Download - Physics Mechanics
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UNIT I
LESSON 1 CONTENTS
1.0 Aims and Objectives
1.1 Introduction
1.2 Co-efficient of thermal conductivity
1.3 Cylindrical flow of heat
1.4 Determination of thermal conductivity (K), of bad conductor by
Lees disc method.
1.5 Convection currents in atmosphere Change of pressure with height Lapse rate.
1.6 Calculation of Lapse Rate (For Dry Adiabatic)
1.7 Let us Sum up
1.8 Check your progress
1.9 Lesson end activities
1.10 Points for discussion
1.11 References
1.0 AIMS AND OBJECTIVES
In this lesson you will learn about the processes, namely (i) conduction (ii) convection and
(iii) radiation by which transference of heat takes place. You will also learn to know how to
determine the coefficient of thermal conductivity for a good conductor and bad conductor.
1.1 INTRODUCTION
(i) CONDUCTION: In this process heat is transmitted from one point to the other through the
substance without the actual motion of the particles. When one end of a brass rod is heated in
flame the other end gets heated in course of time. In this case the molecules at the hot end vibrate
with higher amplitude (K.E) and transmit the heat energy from one particle to the next and so on.
Heat is said to be conducted through the rod. However, particles in the body remain in their
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position and so not move. Thus conduction is the transference of heat from the hotter part of a
body to the colder part without the motion of the particles in the body.
Metals are good conductors of heat and wood, glass, brick, cotton, wool, rubber are bad
conductors of heat. For example thick brick walls are used in the construction of a cold storage.
Brick is a bad conductor of heat and does not allow outside heat to flow inside the cold storage.
Also a steel blade appears colder than a wooden handle in winter. Steel is a good conductor of
heat. As soon as a person touches the blade heat flows from the hand (higher temperature) to the
blade to low temperature. Therefore it appears colder. Since wood is a bad conductor of heat,
does not allow heat to flow to the handle.
(ii) CONVECTION: It is the process in which heat is transmitted from one place to another by
the actual movement of the heated particles. It is prominent in the case of liquids and gases. Land
and sea breezes and trade winds are formed due to convection. Suppose water in a container is
heated from below. The layer of water in the bottom gets heated, its density decreases and it
comes up transferring heat. In this case, the transference of heat from the bottom of the vessel to
the top of the vessel is by convection. It is used in ventilation. Rooms are provided with
ventilators near the ceiling. Air in the room gets warmer due to respiration of persons in the
room. Warm air containing more of CO2 (gas) and water vapour has less density and moves
upwards. Fresh air from outside enters the room through the doors and windows. The impure air
moves outside through the ventilators. This phenomenon is continuous.
(iii)RADIATION:
It is the process in which heat in transmitted from one place to the other directly without the
necessity of the intervening medium. We get heat radiations directly from the sun without
affecting the intervening medium. Heat radiation can pass through vacuum. Also it forms a part
of electromagnetic spectrum.
Applications of heat radiations:
(1) White cloths are preferred in summer and dark colored clothes in winter.
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x
T2
T1
Fig. 1.1
Reason: When heat radiations fall on white clothes, they are reflected back. No heat is
absorbed by the clothes and a person does not get heat from outside in summer. Dark
clothes in winter will absorb the heat radiations falling on them and keep the body
warm.
(2) Polished reflectors are used in electric heaters to reflect maximum heat in the room.
1.2 Co-efficient of thermal conductivity:
Suppose there is a slab of material of area of
cross-section A. Let the opposite faces be maintained at temperatures T1 0C and T2 0C (T1>T2).
Let x be the distance between the faces. It is found that the quantity of Q of heat conducted in a
time t is directly proportional,
to A
to (T1-T2)
to t (time) and
inversely proportional to X.
Therefore, x
tTTAQ 21
or
x
tTTkAQ 21
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Where K is called the co-efficient of thermal conductivity of the material of the slab.
Definition: (Thermal Conductivity)
It is defined as the quantity of heat conducted in one second from one face to the opposite
face of a slab of area of cross-section 1sq.Cm, when the distance between the faces is equal to
1cm and the difference in temperature between the faces is equal to 10C.
