
)
I
= (0.15 kg) (4186 J kg
–1
K
–1
) (6.7 °C – 0
°C)
= 4206.93 J
Heat lost = heat gained
54376.14 J = (0.15 kg) L
f
+ 4206.93 J
L
f
= 3.34×10
5
J kg
–1
. t
Table 11.5 Temperatures of the change of state and latent heats for various substances at
1 atm pressure
Substance Melting L
f
Boiling L
v
Point (
°
C) (10
5
J kg
–1
) Point (
°
C) (10
5
J kg
–1
)
Ethanol –114 1.0 78 8.5
Gold 1063 0.645 2660 15.8
Lead 328 0.25 1744 8.67
Mercury –39 0.12 357 2.7
Nitrogen –210 0.26 –196 2.0
Oxygen –219 0.14 –183 2.1
Water 0 3.33 100 22.6
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290 PHYSICS
t
Example 11.5 Calculate the heat required
to convert 3 kg of ice at –12 °C kept in a
calorimeter to steam at 100
°C at
atmospheric pressure. Given specific heat
capacity of ice = 2100 J kg
–1
K
–1
, specific heat
capacity of water = 4186 J kg
– 1
K
–1
, latent
heat of fusion of ice = 3.35 × 10
5
J kg
–1
and latent heat of steam = 2.256 ×10
6
J kg
–1
.
Answer We have
Mass of the ice, m = 3 kg
specific heat capacity of ice, s
ice
= 2100 J kg
–1
K
–1
specific heat capacity of water, s
water
= 4186 J kg
–1
K
–1
latent heat of fusion of ice, L
f ice
= 3.35 × 10
5
J kg
–1
latent heat of steam, L
steam
= 2.256 × 10
6
J kg
–1
Now, Q = heat required to convert 3 kg of
ice at –12 °C to steam at 100 °C,
Q
1
= heat required to convert ice at
–12 °C to ice at 0 °C.
= m s
ice
∆T
1
= (3 kg) (2100 J kg
–1.
K
–1
) [0–(–12)]°C
= 75600 J
Q
2
= heat required to melt ice at
0 °C to water at 0 °C
= m L
f ice
= (3 kg) (3.35 × 10
5
J kg
–1
)
= 1005000 J
Q
3
= heat required to convert water
at 0 °C to water at 100 °C.
= ms
w
∆T
2
= (3kg) (4186J kg
–1
K
–1
)
(100 °C)
= 1255800 J
Q
4
= heat required to convert water
at 100 °C to steam at 100 °C.
= m L
steam
= (3 kg) (2.256×10
6
J kg
–1
)
= 6768000 J
So, Q = Q
1
+ Q
2
+ Q
3
+ Q
4
= 75600J + 1005000 J
+ 1255800 J + 6768000 J
= 9.1×10
6
J t
11.9 HEAT TRANSFER
We have seen that heat is energy transfer
from one system to another or from one part
of a system to another part, arising due to
temperature difference. What are the different
ways by which this energy transfer takes
place? There are three distinct modes of heat
transfer: conduction, convection and radiation
(Fig. 11.13).
Fig. 11.13 Heating by conduction, convection and
radiation.
11.9.1 Conduction
Conduction is the mechanism of transfer of heat
between two adjacent parts of a body because
of their temperature difference. Suppose, one end
of a metallic rod is put in a flame, the other end
of the rod will soon be so hot that you cannot
hold it by your bare hands. Here, heat transfer
takes place by conduction from the hot end of
the rod through its different parts to the other
end. Gases are poor thermal conductors, while
liquids have conductivities intermediate between
solids and gases.
Heat conduction may be described
quantitatively as the time rate of heat flow in a
material for a given temperature difference.
Consider a metallic bar of length L and uniform
cross-section A with its two ends maintained at
different temperatures. This can be done, for
example, by putting the ends in thermal contact
with large reservoirs at temperatures, say, T
C
and
T
D
, respectively (Fig. 11.14). Let us assume the
ideal condition that the sides of the bar are fully
insulated so that no heat is exchanged between
the sides and the surroundings.
After sometime, a steady state is reached; the
temperature of the bar decreases uniformly with
distance from T
C
to T
D
; (T
C
>T
D
). The reservoir at
C supplies heat at a constant rate, which
transfers through the bar and is given out at
the same rate to the reservoir at D. It is found
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THERMAL PROPERTIES OF MATTER 291
t
experimentally that in this steady state, the rate
of flow of heat (or heat current) H is proportional
to the temperature difference (T
C
– T
D
) and the
area of cross-section
A and is inversely
proportional to the length L :
H = KA
–
C D
T T
L
(11.14)
The constant of proportionality K is called the
thermal conductivity of the material. The
greater the value of K for a material, the more
rapidly will it conduct heat. The SI unit of K is
J s
–1
m
–1
K
–1
or W m
–1
K
–1
. The thermal
conductivities of various substances are listed
in Table 11.6. These values vary slightly with
temperature, but can be considered to be
constant over a normal temperature range.
