2.

Heat

Thermal Conductivity – experimental determination of

thermal conductivity of good and bad conductors –
Forbe’s method – theory and experiment – Lee’s disc
method for bad conductors.

Introduction

The study of thermal physics is important because of its

applications in various fields of engineering.

 Mechanical Engineering

In designs of internal and external combustion engines,

refrigeration and air-conditioning plants, heat-exchangers,
coolers, condensers, furnaces, preheaters, etc.

 Electrical Engineering

In designing cooling system for motors, transformers and

generators.

 Civil Engineering

Control of heat transfer in dams, structures, buildings and

tunnels.

 Chemical Engineering

The application of heat transfer is in freezing, boiling,

evaporation and condensation processes.

Transfer of Heat Energy

Heat transfer is a discipline of thermal engineering that

concerns with exchange of thermal energy or heat between
physical systems.
2.2 Engineering Physics

Fig. 2.1 Modes of transfer heat of energy

When the temperature of a body increases, the energy

supplied to the body is being stored in the form of thermal or
heat energy.
In the normal process, the transmission of heat takes place
from a region of higher temperature to a region of lower
temperature. There are three modes of transmission of heat
(Fig 2.1).

They are

(i) Conduction
(ii) Convection
(iii) Radiation

Thermal Conduction, Convection and Radiation

Thermal Conduction:
It is the process in which heat is transferred from
one point to another through the substance without the
actual motion of the particles.

Thermal Convection:
It is the process in which heat is transmitted from
one place to another by the actual motion of the heated
particles.
Heat 2.3

Example: Let us consider a beaker of water heated by a

flame as shown in figure 2.2. The water in the central portion
at the bottom of beaker gets heated first.

Fig. 2.2 Process of thermal convection

It rises up and the water from the top comes down along
the sides to get heated. This upward and downward motion can
be made visible by placing a crystal of potassium permanganate
at the bottom of the beaker.

The hot air furnace, hot water heating system and the flow
of blood in the body are examples of convection.

This takes places in case of liquids or gases only.

Thermal Radiation
It is the process in which heat is transmitted from one
place to the other directly, without any material medium.

Example: The heat radiation from the sun reaches us

with an enormous velocity of light without any intervening
medium as shown in fig. 2.3.

Fig. 2.3 Process of Thermal Radiation

2.4 Engineering Physics

Though the sun is millions of miles away from the earth

and there is no material medium for the greater part of the
distance, the heat reaches us with the velocity of light.
Thus, heat radiations can pass through vacuum. The
properties of heat radiations are similar to light radiation.
Heat Conduction in S olids
It is a well known fact that heat is conducted through the
material of a body.
The heat is transmitted from a body of higher temperature
to that of lower temperature.
As an example, when a metal rod is heated at one end,
heat gradually flows along the length of the rod and other end
of the rod also becomes hot after some time. (Fig. 2.4)

Fig. 2.4 Heat conduction in a solid

This shows that heat has travelled through the molelcules

of the rod from one end to other. The molecules in the rod
remain fixed in their mean positions.

On heating, the energy of molecules increases and they

start vibrating more about their mean positions. They collide
with the neighbouring molecules. Because of this collision, the
neighbouring molecules are set into vibration.
Heat 2.5

Each molecule thus transfers some of heat it receives from its

predecessor to its successor. Thus, the transmission of heat takes
place by molecular vibrations in the case of conduction.
Note
Heat conduction always requires some material medium. The
material medium must be solid. As it requires material medium,
the heat conduction process never takes place in vacuum.

2.1 THERMAL CONDUCTIVITY

The ability of a substance to conduct heat energy is
measured by its thermal conductivity.

Expression for Thermal Conductivity

Consider a slab of material of length x metre (thickness)
and area of cross-section A as shown in fig. 2.5.

Fig. 2.5 Heat conduction in a slab of material

One end of the slab is maintained at a higher temperature

1 (hot end) and the other end at a lower temperature 2 (cold
end). Now, heat flows from hot end to cold end.