Temperature gradient:
The quantity (T1-T2)/X represents the rate of fall of temperature with respect to distance.
The quantity (dT/dx) represents the rate of change of temperature with respect to the distance. As
temperature decreases with in crease in distance from the hot end, the quantity (dT/dx) is
negative and is called the temperature gradient.
Therefore, tdxdTKAQ
1.3 Cylindrical flow of heat:
Description:
Consider a cylindrical tube of length l, inner radius r1 and outer radius r2. After the steady
state is reached, the temperature on the inner surface is 1 and on the outer surface in 2 in such a
way 1>2. Heat is conducted radially across the wall of the tube. Consider an element of
thickness dr and length l at a distance r from the axis.
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Fig.1.2
r1
r r2
1
2 dr +d
Calculation:
The quantity of heat flowing per second across the element.
drdKAQ
Here A=2rl
Therefore drdKrlQ 2 (1)
Q is a constant after the steady state is reached.
Integrating equation (1) we get,
2
1
2
1
2
dKlrdrQ
r
r
111
2 21
log
Kl
rQ e
121
2 2log
Kl
rr
Q e
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B
A
T1
T2
Fig.1.3
C
211
2
2
log
rr
QK
e
(2)
211
210
2
log3026.2
l
rr
QK
1.4 Determination of thermal conductivity (K), of a bad conductor (card board) by Lees
disc method.
This apparatus consists of three parts A, B and C.
Description:
A is a solid brass plate of mass m, thickness d, specific heat s and radius
r. It is held horizontal by three threads from a stand attached to three hooks at its sides. The
cardboard C is cut to the same size and is placed above A. B is a hollow brass cylinder placed
above C through which steam is sent. C is of the same cross-section as A or B. Two
thermometers T1 and T2 measure the temperatures of B and A respectively. The thickness t of the
cardboard is first found. Steam is passed through B and its temperature 1 k is measured by the
thermometer T1. Cardboard conducts heat slowly and the temperature gradient along the
cardboard is
tdx
d 21
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Now the cardboard is removed and B is kept directly on A. The temperature of A rises beyond
2. Then B is taken out. A is allowed to cool and its temperature at various intervals of time are
noted. Cooling curve is drawn for about 5 K on either side of 2. The rate of cooling (d/dt) at 2
is found from the graph. Heat lost by lower disc per sec is equal to ms(d/dt). This heat is lost by
the top surface, bottom flat surface and the curved side of total area (2r2 + 2rd), But at the
steady temperature of A, it loses heat through the bottom and sides of area (r2 + 2rd)
Hence the rate of flow of heat is,
rdrrdr
dtdms
dtd
2
)22(2
2
dr
drdtdms
222
But we have
dxdKA
dtd
or
drdr
dtdms
trK
222
2212
from which K, the thermal conductivity of cardboard is calculated. The unit is watt per metre
per degree Kelvin.
Time (t)
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T steam
water
C
For various values of , (d/dt) is determined. This is done by drawing tangents to the curve at
various points on the curve.
Note: For experimental determination Ref. Annexure
Determination of thermal conductivity (K) of rubber:
It can be determined in the laboratory applying the principle of cylindrical flow of heat.
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Description:
A known quantity of water is taken in calorimeter C. A rubber tubing whose inner and
outer radii arc r1 and r2 is taken and a known length (50 cm) of it is immersed in water as shown
in the above figure. The initial temperature of water is noted. Let it be 1. Steam is passed
through the rubber tubing for a known time t seconds. Let the final temperature of water be 2
after applying radiation correction.
K of rubber:
Let the temperature of steam is 3. The average temperature on the outer surface of rubber
tubing is,
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Calculation:
Let the mass of water = m.
Water equivalent of the calorimeter (ws)= W
Rise in temperature (2-1)
Heat gained by water = (m+w) (2-1)
Quantity of heat flowing per second,
twmQ 12 (2)
But 211
2
10
2
log3026.2
l
rr
QK (3)
Substituting the values of equation (1) and(2) in (3),
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tl
rr
rr
wmK
22
1log3026.2
213
2
101
212
Thus K for rubber can be calculated.
