1. science
2. physics
3. thermodynamics

Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
Introduction.
(1) Thermodynamics : It is a branch of science which deals with exchange of heat energy between bodies and
conversion of the heat energy into mechanical energy and vice-versa.
(2) Thermodynamic system : A collection of an extremely large number of atoms or molecules confined with in
certain boundaries such that it has a certain value of pressure, volume and temperature is called a thermodynamic
system. Anything outside the thermodynamic system to which energy or matter is exchanged is called its
surroundings.
Example : Gas enclosed in a cylinder fitted with a piston forms the thermodynamic system but the atmospheric air
around the cylinder, movable piston, burner etc. are all the surroundings.
Thermodynamic system may be of three types
(i) Open system : It exchange both energy and matter with the surrounding.
(ii) Closed system : It exchange only energy (not matter) with the surroundings.
(iii) Isolated system : It exchange neither energy nor matter with the surrounding.
(3) Thermodynamic variables and equation of state : A thermodynamic system can be described by specifying
its pressure, volume, temperature, internal energy and the number of moles. These parameters are called
thermodynamic variables. The relation between the thermodynamic variables (P, V, T) of the system is called equation
of state.
For  moles of an ideal gas, equation of state is PV = RT and for 1 mole of an it ideal gas is PV = RT

a 2
For  moles of a real gas, equation of state is  P  2
V


a 
 (V   b )   RT and for 1 mole of a real gas it is  P 
 (V  b)  RT

V2 


(4) Thermodynamic equilibrium : When the thermodynamic variables attain a steady value i.e. they are
independent of time, the system is said to be in the state of thermodynamic equilibrium. For a system to be in
thermodynamic equilibrium, the following conditions must be fulfilled.
(i) Mechanical equilibrium : There is no unbalanced force between the system and its surroundings.
(ii) Thermal equilibrium : There is a uniform temperature in all parts of the system and is same as that of surrounding.
(iii) Chemical equilibrium : There is a uniform chemical composition through out the system and the surrounding.
(5) Thermodynamic process : The process of change of state of a system involves change of thermodynamic
variables such as pressure P, volume V and temperature T of the system. The process is known as thermodynamic
process. Some important processes are
(i) Isothermal process
(ii) Adiabatic process
(iii) Isobaric process
(iv) Isochoric (isovolumic process)
(v) Cyclic and non-cyclic process (vi) Reversible and irreversible process
Zeroth Law of Thermodynamics.
If systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium
with each other.
(1) The zeroth law leads to the concept of temperature. All bodies in thermal equilibrium must have a common
property which has the same value for all of them. This property is called the temperature.
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Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
(2) The zeroth law came to light long after the first and seconds laws of thermodynamics had been discovered and
numbered. Because the concept of temperature is fundamental to those two laws, the law that establishes temperature
as a valid concept should have the lowest number. Hence it is called zeroth law.
Quantities Involved in First Law of Thermodynamics.
(1) Heat (Q) : It is the energy that is transferred between a system and its environment because of the
temperature difference between them. Heat always flow from a body at higher temperature to a body at lower
temperature till their temperatures become equal.
(2) Work (W) : Work can be defined as the energy that is transferred from one body to the other owing to a force
that acts between them
If P be the pressure of the gas in the cylinder, then force exerted by the gas
on the piston of the cylinder F = PA
dx
In a small displacement of piston through dx, work done by the gas
F=PA
dW  F.dx  PA dx  P dV

 Total amount of work done W  dW 

Vf
Vi
P dV  P(V f  Vi )
(3) Internal energy (U) : Internal energy of a system is the energy possessed by the system due to molecular
motion and molecular configuration.
The energy due to molecular motion is called internal kinetic energy UK and that due to molecular configuration
is called internal potential energy UP.
i.e. Total internal energy U  U K  U P
(i) For an ideal gas, as there is no molecular attraction U p  0
i.e. internal energy of an ideal gas is totally kinetic and is given by U  U K 
and change in internal energy U 
3
RT
2
3
R T
2
(ii) In case of gases whatever be the process
U  
 R(T f  Ti ) RT f  RT i (Pf V f  Pi Vi )
f
R
R T  C V T  

T 

2
 1
(  1)
 1
 1
(iii) Change in internal energy does not depends on the path of the process. So it is called a point function i.e. it
depends only on the initial and final states of the system, i.e. U  U f  U i
(iv) Change in internal energy in a cyclic process is always zero as for cyclic process U f  U i
So
U  U f  U i  0
Joule's Law.
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Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
Whenever heat is converted into mechanical work or mechanical work is converted into heat, then the ratio of
work done to heat produced always remains constant.
W  Q or
i.e.
W
J
Q
This is Joule‟s law and J is called mechanical equivalent of heat.
First Law of Thermodynamics.
It is a statement of conservation of energy in thermodynamical process.
According to it heat given to a system (Q) is equal to the sum of increase in its internal energy (U) and the work
done (W) by the system against the surroundings.
Q  U  W
Isothermal Process.
When a thermodynamic system undergoes a physical change in such a way that its temperature remains
constant, then the change is known as isothermal changes.
In this process, P and V change but T = constant i.e. change in temperature T = 0
(1) Essential condition for isothermal process
(i) The walls of the container must be perfectly conducting to allow free
exchange of heat between the gas and its surrounding.
Conducting
walls
Gas
(ii) The process of compression or expansion should be so slow so as to
provide time for the exchange of heat.
Since these two conditions are not fully realised in practice, therefore, no process is perfectly isothermal.
(2) Equation of state : From ideal gas equation PV = RT
If temperature remains constant then PV= constant i.e. in all isothermal process Boyle‟s law is obeyed.
Hence equation of state is PV = constant.
(3) Example of isothermal process
(i) Melting process [Ice melts at constant temperature 0°C]
(ii) Boiling process [water boils at constant temperature 100°C].
(4) Indicator diagram
P
P
P
T1<T2< T3
T3
T2
T1