Compare the relatively large thermal
conductivities of good thermal conductors and,
metals, with the relatively small thermal
conductivities of some good thermal insulators,
such as wood and glass wool. You may have
noticed that some cooking pots have copper
coating on the bottom. Being a good conductor
of heat, copper promotes the distribution of heat
over the bottom of a pot for uniform cooking.
Plastic foams, on the other hand, are good
insulators, mainly because they contain pockets
of air. Recall that gases are poor conductors,
and note the low thermal conductivity of air in
the Table 11.5. Heat retention and transfer are
important in many other applications. Houses
made of concrete roofs get very hot during
summer days because thermal conductivity of
concrete (though much smaller than that of a
metal) is still not small enough. Therefore, people,
usually, prefer to give a layer of earth or foam
insulation on the ceiling so that heat transfer is
prohibited and keeps the room cooler. In some
situations, heat transfer is critical. In a nuclear
reactor, for example, elaborate heat transfer
systems need to be installed so that the
enormous energy produced by nuclear fission
in the core transits out sufficiently fast, thus
preventing the core from overheating.
Table 11.6 Thermal conductivities of some
material
Material Thermal conductivity
(J s
–1
m
–1
K
–1
)
Metals
Silver 406
Copper 385
Aluminium 205
Brass 109
Steel 50.2
Lead 34.7
Mercury 8.3
Non-metals
Insulating brick 0.15
Concrete 0.8
Body fat 0.20
Felt 0.04
Glass 0.8
Ice 1.6
Glass wool 0.04
Wood 0.12
Water 0.8
Gases
Air 0.024
Argon 0.016
Hydrogen 0.14
Example 11.6 What is the temperature of
the steel-copper junction in the steady
state of the system shown in Fig. 11.15.
Length of the steel rod = 15.0 cm, length
of the copper rod = 10.0 cm, temperature
of the furnace = 300 °C, temperature of
the other end = 0 °C. The area of cross
section of the steel rod is twice that of the
copper rod. (Thermal conductivity of steel
= 50.2 J s
–1
m
–1
K
–1
; and of copper
= 385 J s
–1
m
–1
K
–1
).
Fig. 11.14 Steady state heat flow by conduction in
a bar with its two ends maintained at
temperatures T
C
and T
D
; (T
C
> T
D
).
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292 PHYSICS
t
Fig. 11.15
Answer The insulating material around the
rods reduces heat loss from the sides of the rods.
Therefore, heat flows only along the length of
the rods. Consider any cross section of the rod.
In the steady state, heat flowing into the element
must equal the heat flowing out of it; otherwise
there would be a net gain or loss of heat by the
element and its temperature would not be
steady. Thus in the steady state, rate of heat
flowing across a cross section of the rod is the
same at every point along the length of the
combined steel-copper rod. Let T be the
temperature of the steel-copper junction in the
steady state. Then,
(
)
(
)
1 1 2 2
1 2
300 0
K A T K A T –
=
L L
−
where 1 and 2 refer to the steel and copper rod
respectively. For A
1
= 2 A
2
, L
1
= 15.0 cm,
L
2
= 10.0 cm, K
1
= 50.2 J s
–1
m
–1
K
–1
, K
2
= 385 J
s
–1
m
–1
K
–1
, we have
(
)
50.2 2 300
385
15 10
T
T
=
× −
which gives T = 44.4 °C t
Example 11.7 An iron bar (L
1
= 0.1 m, A
1
= 0.02 m
2
, K
1
= 79 W m
–1
K
–1
) and a
brass bar (L
2
= 0.1 m, A
2
= 0.02 m
2
,
K
2
= 109 W m
–1
K
–1
) are soldered end to end
as shown in Fig. 11.16. The free ends of
the iron bar and brass bar are maintained
at 373 K and 273 K respectively. Obtain
expressions for and hence compute (i) the
temperature of the junction of the two bars,
(ii) the equivalent thermal conductivity of
the compound bar, and (iii) the heat
current through the compound bar.
Fig 11.16
Answer
Given, L
1
= L
2
= L = 0.1 m, A
1
= A
2
= A= 0.02 m
2
K
1
= 79 W m
–1
K
–1
, K
2
= 109 W m
–1
K
–1
,
T
1
= 373 K, and T
2
= 273 K.