It is found that the amount of heat Q conducted from

one end to the other end is

 directly proportional to area of cross-section A.

 directly proportional to temperature difference between
the ends 1  2.
2.6 Engineering Physics

 directly proportional to time of conduction t.

 inversely proportional to length x between the faces
(shaded in fig. 2.5)
i.e., Q  A

Q  1  2

Q  t

1
Q 
x

Combining all these factors, we have

A 1  2 t ...(1)

Q 
x

K A 1  2 t
Q 
x ...(2)

where K is a proportionality constant. It is known as coefficient

of thermal conductivity or simply thermal conductivity.
Its value depends on the nature of material.

Qx
K 
A 1  2 t ...(3)

2
If A  1 m 1  2  1 kelvin

x  1 metre t  1 second

Then, K  Q

This condition defines the coefficient of thermal

conductivity.
Heat 2.7

Definition
It is defined as the amount of heat conducted per
second normally across unit area of cross - section of the
material per unit temperature difference per unit length.

1  2
The quantity denotes the rate of fall of
x
temperature with respect to distance. It is known as
temperature gradient.

1  2 d
For the smaller values, is written as
x dx

Rewriting the expression (2), we have

d ...(3)
Q  KA t
dx

d
To indicate is negative, a negative sign is included in
dx
the R.H.S of the equation, since it signifies that the temperature
decreases with distance.

Unit:
Qx
We know that K 
A 1  2 t

Substituting the corresponding units, we have

joule  metre
 2
metre  kelvin  second
joule

second  metre  kelvin
watt W . . 1
   . W  Js 
metre  kelvin mK
1 1
Wm K
1 1
Therefore, the unit of thermal conductivity is W m K .
2.8 Engineering Physics

Note
Thermal conductivity denotes the heat conducting
characteristics of the substances. Generally metals are good
conductors of heat (e.g., silver, copper) and non-metals are bad
conductors of heat (Air, glass, wood).
Thermal conductivities of some common materials are given
in the table. 2.1.
Table 2.1

Thermal conductivity
S.No. Material
Wm  1 K  1
1. Copper 385

2. Aluminium 201

3. Silver 419

4. Wood 0.15

5. Glass 1.0

6. Card board 0.04

The knowledge about thermal conductivity of the

material is very much essential in selecting the materials
for suitable engineering applications (design and
construction).

UNIVERSITY SOLVED PROBLEMS

Problem 2.1
4 2
A rod 0.25 m long and 0.892  10 m area of cross
section is heated at one end through 393 K while the
other end is kept at 323 K. The quantity of heat which
3
will flow in 15 minutes along the rod is 8.811  10 joule.
Calculate thermal conductivity of the rod.
(U.Q. Dec 2010)
Heat 2.9

Given data
4 2
Area of cross section of the rod A  0.892  10 m

Distance between two ends of the rod x  0.25 m

Temperature difference 1  2  393 K  323 K  70 K

3
Quantity of heat conducted Q  8.811  10 joule

Time of flow of heat t  15 minutes  15  60

 900 second

Solution
Qx
We know that K 
A 1  2 t

Substituting the given values, we have

3
8.811  10  0.25
K  4
0.892  10  70  900
1 1
Thermal conductivity K  392 Wm K

Problem 2.2

How much heat will be conducted through a slab of area

4 2 –3
90  10 m and thickness 1.2  10 m in one second
when its opposite faces are maintained at difference in
temperature of 20 K. The coefficient of thermal
1 1
conductivity of that material is 0.04 Wm K
[U.Q. April 2011]

Given data
4 2
Area of the slab A  90  10 m
3
Thickness of the slab x  1.2  10 m
2.10 Engineering Physics

Temperature difference 1  2  20 K

1 1
Thermal conductivity K  0.04 Wm K

Time taken t  1 second

Solution:
Amount of heat conducted

K A 1  2 t
Q 
x

Substituting the given values, we have

4
0.04  90  10  20  1
Q  3
1.2  10

Q  6 joule

Amount of heat conducted in one second

= 6 joule.