Work
V
V
V
(i) Curves obtained on PV graph are called isotherms and they are hyperbolic in nature.
(ii) Slope of isothermal curve : By differentiating PV = constant. We get
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Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
 P dV  V dP
P dV  V dP  0

tan  

dP
P

dV
V
dP
P

dV
V
(iii) Area between the isotherm and volume axis represents the work done in isothermal process.
If volume increases W = + Area under curve and if volume decreases W = – Area under curve
(5) Specific heat : Specific heat of gas during isothermal change is infinite.
As
C
Q
Q


m T m  0
[As T = 0]
(6) Isothermal elasticity : For isothermal process PV = constant
Differentiating both sides PdV  VdP  0  P dV  V dP  P 

dP
Stress

 E
 dV / V Strain
E  P i.e. isothermal elasticity is equal to pressure
At N.T.P. isothermal elasticity of gas = Atmospheric pressure = 1.01  10 5 N / m 2
(7) Work done in isothermal process
W 

Vf
Vi
P dV 

Vf
Vi
RT
V
dV
[As PV = RT]
 Vf 
 Vf 
W  RT log e    2 .303 RT log 10  
 Vi 
 Vi 
or
P
W  RT log e  i
P
 f


  2 . 303 RT log 10  Pi

P

 f




(8) FLTD in isothermal process
Q  U  W but
U  T

U  0
[As T = 0]

Q  W i.e. heat supplied in an isothermal change is used to do work against external surrounding.
or if the work is done on the system than equal amount of heat energy will be liberated by the system.
Adiabatic Process.
When a thermodynamic system undergoes a change in such a way that no exchange of heat takes place between it
and the surroundings, the process is known as adiabatic process.
In this process P, V and T changes but Q = 0.
Insulating
walls
(1) Essential conditions for adiabatic process
(i) There should not be any exchange of heat between the system and its
surroundings. All walls of the container and the piston must be perfectly
insulating.
Gas
(ii) The system should be compressed or allowed to expand suddenly so that there is no time for the exchange of
heat between the system and its surroundings.
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Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
Since, these two conditions are not fully realised in practice, so no process is perfectly adiabatic.
(2) Example of some adiabatic process
(i) Sudden compression or expansion of a gas in a container with perfectly non-conducting walls.
(ii) Sudden bursting of the tube of bicycle tyre.
(iii) Propagation of sound waves in air and other gases.
(iv) Expansion of steam in the cylinder of steam engine.
(3) FLTD in adiabatic process : Q  U  W
but for adiabatic process Q  0  U  W  0
If W = positive then U = negative so temperature decreases i.e. adiabatic expansion produce cooling.
If W = negative then U = positive so temperature increases i.e. adiabatic compression produce heating.
(4) Equation of state : As in case of adiabatic change first law of thermodynamics reduces to,
U  W  0 ,
i.e., dU  dW  0
….. (i)
But as for an ideal gas dU  C V dT and dW  PdV
Equation (i) becomes Cv dT  P dV  0
….. (ii)
But for a gas as PV = RT, P dV  V dP  R dT
….. (iii)
So eliminating dT between equation (ii) and (iii) C v
or
(P dV  V dP )
 P dV  0
(  1)
or
P dV  V dP  0 i.e.,
(P dV  V dP )
 P dV  0
R

R 
as C v 

(  1) 


dV dP

0
V
P
Which on integration gives
 log e V  log e P  C ,
log( PV  )  C
i.e.,
PV   constant
or
….. (iv)
Equation (iv) is called equation of state for adiabatic change and can also be re-written as
TV  1  constant
and
T
P
 1
 constant
[as P = (RT/V)]
RT 

as V  P 


(5) Indicator diagram
….. (iv)
….. (vi)
P
(i) Curve obtained on PV graph are called adiabatic curve.
(ii) Slope of adiabatic curve : From PV   constant
By differentiating, we get
dP V   PV  1 dV  0

V
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Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
dP
PV  1
P
 
   

dV
V
V 
P
 Slope of adiabatic curve tan     
V 
But we also know that slope of isothermal curve tan  
So,
P
V
Cp
  (P / V )
Slope of adiabatic curve

 
1
Slope of isothermal curve
 (P / V )
Cv
(6) Specific heat : Specific heat of a gas during adiabatic change is zero
As
Q
0

0
m T m T
C
[As Q = 0]
(7) Adiabatic elasticity : For adiabatic process PV   constant
Differentiating both sides d PV   PV  1 dV  0
P 
dP
Stress

 E
 dV / V Strain
E  P
i.e. adiabatic elasticity is  times that of pressure but we know isothermal elasticity E  P
So
E
E

Adiabatic elasticity P


Isothermal elasticity
P
i.e. the ratio of two elasticities of gases is equal to the ratio of two specific heats.
(8) Work done in adiabatic process
W 
or

or

or

So


Vf
Vi
P dV 

Vf
Vi
K
V

dV
1  K
K 
  1   1 
[1   ]  V f
Vi 

[Pf V f  Pi Vi ]
R
[T f  Ti ]
(1   )
(  1)
K 

 P   
V 


V  1 
-
 As V dV 

(  1) 


[As K  PV   P f V f  Pi Vi ]
(1   )
[Pi Vi  Pf V f ]