Under steady state condition, the heat
current (H
1
) through iron bar is equal to the
heat current (H
2
) through brass bar.
So, H = H
1
= H
2
=
(
)
1 1 1 0
2 2 0 2
1 2
–
( – )
K A T T
K A T T
L L
=
For A
1
= A
2
= A and L
1
= L
2
= L, this equation
leads to
K
1
(T
1
– T
0
) = K
2
(T
0
– T
2
)
Thus, the junction temperature T
0
of the two
bars is
T
0
=
(
)
( )
1 1 2 2
1 2
K T K T
K K
+
+
Using this equation, the heat current H through
either bar is
H =
(
)
1 1 0
2 0 2
–
( – )
K A T T
K A T T
L L
=
Using these equations, the heat current H′
through the compound bar of length L
1
+ L
2
= 2L
and the equivalent thermal conductivity K′, of
the compound bar are given by
(
)
1 2
–
2
K A T T
H H
L
′
= =′
1 2
1 2
2
=′
+
K K
K
K K
(i)
(
)
( )
1 1 2 2
0
1 2
K T K T
T
K K
+
=
+
(
)
(
)
(
)
(
)
–1 –1 –1 –1
–1 –1 –1 –1
79 m K 373 K 109 W m K 273 K
79 W m K 109 W m K
W +
=
+
= 315 K
(ii)
1 2
1 2
2
=
K K
K
K K
′
+
=
–1 –1 –1 –1
–1 –1 –1 –1
2×(79 W m K ) ×(109 W m K )
79 W m K +109 W m K
= 91.6 W m
–1
K
–1
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THERMAL PROPERTIES OF MATTER 293
(iii)
(
)
1 2
–
2
K A T T
H H
L
′
= =′
(
)
(
)
(
)
( )
–1 –1 2
91.6 W m K × 0.02 m × 373 K–273 K
2× 0.1 m
=
= 916.1 W t
11.9.2 Convection
Convection is a mode of heat transfer by actual
motion of matter. It is possible only in fluids.
Convection can be natural or forced. In natural
convection, gravity plays an important part.
When a fluid is heated from below, the hot part
expands and, therefore, becomes less dense.
Because of buoyancy, it rises and the upper
colder part replaces it. This again gets heated,
rises up and is replaced by the relatively colder
part of the fluid. The process goes on. This mode
of heat transfer is evidently different from
conduction. Convection involves bulk transport
of different parts of the fluid.
In forced convection, material is forced to move
by a pump or by some other physical means. The
common examples of forced convection systems
are forced-air heating systems in home, the
human circulatory system, and the cooling
system of an automobile engine. In the human
body, the heart acts as the pump that circulates
blood through different parts of the body,
transferring heat by forced convection and
maintaining it at a uniform temperature.
Natural convection is responsible for many
familiar phenomena. During the day, the
ground heats up more quickly than large bodies
of water do. This occurs both because water has
a greater specific heat capacity and because
mixing currents disperse the absorbed heat
throughout the great volume of water. The air
in contact with the warm ground is heated by
conduction. It expands, becoming less dense
than the surrounding cooler air. As a result, the
warm air rises (air currents) and the other air
moves (winds) to fill the space-creating a sea
breeze near a large body of water. Cooler air
descends, and a thermal convection cycle is set
up, which transfers heat away from the land.
At night, the ground loses its heat more quickly,
and the water surface is warmer than the land.
As a result, the cycle is reveresed (Fig. 11.17).
The other example of natural convection is
the steady surface wind on the earth blowing
in from north-east towards the equator, the
so-called trade wind. A resonable explanation
is as follows: the equatorial and polar regions of
the earth receive unequal solar heat. Air at the
earth’s surface near the equator is hot, while
the air in the upper atmosphere of the poles is
cool. In the absence of any other factor, a
convection current would be set up, with the
air at the equatorial surface rising and moving
out towards the poles, descending and
streaming in towards the equator. The rotation
of the earth, however, modifies this convection
current. Because of this, air close to the equator
has an eastward speed of 1600 km/h, while it
is zero close to the poles. As a result, the air
descends not at the poles but at 30
°
N (North)
latitude and returns to the equator. This is
called trade wind.
Fig. 11.17 Convection cycles.
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294 PHYSICS
11.9.3 Radiation
Conduction and convection require some
material as a transport medium. These modes
of heat transfer cannot operate between bodies
separated by a distance in vacuum. But the
earth does receive heat from the Sun across a
huge distance. Similarly, we quickly feel the
warmth of the fire nearby even though air
conducts poorly and before convection takes
some time to set in. The third mechanism for
heat transfer needs no medium; it is called
radiation and the energy so transferred by
electromagnetic waves is called radiant energy.