1.5 Convection currents in atmosphere Change of pressure with height lapse rate:
The pressure of the atmosphere decreases with increase in height from the sea level. The
density of air is not the same at all levels. If the density of air were the same at all levels, the
atmospheric pressure at sea level would correspond to a vertical column of 8km of air only. But
it has been found that air exists even at a height of 20km.
Consider that the pressure of air at a height h = P. For an increase in height dh, the decrease in
pressure in dp.
i.e. dp = -(dh)g ( )ghP (1)
where is the density
[negative sign shows that the pressure decreases with increase in height]
We know, VM
[density = (mass/volume)]
Substituting for ,
gVMdhdp
But for a perfect dry air,
PV = RT (Gas equation)
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or
P
RTV
Hence RTgM
(-dh) dP
dhRTMg
pdp
(2)
Integrating,
ChRTMgP log
Where C is a constant.
When h = 0; P = P0; C = log P0
Substituting,
p
p-dp dp
h
dh
Earth
Fig.1.4
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0loglog PhRTMgP
hRTMg
PP
e 0
log (3)
i.e; h
RTMg
ePP
0 (4)
Equation (2) will be true only under isothermal conditions. But air temperature decreases
uniformly with height. Assuming that the lapse rate is in the lower portion of the atmosphere
(troposphere).
T = T - h
Where T in the temperature at the ground.
Substituting in equation (2) we get,
hT
dhR
Mgp
dp
KRTRMgP 0
loglog (5)
when h=0; P = P0
KTRMgP 00 loglog
00 loglog TRMgPK
Substituting this value of K in equation (5),
000 loglogloglog TRMgPhT
RMgP
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0
0
0
loglogT
dhTRMg
PP
(6)
i.e,
00log
0T
hTRMg
ePP
(7)
1.6 CALCULATION OF LAPSE RATE (For Dry Adiabatic):
The relation between pressure and temperature for an adiabatic process is,
Tp 1
constant
or TP 1 constant Differentiating
(-1) P -2 dP .T + P -1 (-) T--1dT = 0
Dividing by T P -1
01 TdT
Pdp
tdt
pdp
1 (1)
For an increase in height dh, the decrease in pressure in dp,
dhgVMdhgdP
But PV = RT
P
RTV
dhgRTMPdP
or
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dhRTMg
pdP
(2)
Equating (1) and (2)
dhRTMg
TdT
1
1
RMg
dhdT (3)
The above equation holds good only for perfect dry air. The lapse rate (dT/dR) can be calculated
from equation (3),
Substituting the values,
M=29g; g=981cm/sec2; R=8.31*107erg/grammole/K; =1.4 (ratio of specific heat)
40.1140.1
1031.898129
7dhdT
=10-4 Kper cm or 10-4 C/cm
Thus the lapse rate for a height of,
1Km=10-4105=100C
But it has been found that the average lapse rate is lower than this value. Under average
conditions the lapse rate has been found to be between 50 C and 6.5 0C per km. At night negative
lapse rate is set up because the layers of air nearer the surface may cool.
1.7 LET US SUM UP
In this lesson you have learned about the coefficient of thermal conductivity and determination
of thermal conductivity for a good conductor and a bad conductor. Also in this lesson you have
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learned about the change of pressure with height in the atmosphere with its relation to Lapse rate
and its calculations
1.8 CHECK YOUR PROGRESS
1. Define thermal conductivity
2. What is the reason for the atmospheric changes in convection currents ?
3. Define Lapse rate
1.9 LESSON END ACTIVITIES
1. The opposite faces of a metal plate of 0.2 cm thickness are at different of temperature at
100oC and the area of the plate is 100 sq. m. Find the quantity of heat that will flow through the
plate in one minute if k=0.2 CGS units.
[ Hint: Q = (KA(1-2)t)/d ]
2. A room is maintained at 20oC by a heater of resistance 20 ohms connected to 200 volts mains.
The temperature is uniform throughout the room and the heat is transmitted through a glass
window area 1 m2 and thickness 0.2 cm. Calculate the temperature outside. K=0.2 cal /m-Co-s.