 As


[As Pf V f  RT f and Pi Vi  RT i ]
R(Ti  T f )
(  1)
(9) Free expansion : Free expansion is adiabatic process in which no work is performed on or by the system.
Consider two vessels placed in a system which is enclosed with thermal insulation (asbestos-covered). One vessel
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Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
contains a gas and the other is evacuated. The two vessels are connected by a stopcock. When suddenly the stopcock is
opened, the gas rushes into the evacuated vessel and expands freely. The process is adiabatic as the vessels are placed
in thermal insulating system (dQ = 0) moreover, the walls of the vessel are rigid and hence no external work is
performed (dW = 0).
Now according to the first law of thermodynamics dU = 0
If U i and U f be the initial and final internal energies of the gas
then
U f  Ui  0
[As U f  U i ]
Thus the final and initial energies are equal in free expansion.
(10) Special cases of adiabatic process
PV

 constant
 P
Type of gas
Monoatomic  = 5/3
Diatomic  = 7/5
Polyatomic  = 4/3
1
V


,

PT 1   constant
P
 P  T  1 and
1
V

1
P
V
5/3
V
7/5
1
P
1
P
V
PT
1
TV  1  constant  T 

 1
T
P  T5/2
T
P  T7/2
T
P  T4
T
4 /3
V  1
1
V  1
1
V2/3
1
V 2/5
1
V1/ 3
(11) Comparison between isothermal and adiabatic process
(i) Compression : If a gas is compressed isothermally and adiabatically from
volume Vi to Vf then from the slope of the graph it is clear that graph 1 represents
adiabatic process where as graph 2 represent isothermal process.
Work done
Wadiabatic > Wisothermal
Final pressure
Padiabatic > Pisothermal
Final temperature
Tadiabatic > Tisothermal
P
1
2
Vf
V
Vi
(ii) Expansion : If a gas expands isothermally and adiabatically from volume Vi to Vf then from the slope of the
graph it is clear that graph 1 represent isothermal process, graph 2 represent
P
adiabatic process.
Work done
Wisothermal > Wadiabatic
Final pressure
Pisothermal > Padiabatic
Final temperature
Tisothermal > Tadiabatic
1
2
Vi
Vf
V
2
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Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
Isobaric Process.
When a thermodynamic system undergoes a physical change in such a way that its pressure remains constant,
then the change is known as isobaric process.
In this process V and T changes but P remains constant. Hence Charle‟s law is obeyed in this process.
(1) Equation of state : From ideal gas equation PV = RT
If pressure remains constant V  T
P
V1 V2

 constant
T1 T2
or
P1
(2) Indicator diagram : Graph I represent isobaric expansion, graph II
represent isobaric compression.
Slope of indicator diagram
P2
A
I
B
II
D
C
dP
0
dV
V
f

(3) Specific heat : Specific heat of gas during isobaric process C P    1  R
2


(4) Bulk modulus of elasticity : K 
P
0
 V / V
(5) Work done in isobaric process : W 


Vf
Vi
[As P = 0]
P dV  P

Vf
Vi
dV  P[V f  Vi ]
[As P = constant]
W  P(V f  Vi )  R[T f  Ti ]  R T
(6) FLTD in isobaric process : U   C V T  
From FLTD
Q  U  W

Q  
R
T
(  1)
and
W  R T
 1

  

R
 R T
T  R T  R T 
 1  R T
  
(  1)
 1
  1

 1
Q   C P T
Isochoric or Isometric Process.
When a thermodynamic process undergoes a physical change in such a way that its volume remains constant,
then the change is known as isochoric process.
In this process P and T changes but V = constant. Hence Gay-lussac‟s law is obeyed in this process.
(1) Equation of state
From ideal gas equation PV = RT
If volume remains constant P  T or
P1
P
 2  constant
T1
T2
D
P
(2) Indicator diagram : Graph I and II represent isometric decrease in
pressure at volume V1 and isometric increase in pressure at volume V2
A
I
II
C
B
V1
V2
V
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Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
respectively and slope of indicator diagram
dP

dV
(3) Specific heat : Specific heat of gas during isochoric process C V 
(4) Bulk modulus of elasticity : K 
P
P


 V / V
0
(5) Work done in isobaric process : W  PV  P[V f  Vi ]

f
R
2
[As V = 0]
W  0
[As W = 0]
(6) FLTD in isochoric process : Q  U  W  U
Q   C V T  
P f V f  Pi Vi
R
T 
 1
 1
Heat Engine.
Heat engine is a device which converts heat into work continuously through a cyclic process.
The essential parts of a heat engine are
Source : It is a reservoir of heat at high temperature and infinite thermal capacity. Any amount of heat can be
extracted from it.
Working substance : Steam, petrol etc.
Sink : It is a reservoir of heat at low temperature and infinite thermal capacity. Any amount of heat can be given
to the sink.
The working substance absorbs heat Q1 from the source, does an amount of
work W, returns the remaining amount of heat to the sink and comes back to its
original state and there occurs no change in its internal energy.
By repeating the same cycle over and over again, work is continuously
obtained.
Source
Heat
Engine
W = Q1 – Q2
Q2
The performance of heat engine is expressed by means of “efficiency” 
which is defined as the ratio of useful work obtained from the engine to the heat
supplied to it.


Sink
(T2)
Work done
W

Heat input
Q1
But by first law of thermodynamics for cyclic process U = 0.