In an electromagnetic wave, electric and
magnetic fields oscillate in space and time. Like
any wave, electromagnetic waves can have
different wavelengths and can travel in vacuum
with the same speed, namely the speed of light
i.e., 3 × 10
8
m s
–1
. You will learn these matters
in more detail later, but you now know why heat
transfer by radiation does not need any medium
and why it is so fast. This is how heat is
transferred to the earth from the Sun through
empty space. All bodies emit radiant energy,
whether they are solid, liquid or gas. The
electromagnetic radiation emitted by a body by
virtue of its temperature, like radiation by a red
hot iron or light from a filament lamp is called
thermal radiation.
When this thermal radiation falls on other
bodies, it is partly reflected and partly absorbed.
The amount of heat that a body can absorb by
radiation depends on the colour of the body.
We find that black bodies absorb and emit
radiant energy better than bodies of lighter
colours. This fact finds many applications in our
daily life. We wear white or light coloured clothes
in summer, so that they absorb the least heat
from the Sun. However, during winter, we use
dark coloured clothes, which absorb heat from
the sun and keep our body warm. The bottoms of
utensils for cooking food are blackened so that
they absorb maximum heat from fire and transfer
it to the vegetables to be cooked.
Similarly, a Dewar flask or thermos bottle is
a device to minimise heat transfer between the
contents of the bottle and outside. It consists
of a double-walled glass vessel with the inner
and outer walls coated with silver. Radiation
from the inner wall is reflected back to the
contents of the bottle. The outer wall similarly
reflects back any incoming radiation. The space
between the walls is evacuted to reduce
conduction and convection losses and the flask
is supported on an insulator, like cork. The
device is, therefore, useful for preventing hot
contents (like, milk) from getting cold, or
alternatively, to store cold contents (like, ice).
11.9.4 Blackbody Radiation
We have so far not mentioned the wavelength
content of thermal radiation. The important
thing about thermal radiation at any
temperature is that it is not of one (or a few)
wavelength(s) but has a continuous spectrum
from the small to the long wavelengths. The
energy content of radiation, however, varies for
different wavelengths. Figure 11.18 gives the
experimental curves for radiation energy per unit
area per unit wavelength emitted by a blackbody
versus wavelength for different temperatures.
Fig. 11.18: Energy emitted versus wavelength
for a blackbody at different
temperatures
Notice that the wavelength λ
m
for which energy
is the maximum decreases with increasing
temperature. The relation between λ
m
and T is
given by what is known as Wien’s Displacement
Law:
λ
m
T = constant (11.15)
The value of the constant (Wien’s constant)
is 2.9 × 10
–3
m K. This law explains why the
colour of a piece of iron heated in a hot flame
first becomes dull red, then reddish yellow, and
finally white hot. Wien’s law is useful for
estimating the surface temperatures of celestial
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THERMAL PROPERTIES OF MATTER 295
bodies like, the moon, Sun and other stars. Light
from the moon is found to have a maximum
intensity near the wavelength 14 µm. By Wien’s
law, the surface of the moon is estimated to have
a temperature of 200 K. Solar radiation has a
maximum at λ
m
= 4753 Å. This corresponds to
T = 6060 K. Remember, this is the temperature
of the surface of the sun, not its interior.
The most significant feature of the
blackbody radiation curves in Fig. 11.18 is that
they are universal. They depend only on the
temperature and not on the size, shape or
material of the blackbody. Attempts to explain
blackbody radiation theoretically, at the
beginning of the twentieth century, spurred the
quantum revolution in physics, as you will
learn in later courses.
Energy can be transferred by radiation over
large distances, without a medium (i.e., in
vacuum). The total electromagnetic energy
radiated by a body at absolute temperature T
is proportional to its size, its ability to radiate
(called emissivity) and most importantly to its
temperature. For a body, which is a perfect
radiator, the energy emitted per unit time (H)
is given by
H = A
σ
T
4
(11.16)
where A is the area and T is the absolute
temperature of the body. This relation obtained
experimentally by Stefan and later proved
theoretically by Boltzmann is known as Stefan-
Boltzmann law and the constant
σ
is called
Stefan-Boltzmann constant. Its value in SI units
is 5.67 × 10
–8
W m
–2
K
–4
. Most bodies emit only a
fraction of the rate given by Eq. 11.16. A substance
like lamp black comes close to the limit. One,
therefore, defines a dimensionless fraction e
called emissivity and writes,
H = Ae
σ
T
4
(11.17)
Here, e = 1 for a perfect radiator. For a tungsten
lamp, for example, e is about 0.4. Thus, a tungsten
lamp at a temperature of 3000 K and a surface
area of 0.3 cm
2
radiates at the rate H = 0.3 ×
10
–4
× 0.4 × 5.67 × 10
–8
× (3000)
4
= 60 W.