[ Hint: H = (KA(1-2)t)/d where H = (v2) /4.2R ]
1.10 POINTS FOR DISCUSSIONS
(1) (i) Give the theory of cylindrical flow of heat.
(ii) Describe Lees disc method to find the coefficient of thermal conductivity of a
. poor conductor.
(2) (i) Explain coefficient of thermal conductivity. What is temperature gradient?
(ii) Describe with a neat diagram explain how you would determine the thermal
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conductivity of rubber.
1.11 REFERENCES
(1) Heat and Thermodynamics by Brij-lal and Subramanyam
(2) Heat and Thermodynamics by Anantha Krishnan
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UNIT-I
LESSON 2 CONTENTS
2.0 Aims and Objectives
2.1 Stability of the Atmosphere
2.2 Green House Gases
2.3 Newtons Law of Cooling 2.4 Black Body
2.5 Weins Displacement Law
2.6 Let us Sum up
2.7 Check your progress
2.8 Lesson end Activities
2.9 Points for Discussion
2.10 References
2.0 AIMS AND OBJECTIVES
In this lesson you will learn about the stability of atmosphere, troposphere, and
stratosphere. Also you will understand about green house gases, green house effect,
Newtons law of cooling. You will learn about what is a black body and Wiens
displacement law.
2.1 STABILITY OF THE ATMOSPHERE:
The atmospheric temperature decreases with altitude upto a height of about 20km. The
rate of fall of temperature is about 50 C per km height and is known as the lapse rate.
The atmosphere is divided into two regions,
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(1) Troposphere or the convective zone:
It is the region in which the temperature falls with increase in height from the ground.
(2) Stratosphere or advective zone:
It is the region in which the decrease in temperature does not take place with increase in
height.
The layer that separates these two regions is known as tropopause. The height of tropopause
is different at different places on the earth. It is about 14 km at the equator and about 10 km at
the poles. The fall in temperature with altitude in the troposphere is due to convection. When the
heat radiations from the sun fall on the fall on the surface of the earth, it gets heated. The
atmospheric air surrounding the earth gets heated by conduction and radiation. The heated air
moves up and convection currents are set up. The heated air that moves up from the regions of
higher to lower pressure is adiabatically cooled (ie it can neither give heat nor take heat from the
surroundings). Similarly when the colder air moves down from lower to higher pressure regions
it is adiabatically heated. Thus a convective equilibrium is set up and there is a gradual fall of
temperature with increase in height.
2.2 GREEN HOUSE GASES:
Sunlight warms the earths surface during day time and earths surface radiates heat back
to the space. Certain gases in the atmosphere absorb this radiant energy and re-emit the heat back
to the space. Those gases which are capable of absorbing and re-emitting the heat radiation are
called green house gases. The heat from the surface of the earth warms up the green house gases.
These gases in turn emit some heat into space and some back down to the surface. This fraction
of the heat provides global warming in addition to the suns direct heat. Without any green house
gases in the atmosphere, the average surface temperature would be very cold ie., around -18 0C.
Todays greenhouse gases radiate sufficient heat beck to earth to give an average global
temperature of +15 0C. In if future concentration of greenhouse gases increases, there will be
additional global warming. It is therefore necessary to maintain the present level of greenhouse
gases to avoid any drastic change in the climate.
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Greenhouse effect:
The heating up of earths atmosphere due to infrared rays which are reflected from the
earths surface by the carbon-di-oxide layer in the atmosphere is called greenhouse effect.
If green house gases continue to increase at the present ratio, it is predicted that the
earths average surface temperature may go up by 1 to 5 0C, before the end of twenty first
century. This global warming will not be evenly distributed ie., the equatorial region will warm
by 1 0C to 2 0C, the polar regions will warm up most rapidly to an increase of 6 0C to 120C.
Global warming will have two major effects on climate,
I. The polar ice caps will begin to melt and the sea level throughout the world will increase.
This will cause many low-lying coastal regions to submerge. As the polar ice melts, less
sunlight will be reflected back from the snow or more sunlight will be absorbed which will
contribute to the additional global warming.