(T1)
Q1
 Q = W so W  Q1  Q 2
Q1  Q 2
Q
1 2
Q1
Q1
A perfect heat engine is one which converts all heat into work i.e. W  Q1 so that Q 2  0 and hence   1 .
But practically efficiency of an engine is always less than
Refrigerator or Heat Pump.
A refrigerator or heat pump is basically a heat engine run in reverse direction.
It essentially consists of three parts
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Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
Source : At higher temperature T1.
Working substance : It is called refrigerant liquid ammonia and freon works as a working substance.
Sink : At lower temperature T2.
The working substance takes heat Q2 from a sink (contents of refrigerator) at lower temperature, has a net
amount of work done W on it by an external agent (usually compressor of
Source
refrigerator) and gives out a larger amount of heat Q1 to a hot body at temperature
(Atmosphere) (T1)
Q1
T1 (usually atmosphere). Thus, it transfers heat from a cold to a hot body at the
Heat
expense of mechanical energy supplied to it by an external agent. The cold body is
W = Q1 – Q2
Engine
thus cooled more and more.
Q
2
The performance of a refrigerator is expressed by means of “coefficient of
performance”  which is defined as the ratio of the heat extracted from the cold
body to the work needed to transfer it to the hot body.
i.e.
 
Q
Q2
Heat extracted
 2 
work done
W
Q1  Q 2


Q2
Q1  Q 2
Sink
(Contents of refrigerator) (T2)
A perfect refrigerator is one which transfers heat from cold to hot body without doing work
i.e. W = 0 so that Q1  Q 2 and hence   
(1) Carnot refrigerator
For Carnot refrigerator
Q1
T
 1
Q 2 T2
So coefficient of performance  

Q 1  Q 2 T1  T2
Q2
T2
or


Q2
T2
Q 1  Q 2 T1  T2
T2
T1  T2
where T1 = temperature of surrounding, T2 = temperature of cold body
It is clear that  = 0 when T2 = 0
i.e. the coefficient of performance will be zero if the cold body is at the temperature equal to absolute zero.
(2) Relation between coefficient of performance and efficiency of refrigerator
We know  
Q2
Q 2 / Q1
or  
Q1  Q 2
1  Q 2 / Q1
But the efficiency   1 
Q2
Q
or 2  1  
Q1
Q1
From (i) and (ii) we get  
….. (i)
…..(ii)
1 

Carnot Engine.
11
House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362
Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
Carnot designed a theoretical engine which is free from all the defects of a practical engine. This engine cannot be
realised in actual practice, however, this can be taken as a standard against which the performance of an actual engine
can be judged.
It consists of the following parts
(i) A cylinder with perfectly non-conducting walls and a perfectly conducting base containing a perfect gas as
working substance and fitted with a non-conducting frictionless piston
(ii) A source of infinite thermal capacity maintained at constant higher temperature T1.
(iii) A sink of infinite thermal capacity maintained at constant lower temperature T2.
(iv) A perfectly non-conducting stand for the cylinder.
(1) Carnot cycle : As the engine works, the working substance of the engine undergoes a cycle known as Carnot
cycle. The Carnot cycle consists of the following four strokes
(i) First stroke (Isothermal expansion) (curve AB) :
The cylinder containing ideal gas as working substance allowed
to expand slowly at this constant temperature T1.
Q1
W = Q1 – Q2
Work done = Heat absorbed by the system
W1  Q1 

V2
V1
V
P dV  RT1 log e  2
 V1

  Area ABGE

Q2
(ii) Second stroke (Adiabatic expansion) (curve BC) :
The cylinder is then placed on the non conducting stand and the gas is allowed to expand adiabatically till the
temperature falls from T1 to T2.
W2 

V3
V2
P dV =
R
[T1  T2 ]  Area BCHG
(  1)
(iii) Third stroke (Isothermal compression) (curve CD) :
The cylinder is placed on the sink and the gas is compressed at constant temperature T2.
Work done = Heat released by the system
W3  Q 2  

V4
V3
P dV   RT 2 log e
V4
V
 RT 2 log e 3  Area CDFH
V3
V4
(iv) Fourth stroke (adiabatic compression) (curve DA) : Finally the cylinder is again placed on non-conducting
stand and the compression is continued so that gas returns to its initial stage.
W4  
i.e.

V1
V4
P dV  
R
R
(T2  T1 ) 
(T1  T2 )  Area ADFE
 1
 1
(2) Efficiency of Carnot cycle : The efficiency of engine is defined as the ratio of work done to the heat supplied
work done
W