Problem 2.3
2
The total area of the glass window is 0.5 m . Calculate
how much heat is conducted per hour through the glass
–3
window if thickness of the glass is 7  10 m the
o
temperature of the inside surface is 25 C and of the
o
outside surface is 40 C. Thermal conductivity of glass is
–1 –1
1.0 Wm K [U.Q. Jan 2011]

Given data
2
Area of glass window A  0.5 m
3
Thickness of the glass x  7  10 m
1  40C  273  40  313 K

2  25C  273  25  298 K

Heat 2.11

1 1
Thermal conductivity K  1 Wm K

t  1 hour  60  60 second

 3600 second

Solution

Amount of heat conducted

K A 1  2 t
Q 
x

1  0.5  313  298  3600

Q  3
7  10

27000
Q  3
7  10
3
Q  3857  10 J
6
Q  3.857  10 J

2.2 EXPERIMENTAL DETERMINATION OF THERMAL

CONDUCTIVITY OF GOOD AND BAD CONDUCTORS

Methods to determine thermal Conductivity

The thermal conductivity of a material is determined by

various methods

1. Searle’s method - for good conductors like

metallic rods
2. Forbe’s method - for determining the absolute
conductivity of metals.
3. Lee’s disc method - for bad conductors
4. Radial flow method - for bad conductors.
2.12 Engineering Physics

2.3 FORBES METHOD - THEORY AND EXPERIMENT

This is one of the earliest method to find the absolute
thermal conductivity of metals.

Theory of the experiment

Consider a long rod. This rod is heated at one end and a
steady state is reached after some time.

Amount of heat flowing per second across the cross-section

A at the point B

 d 
 KA 
 dx B

where K – thermal conductivity

A – cross sectional area
 d 
 dx  – temperature gradient at B.
 B

This amount of heat flowing across the section B is equal

to the heat lost by radiation by the rod beyond the section B.

Consider an element of thickness dx of the rod.

Mass of the element  A dx 

where  - density of the rod

[Mass  Volume  density]

Heat lost by the element per second

 Mass  specific heat capacity  rate of fall of temperature
d
 Adx   S 
dt
where d
– rate of fall of temperature of the element
dt
S – specific heat capacity of the rod
Heat 2.13

Total heat lost by the portion of the rod between section

B and the end C

C
d
  Adx  S
dt
B

Amount of heat flowing per Heat lost by radiation by

second across the cross = the rod beyond the
section at the point B. section B.

C
 d  d
 KA     Adx  S
 dx B B
dt

C
d
S  dt
dx
B
or K 
 d 
 dx 
 B

Experiment consists of two parts

 d 
1. Static experiment to find  
 dx B

C
 d  d
2. Dynamic experiment to find   and  dx
 dt  B
dt

1. Static experiment

The specimen metal is taken in the form of a long rod.

One end of this rod is heated by a steam chamber. The rod
has a series of holes into which thermometers are fitted. These
thermometers record temperatures at different points along the
rod. (Fig. 2.6)
2.14 Engineering Physics

Fig. 2.6 Forbe’s method - Static experiment

When the steady state is reached, the temperature 

shown by the thermometers of the rod and their respective
distances x from the hot end are noted.
A graph is drawn between the temperature  and ...(5)
the
distance x from the hot end (fig. 2.7).

Fig. 2.7 Graph between temperature  and

distances from the hot end x

 d 
The value of   is obtained by drawing a tangent to
 dx B
the curve at a point B.
If this tangent makes an angle  with the x axis, then
from the graph.
 d  AB
 dx   BC  tan 
 B
Heat 2.15

2. Dynamic experiment
A piece of the original rod is heated to the same
temperature as that of the hot end in the static experiment.
The heated piece of the rod is suspended in air.
Now, it is allowed to cool. Its temperature is noted at
regular intervals of time by a thermometer placed in a hole at
the centre. (Fig. 2.8)

Fig. 2.8 Forbe’s method - Dynamic experiment

A graph is drawn between temperature  and time t

(fig. 2.9).