A body at temperature T, with surroundings
at temperatures T
s
, emits, as well as, receives
energy. For a perfect radiator, the net rate of
loss of radiant energy is
H =
σ
A (T
4
– T
s
4
)
For a body with emissivity e, the relation
modifies to
H = e
σ
A (T
4
– T
s
4
) (11.18)
As an example, let us estimate the heat
radiated by our bodies. Suppose the surface area
of a person’s body is about 1.9 m
2
and the room
temperature is 22°C. The internal body
temperature, as we know, is about 37 °C. The
skin temperature may be 28°C (say). The
emissivity of the skin is about 0.97 for the
relevant region of electromagnetic radiation. The
rate of heat loss is:
H = 5.67 × 10
–8
× 1.9 × 0.97 × {(301)
4
– (295)
4
}
= 66.4 W
which is more than half the rate of energy
production by the body at rest (120 W). To
prevent this heat loss effectively (better than
ordinary clothing), modern arctic clothing has
an additional thin shiny metallic layer next to
the skin, which reflects the body’s radiation.
11.9.5 Greenhouse Effect
The earth’s surface is a source of thermal
radiation as it absorbs energy received from the
Sun. The wavelength of this radiation lies in the
long wavelength (infrared) region. But a large
portion of this radiation is absorbed by
greenhouse gases, namely, carbon dioxide
(CO
2
); methane (CH
4
); nitrous oxide (N
2
O);
chlorofluorocarbon (CF
x
Cl
x
); and tropospheric
ozone (O
3
). This heats up the atmosphere which,
in turn, gives more energy to earth, resulting in
warmer surface. This increases the intensity of
radiation from the surface. The cycle of
processes described above is repeated until no
radiation is available for absorption. The net
result is heating up of earth’s surface and
atmosphere. This is known as Greenhouse
Effect. Without the Greenhouse Effect, the
temperature of the earth would have been –18°C.
Concentration of greenhouse gases has
enhanced due to human activities, making the
earth warmer. According to an estimate, average
temperature of earth has increased by 0.3 to
0.6°C, since the beginning of this century
because of this enhancement. By the middle of
the next century, the earth’s global temperature
may be 1 to 3°C higher than today. This global
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296 PHYSICS
warming may cause problem for human life,
plants and animals. Because of global warming,
ice caps are melting faster, sea level is rising,
and weather pattern is changing. Many coastal
cities are at the risk of getting submerged. The
enhanced Greenhouse Effect may also result in
expansion of deserts. All over the world, efforts
are being made to minimise the effect of global
warming.
11.10 NEWTON’S LAW OF COOLING
We all know that hot water or milk when left on
a table begins to cool, gradually. Ultimately it
attains the temperature of the surroundings. To
study how slow or fast a given body can cool on
exchanging heat with its surroundings, let us
perform the following activity.
Take some water, say 300 mL, in a
calorimeter with a stirrer and cover it with a
two-holed lid. Fix the stirrer through one hole
and fix a thermometer through another hole
in the lid and make sure that the bulb of
thermometer is immersed in the water. Note
the reading of the thermometer. This reading
T
1
is the temperature of the surroundings.
Heat the water kept in the calorimeter till it
attains a temperature, say 40 °C above room
temperature (i.e., temperature of the
surroundings). Then, stop heating the water
by removing the heat source. Start the
stop-watch and note the reading of the
thermometer after a fixed interval of time, say
after every one minute of stirring gently with
the stirrer. Continue to note the temperature
(T
2
) of water till it attains a temperature about
5 °C above that of the surroundings. Then, plot
a graph by taking each value of temperature
∆T = T
2
– T
1
along y-axis and the coresponding
value of t along x-axis (Fig. 11.19).
Fig. 11.19 Curve showing cooling of hot water
with time.
From the graph you can infer how the cooling
of hot water depends on the difference of its
temperature from that of the surroundings. You
will also notice that initially the rate of cooling
is higher and decreases as the temperature of
the body falls.
The above activity shows that a hot body loses
heat to its surroundings in the form of heat
radiation. The rate of loss of heat depends on
the difference in temperature between the body
and its surroundings. Newton was the first to
study, in a systematic manner, the relation
between the heat lost by a body in a given
enclosure and its temperature.