Heat input Q1
So efficiency of Carnot engine   1 
T2
T1
2
House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362
Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
(i) Efficiency of a heat engine depends only on temperatures of source and sink and is independent of all other factors.
(ii) All reversible heat engines working between same temperatures are equally efficient and no heat engine can
be more efficient than Carnot engine (as it is ideal).
(iii) As on Kelvin scale, temperature can never be negative (as 0 K is defined as the lowest possible temperature)
and Tl and T2 are finite, efficiency of a heat engine is always lesser than unity, i.e., whole of heat can never be converted
into work which is in accordance with second law.
(3) Carnot theorem : The efficiency of Carnot‟s heat engine depends only on the temperature of source (T1) and
T
temperature of sink (T2), i.e.,   1  2 .
T1
Carnot stated that no heat engine working between two given temperatures of source and sink can be more
efficient than a perfectly reversible engine (Carnot engine) working between the same two temperatures. Carnot's
reversible engine working between two given temperatures is considered to be the most efficient engine.
Conduction.
The process of transmission of heat energy in which the heat is transferred from one particle to other particle without
dislocation of the particle from their equilibrium position is called conduction.
(i) Conduction is a process which is possible in all states of matter.
(ii) In solids only conduction takes place.
(iii) In non-metallic solids and fluids the conduction takes place only due to vibrations of molecules, therefore they are poor
conductors.
(iv) In metallic solids free electrons carry the heat energy, therefore they are good conductor of heat.
(1) Variable and steady state
When one end of a metallic rod is heated, heat flows by conduction from the hot end to the cold end.
In the process of conduction each cross-section of the rod receives heat from the adjacent cross-section towards the hot end.
A part of this heat is absorbed by the cross-section itself whose temperature increases, another part is lost into atmosphere by
convection & radiation and the rest is conducted away to the next cross-section.
Because in this state temperature of every cross-section of the rod goes on increasing, hence rod is said to exist in variable
Temperature
state.
Hot end
Cold end
Metallic rod
Distance from hotter end
After sometime, a state is reached when the temperature of every cross-section of the rod becomes constant. In this state, no
heat is absorbed by the rod. The heat that reaches any cross-section is transmitted to the next except that a small part of heat is
lost to surrounding from the sides by convection & radiation. This state of the rod in which no part of rod absorbs heat is called
steady state.
(2) Isothermal surface
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House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362
Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
Any surface (within a conductor) having its all points at the same temperature, is called isothermal surface. The direction of
flow of heat through a conductor at any point is perpendicular to the isothermal surface passing through that point.
(i) If the material is rectangular or cylindrical rod, the isothermal surface is a plane surface.
(ii) If a point source of heat is situated at the centre of a sphere the isothermal surface will be spherical,
(iii) If steam passes along the axis of the hollow cylinder, heat will flow through the walls of the cylinder so that in this
condition the isothermal surface will be cylindrical.
Q
Q
Q
Q
S
Cylindrical isothermal surface
Spherical isothermal surface
Plane isothermal surfaces
(3) Temperature Gradient
The rate of change of temperature with distance between two isothermal surfaces is called temperature gradient.
If the temperature of two isothermal surfaces be  and (    ) , and the perpendicular distance between them be x
then Temperature gradient =
(    )  
x
=
 
x
The negative sign show that temperature  decreases as the distance x
increases in the direction of heat flow.

( –  )
Q
Unit : K/m (S.I.) and Dimensions : [L1 ]
x
(4) Coefficient of thermal conductivity
If L be the length of the rod, A the area of cross-section and 1 and 2 are the temperature of its two faces, then the amount of
heat flowing from one face to the other face in time t is given by
Q
KA ( 1   2 ) t
l
Where K is coefficient of thermal conductivity of material of rod. It is the measure of
the ability of a substance to conduct heat through it.
1
A
2
Q
l
If A = 1m2, (1 – 2) = 1oC, t = 1 sec and l = 1m, then Q = K.
Thus, thermal conductivity of a material is the amount of heat flowing per second during steady state through its rod of
length 1 m and cross-section 1 m2 with a unit temperature difference between the opposite faces.
(i) Units : Cal/cm-sec oC (in C.G.S.), kcal/m-sec-K (in M.K.S.) and W/m- K (in S.I.)
(ii) Dimension : [MLT 3 1 ]
(iii) The magnitude of K depends only on nature of the material.
(iv) For perfect conductors, K   and for perfect insulators, K  0
(v) Substances in which heat flows quickly and easily are known as good conductor of heat. They possesses large thermal
conductivity due to large number of free electrons. Example : Silver, brass etc.
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House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362
Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
(vi) Substances which do not permit easy flow of heat are called bad conductors. They possess low thermal conductivity due
to very few free electrons. Example : Glass, wood etc.
(vii) The thermal conductivity of pure metals decreases with rise in temperature but for alloys thermal conductivity
increases with increase of temperature.
(viii) Human body is a bad conductor of heat (but it is a good conductor of electricity).
(6) Relation between temperature gradient and thermal conductivity
In steady state, rate of flow of heat
If
dQ
d
  KA
dt
dx
= – KA (Temperature gradient)
1
dQ
is constant then temperature gradient 
dt
K
Temperature difference between the hot end and the cold end in steady state is inversely proportional to K, i.e. in case of
good conductors temperature of the cold end will be very near to hot end.
In ideal conductor where K = , temperature difference in steady state will be zero.
(7) Wiedmann-Franz law
At a given temperature T, the ratio of thermal conductivity to electrical conductivity is constant i.e., (K / T ) = constant, i.e.,
a substance which is a good conductor of heat (e.g., silver) is also a good conductor of electricity. Mica is an exception to above law.
(8) Thermal resistance
The thermal resistance of a body is a measure of its opposition to the flow of heat through it.
It is defined as the ratio of temperature difference to the heat current (= Rate of flow of heat)
Now, temperature difference = ( 1   2 ) and heat current, H =
 Thermal resistance, R 
Unit :
o
1   2
H

1   2
Q/t

Q
t
1   2
l

KA( 1   2 ) / l KA
C  sec / cal or K  sec / kcal and Dimension : [M 1 L2T 3 ]
Electrical Analogy For Thermal Conduction.
It is an important fact to appreciate that there exists an exact similarity between thermal and electrical conductivities of a
conductor.
Electrical conduction
Thermal conduction
Electric charge flows from higher potential to lower potential
Heat flows from higher temperature to lower temperature
The rate of flow of charge is called the electric current,
The rate of flow of heat may be called as heat current
i.e.
I
dq
dt
i.e.
H
dQ
dt
The relation between the electric current and the potential
V  V2
difference is given by Ohm's law, that is I  1
R
Similarly, the heat current may be related with the
 2
temperature difference as H  1
R
where R is the electrical resistance of the conductor
where R is the thermal resistance of the conductor
15
House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362
Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
The electrical resistance is defined as R 
l
A

l
A
The thermal resistance may be defined as R 
where  = Resistivity and  = Electrical conductivity
where K = Thermal conductivity of conductor
V  V2
A
dq
I 1