II. In a warmer world, more water will be returned to the atmosphere by evaporation and
transpiration. There will be an increase in the total rainfall accompanied by stronger wind in
the equatorial region. The present snow regions will have less snow in winter and more heat
in summer. The vegetation may go dry.
2.3 NEWTONS LAW OF COOLING:
Newtons law of cooling states that the rate of cooling of a body is directly proportional to the
difference in temperature between the body and the surroundings. By rate of cooling is meant the
amount of heat radiated by the body per second. For any body, the rate of cooling is directly
proportional to the fall in temperature per second. Thus, if a hot body at a temperature T 1 0C cools to T2 0C in t seconds, when kept in a surrounding at a temperature T 0C, the fall in
temperature per second.
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tTT 21
The mean difference in temperature between the body and the surrounding,
TTT
221
From Newtons law it follows that when the body cools from T 1 0C to T2 0C in t seconds,
TTTt
TT
22121
Specific Heat of a Liquid by the Method of Cooling:
A calorimeter is weighed empty with a stirrer. It is filled to 2/3 of its capacity with hot water
at about 60 0C. It is suspended inside an enclosure maintained at a constant temperature and a
thermometer is placed in it. The calorimeter is kept stirred and the time taken for the calorimeter
to cool from T 1 0C to T2 0C (from 55 0C to 50 0C) is found. The calorimeter is cooled to the
room temperature and weighed. The experiment is repeated by filling the calorimeter with the
liquid up to the same level and finding the time taken to cool through the same range of
temperature. The calorimeter is weighed with the liquid. The specific heat of the liquid is
calculated as follows:
Weight of calorimeter and stirrer = w1 gm.
Time taken to cool from T 1 0C to T2 0C with water = t1 sec.
Weight of calorimeter, stirrer and water = w2 gm.
Time taken to cool from T 1 0C to T2 0C with liquid = T2 gm.
Weight of calorimeter, stirrer and water =w3 gm
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Let S1 be the specific heat of the calorimeter and S the specific heat of the liquid.
Rate of cooling of the calorimeter when filled with water,
1
211211
tTTwwSw
Rate of cooling of the calorimeter when filled with liquid,
2
211311
tTTSwwSw
The rate of cooling is the same in both the cases since the mean difference in temperature
between the calorimeter and surrounding is the same in both cases. Therefore
2
211311
1
211211
tTTSwwSw
tTTwwSw
2
1311
1
1211
tSwwSw
twwSw
from which S is calculated.
2.4 BLACK BODY:
A perfectly blackbody is one which absorbs all the heat radiations (corresponding to all
wavelengths) incident on it. When such a body is placed inside an isothermal enclosure (if a
system is perfectly conducting to the surroundings and the temperature remains constant
throughout the process, it is called isothermal process) it will emit the full radiation of the
enclosure after it is in equilibrium with the enclosure. These radiations are independent of the
nature of the substance. Such radiations in a uniform temperature enclosure are known as
blackbody radiations. Also blackbody completely absorbs heat radiations of all wavelengths.
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Black body emitter
Fig.1.9b
Heat Radiations
Incident radiations
hole
Black body absorber
Fig.1.9a
Hence we can say that the blackbody also emits completely the radiations of all wavelengths at
that temperature. It is a good absorber and emitter.
Illustrations: Absorber
A hollow copper sphere in taken and is coated with lamp black in its inner surface. A fine hole
is made and a projection is made just in front of the hole. When the radiations enter the hole,
they suffer multiple reflections and are completely absorbed. This body acts as a blackbody
absorber.
Emitter:
When this body is placed in a bath at a fixed temperature, the heat radiations come out of the
hole. The hole acts as a blackbody radiator or emitter. It should be noted that only the hole and
not the walls of the body acts as the blackbody radiator.
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Perfectly blackbody:
It is one which absorbs all the heat radiations incident on it. It does not reflect or transmit any
heat radiations. Irrespective of the colour of the incident radiations, the body appears black.
When such a body is heated to a high temperature, it emits the radiations of all wavelengths. The
radiations emitted by a perfectly blackbody depend only upon the temperature of the blackbody
and not on the nature of its material.