According to Newton’s law of cooling, the rate
of loss of heat, – dQ/dt of the body is directly
proportional to the difference of temperature
∆T = (T
2
–T
1
) of the body and the surroundings.
The law holds good only for small difference of
temperature. Also, the loss of heat by radiation
depends upon the nature of the surface of the
body and the area of the exposed surface. We
can write
–
(11.19)
where k is a positive constant depending upon
the area and nature of the surface of the body.
Suppose a body of mass m and specific heat
capacity s is at temperature T
2
. Let T
1
be the
temperature of the surroundings. If the
temperature falls by a small amount dT
2
in time
dt, then the amount of heat lost is
dQ = ms dT
2
∴ Rate of loss of heat is given by
dQ
dt
ms
dT
dt
=
2
(11.20)
From Eqs. (11.15) and (11.16) we have
– ( – )m s
dT
dt
k T T
2
2 1
=
dT
T T
k
ms
dt K dt
2
2 1
–
– –= =
(11.21)
where K = k/m s
On integrating,
log
e
(T
2
– T
1
) = – K t + c (11.22)
or T
2
= T
1
+ C
′
e
–Kt
; where C
′
= e
c
(11.23)
Equation 11.23 enables you to calculate the
time of cooling of a body through a particular
range of temperature.
∆
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THERMAL PROPERTIES OF MATTER 297
t
For small temperature differences, the rate
of cooling, due to conduction, convection, and
radiation combined, is proportional to the
difference in temperature. It is a valid
approximation in the transfer of heat from a
radiator to a room, the loss of heat through the
wall of a room, or the cooling of a cup of tea on
the table.
Fig. 11.20 Verification of Newton’s Law of cooling.
Newton’s law of cooling can be verified with
the help of the experimental set-up shown in
Fig. 11.20(a). The set-up consists of a double-
walled vessel (V) containing water between
the two walls. A copper calorimeter (C)
containing hot water is placed inside the
double-walled vessel. Two thermometers
through the corks are used to note the
temperatures T
2
of water in calorimeter and
T
1
of hot water in between the double walls,
respectively. Temperature of hot water in the
calorimeter is noted after equal intervals of
time. A graph is plotted between log
e
(T
2
–T
1
)
[or ln(T
2
–T
1
)] and time (t). The nature of the
graph is observed to be a straight line having
a negative slope as shown in Fig. 11.20(b). This
is in support of Eq. 11.22.
Example 11.8 A pan filled with hot food
cools from 94 °C to 86 °C in 2 minutes when
the room temperature is at 20 °C. How long
will it take to cool from 71 °C to 69 °C?
Answer The average temperature of 94 °C and
86 °C is 90 °C, which is 70 °C above the room
temperature. Under these conditions the pan
cools 8 °C in 2 minutes.
Using Eq. (11.21), we have
Change in temperature
Time
K T
= ∆
( )
°
°
8 C
= 70 C
2 min
K
The average of 69 °C and 71 °C is 70 °C, which
is 50 °C above room temperature. K is the same
for this situation as for the original.
°
2 C
Time
= K (50 °C)
When we divide above two equations, we
have
8 C/2 min (70 C)
=
2 C/time (50 C)
K
K
° °
° °
Time = 0.7 min
= 42 s t
SUMMARY
1. Heat is a form of energy that flows between a body and its surrounding medium by
virtue of temperature difference between them. The degree of hotness of the body is
quantitatively represented by temperature.
2. A temperature-measuring device (thermometer) makes use of some measurable property
(called thermometric property) that changes with temperature. Different thermometers
lead to different temperature scales. To construct a temperature scale, two fixed points
are chosen and assigned some arbitrary values of temperature. The two numbers fix
the origin of the scale and the size of its unit.
3. The Celsius temperature (t
C
) and the Farenheit temperare (t
F
)are related by
t
F
= (9/5) t
C
+ 32
4. The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T)
is :
PV =
µ
RT
where
µ
is the number of moles and R is the universal gas constant.
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298 PHYSICS
5. In the absolute temperature scale, the zero of the scale corresponds to the temperature
where every substance in nature has the least possible molecular activity. The Kelvin
absolute temperature scale (T ) has the same unit size as the Celsius scale (T
c
), but
differs in the origin :
T
C
= T – 273.15
6. The coefficient of linear expansion (
α
l
) and volume expansion (
α
v
) are defined by the
relations :
l
l
T
l
α
∆
= ∆
V
V
T
V
α
∆
= ∆
where ∆l and ∆V denote the change in length l and volume V for a change of temperature
∆T. The relation between them is :
α
v
= 3
α
l
7. The specific heat capacity of a substance is defined by
s
m
Q
T
=
1
∆
∆
where m is the mass of the substance and ∆Q is the heat required to change its
temperature by ∆T. The molar specific heat capacity of a substance is defined by
1
Q
C
T
µ
∆
=
∆
where
µ
is the number of moles of the substance.