(V1  V2 )
dt
R
l
dQ
   2 KA
H 1

(1   2 )
dt
R
l
l
KA
Combination of Conductors.
(1) Series combination :
Let n slabs each of cross-sectional area A, lengths l1 , l2 , l3 ,......ln and conductivities K1 , K 2 , K 3 ...... K n respectively be
connected in the series
1
Heat current is the same in all the conductors.
2
Q
 H 1  H 2  H 3 .........  H n
i.e.,
t
3
4
n – 1
n
K1
K2
K3
Kn
l1
l2
l3
ln
K 1 A( 1   2 ) K 2 A( 2   3 ) K 3 A( 3   4 )
K A( n 1   n )


 ........  n
l1
l2
l3
ln
(i) Equivalent resistance R  R1  R 2  R 3  .....Rn
(ii) If K s is equivalent conductivity, then from relation R 
l
KA
l1  l 2  l 3  ..... ln
l
l
l
l
 1  2  3  ....  n
Ks
K1 A K 2 A K 3 A
Kn A

Ks 
l1  l 2  l 3  ...... ln
l1
l
l
l
 2  3  ........ n
K1 K 2 K 3
Kn
(iii) Equivalent thermal conductivity for n slabs of equal length K 
For two slabs of equal length, K 
n
1
1
1
1


 .....
K1 K 2 K 3
Kn
2 K1 K 2
K1  K 2
(iv) Temperature of interface of composite bar : Let the two bars are arranged in series as shown in the figure.
Then heat current is same in the two conductors.
i.e.,
Q K1 A(1   ) K 2 A(   2 )


t
l1
l2
1
K1
K
1  2  2
l
l2
By solving we get   1
K1 K 2

l1
l2

K1
l1
2
K2
l2
2
House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362
Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
If (l1  l2  l) then  
K1 1  K 2 2
K1  K 2
(2) Parallel Combination
Let n slabs each of length l, areas A1 , A2 , A3 ,.....An and thermal conductivities K1 , K 2 , K 3 ,..... K n are connected in parallel
then.
1
1
1
1
1



 .....
R R1 R 2 R 3
Rn
(i) Equivalent resistance
(ii) Temperature gradient across each slab will be same.
(iii) Heat current in each slab will be different. Net heat current will be the sum of heat currents through individual slabs. i.e.,
H  H1  H 2  H3  ....Hn
K ( A1  A 2  A 3  .....  A n ) ( 1   2 ) K1 A1 ( 1   2 ) K 2 A 2 ( 1   2 ) K 3 A3 ( 1   2 )
K A (   2 )



.......  n n 1
l
l
l
l
l
 K 
K 1 A1  K 2 A 2  K 3 A 3  ..... K n A n
A1  A 2  A 3  ..... A n
l
1
K  K 2  K 3  ..... K n
For n slabs of equal area K  1
n
Equivalent thermal conductivity for two slabs of equal area K 
K1  K 2
2
2
K1
K2
A1
K3
A3
Kn
A3
A2
Ingen-Hauz Experiment.
It is used to compare thermal conductivities of different materials. If l1 and l2 are the lengths of wax melted on rods, then
the ratio of thermal conductivities is
K1 l12

K 2 l22
i.e., in this experiment, we observe Thermal conductivity  (length)2
Searle's Experiment.
It is a method of determination of K of a metallic rod. Here we are not much interested in the detailed description of the
experimental setup. We will only understand its essence, which is the essence of solving many numerical problems.
In this experiment a temperature difference ( 1   2 ) is maintained across a rod of length l and area of cross section A. If
the thermal conductivity of the material of the rod is K, then the amount of heat transmitted by the rod from the hot end to the
cold end in time t is given by, Q 
KA ( 1   2 ) t
l
......(i)
In Searle's experiment, this heat reaching the other end is utilized to raise the temperature of certain amount of water
flowing through pipes circulating around the other end of the rod. If temperature of the water at the inlet is  3 and at the outlet is
 4 , then the amount of heat absorbed by water is given by, Q  mc ( 4   3 ) ......(ii)
Where, m is the mass of the water which has absorbed this heat and temperature is raised and c is the specific heat of the
water
Equating (i) and (ii), K can be determined i.e., K 
mc ( 4   3 ) l
A ( 1   2 ) t
17
House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362
Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
Note : 
In numericals we may have the situation where the amount of heat travelling to the other end may be
required to do some other work e.g., it may be required to melt the given amount of ice. In that case equation (i)
will have to be equated to mL.
mL 
i.e.
KA ( 1   2 ) t
l
Growth of Ice on Lake.
Water in a lake starts freezing if the atmospheric temperature drops below 0 o C . Let y be the thickness of ice layer in the
lake at any instant t and atmospheric temperature is   o C . The temperature of water in contact with lower surface of ice will be
zero. If A is the area of lake, heat escaping through ice in time dt is
dQ 1 
KA[0  ( )] dt
y
Now, suppose the thickness of ice layer increases by dy in time dt, due to escaping of above heat. Then
dQ 2  mL  (dy A) L
– °C
y
As dQ1  dQ2 , hence, rate of growth of ice will be (dy / dt )  (K / Ly )
So, the time taken by ice to grow to a thickness y is t 
L
K