Fig. 2.9 Graph between temperature  and time t

d
From this graph, the value of for various values of 
dt
are determined by drawing tangents at various points of the
cooling curve.

From the graph between temperature  and the distances

of hot end x (fig. 2.7) for various values of temperature  are
d
obtained. Now, third graph is drawn between and the
dt
corresponding values of x. (fig. 2.10)
2.16 Engineering Physics

d
Fig. 2.10 Graph between and x
dt

The curve is extended to meet the x - axis. Corresponding

to the point B, a point is located on the curve.

The area bounded by the curve, x - axis and the ordinate

C
d
passing through B is  dt
dx.
B

The area of the shaded portion is determined.

C
d
S  dt
dx
B
We know that K 
 d 
 dx 
 B

substituting the values in the above equation, we have

 S  Area of the shaded portion

K 
tan 

Hence, K is determined
Heat 2.17

Merits

 It is one of the earliest method to determine the absolute

thermal conductivity of the material.
 This method is based on the fundamental relation
Q
K which defines thermal conductivity.
 dQ 
A  t
 dx 

Demerits

 It is tedious method and requires a lot of time for the

completion of the experiment and drawing the three
graphs.
 The specific heat capacity of the material of the rod does
not remain constant at different temperatures as assumed.
 The distribution of heat is not uniform along the bar in
the two experiments. Therefore, this experiment is not
accurate.

2.4 LEE’S DISC METHOD FOR BAD CONDUCTORS

The thermal conductivity of bad conductors like
glass, ebonite or card board is determined by this method.

Description
The apparatus consists of a circular metal disc or slab
C (Lee’s disc) of radius r and thickness h suspended by the
strings from a stand (fig. 2.11).
The given bad conductor (such as glass, ebonite) is taken in
the form of a disc (D). This bad conductor has the same diameter
as that of the slab and it is placed on the slab. (Lee’s disc)
A cylindrical hollow steam chamber A having the same
diameter as that of the slab is placed on the bad conductor.
There are holes in steam chamber and slab into which
thermometers T1 and T2 are inserted to record the respective
temperatures.
2.18 Engineering Physics

Fig. 2.11 Lee’s disc method

Working
Steam is passed into the steam chamber until the
temperatures in the chamber and the slab are steady. When
thermometers T1 and T2 show steady temperatures, their
readings 1 and 2 respectively are noted. The radius r and
thickness d of the disc D are also measured.

Observation and calculation

Thickness of the bad conductor  d

Radius of the bad conductor and metal disc  r

Mass of the slab (Lee’s disc)  M

Thickness of the metal disc  h

Steady temperature in the steam chamber  1

Steady temperature in the slab  2

Thermal conductivity of the bad conductor  K

Heat 2.19

Rate of cooling of the slab at 2  R

Specific heat capacity of the slab  S

2
Area of the cross section Ar

Amount of heat conducted through the disc D per second

2
KA 1  2 K  r  1  2 ...(1)
Q  
d d

At this stage all the heat conducted through the bad

conductor is completely radiated by the bottom flat surface and
the curved surface of the slab C.

Amount of heat lost per second by the slab C

Q  Mass  Specific heat capacity  Rate of cooling

Q  MSR ...(2)

At steady state,

Heat conducted through the Heat lost per second


bad conductor per second by the slab C

Hence, equations (1) and (2) are equal

2
K  r 1  2
 MSR
d

MSRd 1 1 ...(3)
K  2
Wm K
 r 1  2

(ii) Determination of rate of cooling R.

The bad conductor is removed and the steam chamber is

placed directly on the slab. The slab is heated to a temperature
about 5C higher than 2. The steam chamber is removed and
the slab alone is allowed to cool.
2.20 Engineering Physics

As the slab cools, the temperatures of the slab are noted

at regular intervals of time (0.5 minute) until the temperature
of the slab falls to about 5C below 2.