8. The latent heat of fusion (L
f
) is the heat per unit mass required to change a substance
from solid into liquid at the same temperature and pressure. The latent heat of
vaporisation (L
v
) is the heat per unit mass required to change a substance from liquid
to the vapour state without change in the temperature and pressure.
9. The three modes of heat transfer are conduction, convection and radiation.
10. In conduction, heat is transferred between neighbouring parts of a body through
molecular collisions, without any flow of matter. For a bar of length L and uniform
cross section A with its ends maintained at temperatures T
C
and T
D
, the rate of flow of
heat H is :
C D
T T
H = K A
L
−
where K is the thermal conductivity of the material of the bar.
11. Newton’s Law of Cooling says that the rate of cooling of a body is proportional to the
excess temperature of the body over the surroundings :
2 1
d
( )
d
Q
= – k T – T
t
Where T
1
is the temperature of the surrounding medium and T
2
is the temperature of
the body.
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THERMAL PROPERTIES OF MATTER 299
POINTS TO PONDER
1. The relation connecting Kelvin temperature (T ) and the Celsius temperature t
c
T = t
c
+ 273.15
and the assignment T = 273.16 K for the triple point of water are exact relations (by
choice). With this choice, the Celsius temperature of the melting point of water and
boiling point of water (both at 1 atm pressure) are very close to, but not exactly equal
to 0 °C and 100 °C respectively. In the original Celsius scale, these latter fixed points
were exactly at 0 °C and 100 °C (by choice), but now the triple point of water is the
preferred choice for the fixed point, because it has a unique temperature.
2. A liquid in equilibrium with vapour has the same pressure and temperature throughout
the system; the two phases in equilibrium differ in their molar volume (i.e. density).
This is true for a system with any number of phases in equilibrium.
3. Heat transfer always involves temperature difference between two systems or two parts
of the same system. Any energy transfer that does not involve temperature difference
in some way is not heat.
4. Convection involves flow of matter within a fluid due to unequal temperatures of its
parts. A hot bar placed under a running tap loses heat by conduction between the
surface of the bar and water and not by convection within water.
EXERCISES
11.1 The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively.
Express these temperatures on the Celsius and Fahrenheit scales.
11.2 Two absolute scales A and B have triple points of water defined to be 200 A and 350
B. What is the relation between T
A
and T
B
?
11.3 The electrical resistance in ohms of a certain thermometer varies with temperature
according to the approximate law :
R = R
o
[1 +
α
(T – T
o
)]
The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the
normal melting point of lead (600.5 K). What is the temperature when the resistance
is 123.4 Ω ?
11.4 Answer the following :
(a) The triple-point of water is a standard fixed point in modern thermometry.
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300 PHYSICS
Why ? What is wrong in taking the melting point of ice and the boiling point of
water as standard fixed points (as was originally done in the Celsius scale) ?
(b) There were two fixed points in the original Celsius scale as mentioned above
which were assigned the number 0 °C and 100 °C respectively. On the absolute
scale, one of the fixed points is the triple-point of water, which on the Kelvin
absolute scale is assigned the number 273.16 K. What is the other fixed point
on this (Kelvin) scale ?
(c) The absolute temperature (Kelvin scale) T is related to the temperature t
c
on
the Celsius scale by
t
c
= T – 273.15
Why do we have 273.15 in this relation, and not 273.16 ?
(d) What is the temperature of the triple-point of water on an absolute scale
whose unit interval size is equal to that of the Fahrenheit scale ?
11.5 Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The
following observations are made :
Temperature Pressure Pressure
thermometer A thermometer B
Triple-point of water 1.250 × 10
5
Pa 0.200 × 10
5
Pa
Normal melting point 1.797 × 10
5
Pa 0.287 × 10
5
Pa
of sulphur
(a) What is the absolute temperature of normal melting point of sulphur as read
by thermometers A and B ?
(b) What do you think is the reason behind the slight difference in answers of
thermometers A and B ? (The thermometers are not faulty). What further
procedure is needed in the experiment to reduce the discrepancy between the
two readings ?
11.6 A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The
length of a steel rod measured by this tape is found to be 63.0 cm on a hot day
when the temperature is 45.0 °C. What is the actual length of the steel rod on that
day ? What is the length of the same steel rod on a day when the temperature is
27.0 °C ? Coefficient of linear expansion of steel = 1.20 × 10
–5
K
–1
.