y
0
y dy 
If the thickness is increased from y 1 to y 2 then time taken t 
L
K

y1
ydy 
Ice
dy
0°C
L 2
y
2 K
y2
Air
Water
4°C
L 2
(y 2  y 12 )
2 K
(i) Take care and do not apply a negative sign for putting values of temperature in formula and also do not convert it to
absolute scale.
(ii) Ice is a poor conductor of heat, therefore the rate of increase of thickness of ice on ponds decreases with time.
(iii) It follows from the above equation that time taken to double and triple the thickness, will be in the ratio of
t1 : t 2 : t3 :: 1 2 : 2 2 : 3 2 , i.e., t1 : t2 : t3 :: 1 : 4 : 9
(iv) The time intervals to change the thickness from 0 to y, from y to 2y and so on will be in the ratio
t1 : t 2 : t3 :: (1 2  0 2 ) : (2 2  1 2 ) : (3 2 : 2 2 ) ; t1 : t2 : t3 :: 1 : 3 : 5
Convection.
Mode of transfer of heat by means of migration of material particles of medium is
called convection. It is of two types.
(1) Natural convection : This arise due to difference of densities at two places
and is a consequence of gravity because on account of gravity the hot light particles rise
up and cold heavy particles try setting down. It mostly occurs on heating a liquid/fluid.
(2) Forced convection : If a fluid is forced to move to take up heat from a hot
body then the convection process is called forced convection. In this case Newton's law of cooling holds good. According to which
rate of loss of heat from a hot body due to moving fluid is directly proportional to the surface area of body and excess temperature
of body over its surroundings
2
House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362
Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
Q
 A(T  T0 )
t
i.e.
Q
 h A(T  T0 )
t
where h = Constant of proportionality called convection coefficient,
T = Temperature of body and T 0 = Temperature of surrounding
Convection coefficient (h) depends on properties of fluid such as density, viscosity, specific heat and thermal conductivity.
(i) Natural convection takes place from bottom to top while forced convection in any direction.
(ii) In case of natural convection, convection currents move warm air upwards and cool air downwards. That is why heating
is done from base, while cooling from the top.
(iii) Natural convection plays an important role in ventilation, in changing climate and weather and in forming land and sea
breezes and trade winds.
(iv) Natural convection is not possible in a gravity free region such as a free falling lift or an orbiting satellite.
(v) The force of blood in our body by heart helps in keeping the temperature of body constant.
(vi) If liquids and gases are heated from the top (so that convection is not possible) they transfer heat (from top to bottom) by
conduction.
(vii) Mercury though a liquid is heated by conduction and not by convection.
Radiation.
The process of the transfer of heat from one place to another place without heating the intervening medium is
called radiation.
Precisely it is electromagnetic energy transfer in the form of electromagnetic wave through any medium. It is possible even
in vacuum.
For example, the heat from the sun reaches the earth through radiation.
Some Definition About Radiations .
Reflectance, Absorptance and transmittance
When thermal radiations (Q) fall on a body, they are partly reflected, partly absorbed and partly transmitted.
(i) Reflectance or reflecting power (r) : It is defined as the ratio of the amount of thermal radiations reflected (Qr) by the body in a
given time to the total amount of thermal radiations incident on the body in that time.
(ii) Absorptance or absorbing power (a) : It is defined as the ratio of the amount of thermal radiations absorbed (Qa) by the body
in a given time to the total amount of thermal radiatons incident on the body in that time.
(iii) Transmittance or transmitting power (t) : It is defined as the ratio of the amount of thermal radiations transmitted (Qt) by the
body in a given time to the total amount of thermal radiations incident on the body in that time.
From the above definitions r 
By adding we get
rat 
Qr
Q
, a a
Q
Q
and t 
Qt
Q
Q r Q a Q t (Q r  Q a  Q t )



1
Q
Q
Q
Q
19
House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362
Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
rat 1

(a) r, a and t all are the pure ratios so they have no unit and dimension.
(b) For perfect reflector
: r = 1, a = 0 and t = 0
(c) For perfect absorber : a = 1, r = 0 and t = 0
(Perfectly black body)
(d) For perfect transmitter : t = 1, a = 0 and r = 0
(e) If body does not transmit any heat radiation, t = 0
 r  a  1 or a  1  r
So if r is more, a is less and vice-versa. It means good reflectors are bad absorbers.
Monochromatic Emittance or Spectral emissive power
For a given surface it is defined as the radiant energy emitted per sec per unit area of the surface with in a unit wavelength


1
1

 to     .
2
2

around  i.e. lying between   
Spectral emissive power (E  ) 
Joule
Unit :
Energy
Area  time  wavelength
and
m  sec  Å
2
Dimension : [ML1 T 3 ]
(5) Total emittance or total emissive power
It is defined as the total amount of thermal energy emitted per unit time, per unit area of the body for all possible
wavelengths.
Unit :
E


0
Joule
m  sec
2
E d
or
Watt
and
m2
Dimension : [MT 3 ]
Monochromatic absorptance or spectral absorptive power
It is defined as the ratio of the amount of the energy absorbed in a certain time to the total heat energy incident upon it in
the same time, both in the unit wavelength interval. It is dimensionless and unit less quantity. It is represented by a.
Total absorptance or total absorpting power : It is defined as the total amount of thermal energy absorbed per unit time,
per unit area of the body for all possible wavelengths.
a


0
a d
It is also unit less and dimensionless quantity.
Emissivity (e) : Emissivity of a body at a given temperature is defined as the ratio of the total emissive power of the body
(Epractical ) to the total emissive power of a perfect black body (Eblack) at that temperature.
i.e.
e
E practical
Eblack
e = 1 for perfectly black body but for practical bodies emissivity (e) lies between zero and one (0 < e < 1).
2
House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362
Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
Perfectly black body : A perfectly black body is that which absorbs completely the radiations of all wavelengths incident on
it. As a perfectly black body neither reflects nor transmits any radiation, therefore the absorptance of a perfectly black body is
unity i.e. t = 0 and r = 0
 a = 1.
We know that the colour of an opaque body is the colour (wavelength) of radiation reflected by it. As a black body
reflects no wavelength so, it appears black, whatever be the colour of radiations incident on it.
When perfectly black body is heated to a suitable high temperature, it emits radiation of all possible wavelengths. For
example, temperature of the sun is very high (6000 K approx.) it emits all possible radiation so it is an example of black body.
Kirchoff's Law.
The ratio of emissive power to absorptive power is same for all surfaces at the same temperature and is equal to the emissive
power of a perfectly black body at that temperature.
Thus if apractical and Epractical represent the absorptive and emissive power of a given surface, while ablack and Eblack for a perfectly
black body, then according to law
E practical
a practical