The temperature-time graph is drawn and the rate of

d
cooling at the steady temperature 2 is determined.
dt
(Fig. 2.12)

Fig. 2.12 Graph between temperature and time

During the first part of the experiment, before the removal

of bad conductor and steam chamber, the top surface of the slab
is covered by the bad conductor.

Radiation is taking place only from the bottom surface area

and curved surface area of the slab.

2
i.e., Total area   r  2 rh   r r  2h ... (4)

where h is the height of the slab C.

In the second part of the experiment, heat is radiated from

top surface area, bottom surface area and curved sides. i.e., over
entire surface area
2 2 2
 r   r  2 r h  2r  2rh  2 r r  h ... (5)

As the rate of cooling is directly proportional to the surfaces

that are exposed (other condition being equal)

R  r r  2h r  2h r  2h ... (6)

  
d 2 r r  h 2 r  h 2r  2h
dt
Heat 2.21

r  2h  d  ...(7)
R   
2r  2h  dt 

Substituting for R in equation (3), we have

 d 
MSd 
 dt  r  2 h ...(8)
K  2
 r     2r  2h
1 2

Thus, thermal conductivity of the bad conductor is

determined.

Problem 2.4

By means of an electric heater of 12 kW, the

2
temperature in a room with 6.0 m of windows is to be
maintained so that the inner surface of the glass is
10C above the outer surface. Ignoring the heat losses
through the walls of the room and assuming that heat
is lost through the window glass of thickness 6 mm,
what is the coefficient of thermal conductivity of glass.
(U.Q. Jan - 2012)

Given data
Heat generated by the electric heater  12 kW
3
 12  10 watt
3
 12  10 joule/second
2
Area of the window A  6 m
Temp difference 1  2  10 K
3
Thickness of glass window x  6 mm  6  10 m

Solution
Heat conducted through the window glass
2.22 Engineering Physics

K A 1  2 t
Q 
x

Q K A 1  2

t x

Q x
K   
t 
  1  2

Q
Here is heat conducted per second = heat generated per
t
second (power)
3 3
12  10  6  10
K 
10

 1.2

1 1
Thermal conductivity of glass  1.2 Wm K .

Problem 2.5

A solid of square of side 50 cm and thickness 10 cm is

in contact with steam at 100C on one side. A block of
ice at 0C rests on the other side of the solid. 5 kg of
ice is melted in one hour. Calculate the thermal
conductivity of the solid. [U.Q. Jan 2012]

Given data
2
Side length of the solid  50 cm  50  10 m

2 2 2
Area of the solid A  50  10  50  10 m

4 2
 2500  10 m

2
Thickness of the solid x  10 cm  10  10 m

Final temperature of the steam  100C  100  273  373 K

Heat 2.23

Initial temperature of ice  0C  0  273  273 K

Temperature difference  373 K  273 K

 100 K

Mass of ice melted  5 kg

Time taken for melting of ice  1 hour

 1  60  60
 3600 second
1
Latent heat of ice  3,36,000 J kg
5 1
 3.36  10 J kg

Solution:
Amount of heat Q passing across the solid

KA 1  2 t

x

Substituting the given values, we have

4
K  2500  10  100  3600
Q  2
10  10
5
Q  9  10 K

Heat gained by ice in one hour Q

 Mass of the ice  Latent heat of ice
5
 5  3.36  10
5
 16.8  10 joule
Heat conducted across the solid = Heat gained by ice
for one hour in one hour
5 5
9  10 K  16.8  10
2.24 Engineering Physics

5
16.8  10
K  5
9  10

1 1
K  1.86 Wm K

UNIVERSITY PART - A
‘2’ Marks Q & A

1. Define heat conduction. [U.Q. Jan 2011]

Conduction is the process of transmission of heat from one

point to another through substance (or some medium) without the
actual motion of the particles (atoms / molecules) of the substance.