11.7 A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the
outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the
wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the
shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of
the steel to be constant over the required temperature range :
α
steel
= 1.20 × 10
–5
K
–1
.
11.8 A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C.
What is the change in the diameter of the hole when the sheet is heated to 227 °C?
Coefficient of linear expansion of copper = 1.70 × 10
–5
K
–1
.
11.9 A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid
supports. If the wire is cooled to a temperature of –39 °C, what is the tension
developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion
of brass = 2.0 × 10
–5
K
–1
; Young’s modulus of brass = 0.91 × 10
11
Pa.
11.10 A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same
length and diameter. What is the change in length of the combined rod at 250 °C, if
the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the
junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of
brass = 2.0 × 10
–5
K
–1
, steel = 1.2 × 10
–5
K
–1
).
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THERMAL PROPERTIES OF MATTER 301
11.11 The coefficient of volume expansion of glycerine is 49 × 10
–5
K
–1
. What is the
fractional change in its density for a 30 °C rise in temperature ?
11.12 A 10 kW drilling machine is used to drill a bore in a small aluminium block of
mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes,
assuming 50% of power is used up in heating the machine itself or lost to the
surroundings. Specific heat of aluminium = 0.91 J g
–1
K
–1
.
11.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and
then placed on a large ice block. What is the maximum amount of ice that can
melt? (Specific heat of copper = 0.39 J g
–1
K
–1
; heat of fusion of water
= 335 J g
–1
).
11.14 In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at
150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing
150 cm
3
of water at 27 °C. The final temperature is 40 °C. Compute the specific
heat of the metal. If heat losses to the surroundings are not negligible, is your
answer greater or smaller than the actual value for specific heat of the metal ?
11.15 Given below are observations on molar specific heats at room temperature of some
common gases.
Gas Molar specific heat (C
v
)
(cal mo1
–1
K
–1
)
Hydrogen 4.87
Nitrogen 4.97
Oxygen 5.02
Nitric oxide 4.99
Carbon monoxide 5.01
Chlorine 6.17
The measured molar specific heats of these gases are markedly different from
those for monatomic gases. Typically, molar specific heat of a monatomic gas is
2.92 cal/mol K. Explain this difference. What can you infer from the somewhat
larger (than the rest) value for chlorine ?
11.16 A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that
lowers fever) which causes an increase in the rate of evaporation of sweat from his
body. If the fever is brought down to 98 °F in 20 minutes, what is the average rate
of extra evaporation caused, by the drug. Assume the evaporation mechanism to
be the only way by which heat is lost. The mass of the child is 30 kg. The specific
heat of human body is approximately the same as that of water, and latent heat of
evaporation of water at that temperature is about 580 cal g
–1
.
11.17 A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities
of cooked food in summer in particular. A cubical icebox of side 30 cm has a
thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice
remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal
conductivity of thermacole is 0.01 J s
–1
m
–1
K
–1
. [Heat of fusion of water = 335 × 10
3
J kg
–1
]
11.18 A brass boiler has a base area of 0.15 m
2
and thickness 1.0 cm. It boils water at the
rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part
of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s
–1
m
–1
K
–1
;
Heat of vaporisation of water = 2256 × 10
3
J kg
–1
.
11.19 Explain why :
(a) a body with large reflectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal
black body radiation gives too low a value for the temperature of a red hot
iron piece in the open, but gives a correct value for the temperature when the
same piece is in the furnace
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302 PHYSICS
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more efficient in warming
a building than those based on circulation of hot water
11.20 A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool
from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.
ADDITIONAL EXERCISES
11.21 Answer the following questions based on the P-T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of
CO
2
co-exist in equilibrium ?
(b) What is the effect of decrease of pressure on the fusion and boiling point of
CO
2
?
(c) What are the critical temperature and pressure for CO
2
? What is their
significance ?
(d) Is CO
2
solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm,
(c) 15 °C under 56 atm ?
11.22 Answer the following questions based on the P – T phase diagram of CO
2
:
(a) CO
2
at 1 atm pressure and temperature – 60 °C is compressed isothermally.
Does it go through a liquid phase ?
(b) What happens when CO
2
at 4 atm pressure is cooled from room temperature
at constant pressure ?
(c) Describe qualitatively the changes in a given mass of solid CO
2
at 10 atm
pressure and temperature –65 °C as it is heated up to room temperature at
constant pressure.
(d) CO
2
is heated to a temperature 70 °C and compressed isothermally. What
changes in its properties do you expect to observe ?
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