But for a perfectly black body ablack = 1 so
E black
a black
E practical
a practical
 E black
 E 

 (E  ) black
 a   practical
If emissive and absorptive powers are considered for a particular wavelength , 
Now since (E)black is constant at a given temperature, according to this law if a surface is a good absorber of a particular
wavelength it is also a good emitter of that wavelength.
This in turn implies that a good absorber is a good emitter (or radiator)
Distribution of Energy in The Spectrum of Black Body.
Langley and later on Lummer and Pringsheim investigated the distribution of energy amongst the different wavelengths in
the thermal spectrum of a black body radiation. The results obtained are shown in figure. From these curves it is clear that
(1) At a given temperature energy is not uniformly distributed among different wavelengths.
(2) At a given temperature intensity of heat radiation increases with
wavelength, reaches a maximum at a particular wavelength and with further
increase in wavelength it decreases.
T3
T3> T2> T1
(3) With increase in temperature wavelength  m corresponding to most intense
radiation decreases in such a way that m  T  constant. [Wien‟s law]
E
(4) For all wavelengths an increase in temperature causes an increase in
intensity.

E  d  will represent the total intensity of
T1
m 3
m 2
m 1
(5) The area under the curve 
T2
radiation at a particular temperature. This area increases with rise in temperature of the
body. It is found to be directly proportional to the fourth power of absolute temperature of the body, i.e.,

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House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362
Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES

E  E d  T 4
[Stefan‟s law]
(6) The energy (Emax) emitted corresponding to the wavelength of maximum emission (m) increases with fifth power of the
absolute temperature of the black body i.e., E max  T 5
Wien's Displacement Law.
When a body is heated it emits radiations of all wavelength. However the intensity of radiations of different wavelength is
different.
m
logm
According to Wien's law the product of wavelength corresponding to maximum
intensity of radiation and temperature of body (in Kelvin) is constant, i.e.
m T  b  constant
3
Where b is Wien's constant and has value 2 . 89  10 m - K .
T
log T
This law is of great importance in „Astrophysics‟ as through the analysis of
radiations coming from a distant star, by finding  m the temperature of the star T( b / m ) is determined.
Stefan's Law.
According to it the radiant energy emitted by a perfectly black body per unit area per sec (i.e. emissive power of black body)
is directly proportional to the fourth power of its absolute temperature,
i.e.
E  T 4 or E  T 4
where  is a constant called Stefan‟s constant having dimension [MT 3 4 ] and value 5.67  10 8 W / m 2 K 4 .
(i) If e is the emissivity of the body then E  eT 4
(ii) If Q is the total energy radiated by the body then E 
Q
 e T 4  Q  A t e T 4
At
(iii) If a body at temperature T is surrounded by a body at temperature T0, then Stefan's law may be put as
E  e  (T 4  T04 )
(iv) Cooling by radiation : If a body at temperature T is in an environment of temperature T0(< T), the body is loosing as well
as receiving so net rate of loss of energy
dQ
 eA  (T 4  T04 )
dt
…..(i)
Now if m is the mass of body and c its specific heat, the rate of loss of heat at temperature T must be
dQ
dT
 mc
dt
dt
From equation (i) and (ii) mc
…..(ii)
dT
 eA  (T 4  T04 )
dt
 Rate of fall of temperature or rate of cooling,
dT eA  4

(T  T04 )
dt
mc
…..(iii)
Newton's Law of Cooling.
If in case of cooling by radiation the temperature T of body is not very different from that of surrounding
i.e.
T  T0  T
2
House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362
Heat And Thermodynamics Notes
ALFA PHYSICS CLASSES
T 
4
 [(T0  T ) 
T04
4
T04 ]

T04
4




T 
4 T
 1 
  1  T04  1 
 1  [Using Binomial theorem]

T0 
T0




 4 T03 T
…..(i)
dT eA  4

[T  T04 ]
By Stefan‟s law,
dt
mc
From equation (i),

0
dT eA 

4 T03 T
dt
mc
dT
 T
dt
So
or
Cooling curve of a
body
0
Time t
d
  0
dt
i.e., if the temperature of body is not very different from surrounding, rate of cooling is proportional to temperature
difference between the body and its surrounding. This law is called Newton‟s law of cooling.
(5) Cooling curves :
Curve between temperature of body  and time.
Curve between rate of cooling (R) and temperature
Temperature
difference between body () and surrounding (0)
R
Time
R  (   0 ) . This is Oa straight line
through origin.
( – passing
0)
   0  Ae kt , which indicates temperature decreases
exponentially with increasing time.
temperature ().
Curve between log( – 0) and time
loge ( – 0)
Curve between the rate of cooling (R) and body
R
K0

O
R  K(   0 )  K  K 0
This is a straight line intercept R-axis at  K 0
As
t
d
d
 (   0 ) 
  Kdt
dt
(   0 )
Integrating log e (   0 )   Kt  C
log e (   0 )   Kt  log e A
This is a straight line with negative slope
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House No 249, Chotti Baradari Part-2, Jalandhar #98152-15362