2. Define coefficient of thermal conductivity and

mention its unit. [U.Q. Dec. 2010]
It is defined as the quantity of heat conducted per second
normally across unit area of cross-section per unit temperature
difference per unit length of the material.
–1 –1
Its unit is Wm K

3. Derive the unit in which thermal conductivity is

measured. [U.Q. April 2009]

Thermal conductivity of material

Qx
K 
A 1  2 t

joule  metre joule

K  2

metre  kelvin  second second  metre  kelvin

watt

metre  kelvin

1 1
 Wm K
Heat 2.25

4. What is basic principle behind Lee’s disc method in

determining thermal conductivity of bad conductor?
[U.Q. Jan 2012]

The given bad conductor is taken in the form of disc is

placed in between the metal disc (Lee’s disc) and steam chamber.
The steam is passed through the steam chamber. Heat conducted
through bad conductor per second is calculated. Amount of heat
lost per second by disc is also calculated.

At steady state

 Heat conducted through 

   Amount of heat lost 
 the bad conductor    
 per second   per second by the disc 
 

From this, thermal conductivity of the bad conductor is

calculated.

5. How are heat conduction and electrical conduction

analogous to each other? [U.Q. Dec. 2008]

Heat conduction Electrical conduction

1. Heat is conducted from a Electricity is conducted from
point of higher temperature a point at higher potential to
to a point of lower a point at lower potential.
temperature.
2. In metals, heat conduction is In metals, electrical
mainly due to free electrons. conduction is due to free
electrons.
3. The ability to conduct heat The ability to conduct
is measured by thermal electricity is measured by
conductivity. electrical conductivity.

6. What are the characteristics of good and bad

conductors? [U.Q. Nov. 2011]

Good conductors Bad conductors

1. They have high electrical & They have very low electrical
thermal conductivity and thermal conductivity
2.26 Engineering Physics

Good conductors Bad conductors

2. They can be easily heated or They cannot be easily heated
cooled or cooled
3. Examples: Metals like iron, Examples: Non metals like
copper glass, wood

7. What is thermal resistance? [U.Q. Jan 2012]

The thermal resistance of a body is a measure of its

opposition to the flow of heat through it.

8. Explain why the specimen used to determine thermal

conductivity of a bad conductor should have a larger
area and smaller thickness. (U.Q. April, 2013)

For a bad conductor with a small thickness and large area

of cross-section, the amount of heat conducted increases (large).
This will increase accuracy of the measurement

9. Mention the methods to determine thermal

conductivity of good and bad conductors.
(U.Q. Jan 2011)

 Searle’s method – for good conductors like metallic rod

 Forbe’s method – for determining the absolute
conductivity of metals
 Lee’s disc method – for poor conductors
 Radial flow method – for bad conductors

University Part - B – ‘6’ Marks Questions

1. Explain thermal conduction, thermal convection and thermal

radiation.
2. Derive an expression for thermal conductivity.

University Part - C – ‘10’ Marks Questions

1. Describe Forbe’s method to determine thermal conductivity

of metals with relevent theory. [U.Q. Dec 2005]
Heat 2.27

2. Describe with relevant theory the method of determining

the coefficient of thermal conductivity of a bad conductor
by Lee’s disc method. [U.Q. Dec 2011, Jan 2014]

ASSIGNMENT PROBLEMS
2
1. A copper rod 19 cm long and of 0.785 cm area of
cross-section which is thermally insulated is heated at one
end to 100C while the other end is kept at 30C. Calculate
the amount of heat that will flow in 10 minutes along the
–1 –1
way. K of copper is 380 W m K . (Ans. 6.594 kJ)

4 2
2. Calculate the thickness of the slab of area 85  10 m
through which 8 joules of heat is flowing through the
opposite faes maintained at a temperature difference of 30
K. The coefficient of thermal conductivity of the material
1 1
of the slab is 0.05 Wm K . The time taken for the heat
flow is 10 seconds. (Ans. 0.0159 m)