Absorptive and Emissive Powers:
Absorptive power:
It is the ratio of the amount of heat absorbed in a given time by the surface to the amount
of heat incident on the surface in the same time.
Emissive power:
It is the ration of the amount of heat radiations emitted by unit area of a surface in one
second to the amount of heat radiated by a perfectly blackbody of unit area in one second under
ideal conditions.
2.5 WEINS DISPLACEMENT LAW:
It states that the product of the wavelength corresponding to maximum energy and
absolute temperature is constant.
max T = Constant =0.2892 cm-k
It also shows that with increase in temperature m decreases. Wein has shown that the energy
E max is directly proportional to the fifth power of the absolute temperature.
ie., E max T5
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E max = Constant T5
Rayleigh Jeans law:
The energy distribution in the thermal spectrum according to Rayleigh is given by,
48
KTE
Where K is the Boltzmanns constant
Note: Weins law holds good only in the region of shorter wavelengths. It does not holds good at
longer wavelengths. The Rayleigh Jeans law holds good in the region of larger wavelengths
but not for shorter wavelengths.
2.6 LET US SUM UP
On going through this lesson you will understand about the stability of the atmosphere, green
house gases that are responsible for green house effect. Also you will come to know about the
Newtons law of cooling, the sources of black body radiation and the connecting law namely
Wiens displacement law.
2.7 CHECK YOUR PROGRESS
1. What are green house gases ?
2. State Wiens displacement law
3. State Newtons law of cooling
4. What is black body radiation ?
2.8 LESSON END ACTIVITIES
1. Calculate the value of the surface temperature of the moon using Weins displacement
law.
[Hint: m = 2892 x 10-6 m-k, Given m = 14.46 x 10-6 m]
2. Calculate the radiant emittance of a black body at a temperature (i) 400 K (ii) 4000 K.
( = 5.67210-8 M.K.S units)
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2.9 POINTS FOR DISCUSSION
1. Write short notes on,
(i) Black body radiation
(ii) Temperature of the sun.
2. How will you determine the specific heat of liquid by Newtons law of cooling.
2.10 REFERENCES
1 Heat and Thermodynamics by Brij-lal and Subramanyam
2 Heat and Thermodynamics by Anantha Krishnan
3 Heat and Thermodynamics by R. Murugesan
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LESSON 3 CONTENTS
3.0 Aims and Objectives
3.1 Stefans Law Also Called as Stefan Boltzmann Law
3.2 Determination of Stefans Constant (Laboratory Method)
3.3 Solar Constant
3.4 Solar Spectrum
3.5 Temperature of Sun
3.6 Let us Sum up
3.7 Check your progress3
3.8 Lesson end Activities
3.9 Points for Discussion
3.10 References
3.0 AIMS AND OBJECTIVES
In this lesson another important law related to black body radiation namely Stefans law will be
discussed. The derivation of it and its determination by experimental method will also be
detailed. Followed by it the Solar constant, solar spectrum and temperature of the sun will be
discussed.
3.1 STEFANS LAW ALSO CALLED AS STEFAN BOLTZMANN LAW:
According to this law, the rate of emission of radiant energy by unit area of a perfectly
blackbody is directly proportional to the fourth power of its absolute temperature,
R T4 or R = T4 (1)
where is called Stefans constant.
If the body is not perfectly black and its emmissivity or relative emittence is e, then
R = e T4 (2)
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Here e varies between zero and one, depending on the nature of the surface. For a perfectly
blackbody e = 1. This law is not only true for emission but also for absorption of radiant energy.
When the body has the same temperature as that of the surroundings, the rate of emission and
absorption are equal.
Hence, if a perfectly blackbody at temperature T1 is surrounded by a wall at temperature T2,
the net rate of loss (or gain) of heat energy per unit area of the surface is given by,
R (T14 - T24)
R = (T14 - T24)
If the body gas an emissivity e then,
R = e (T14 - T24)
Mathematical derivation of Stefans law:
The fact that blackbody radiations exert pressure similar to a gas, helps in applying
thermodynamics to heat radiations.
Let be the energy density of radiations inside a uniform temperature enclosure at
temperature T. P is the pressure and V is the volume.
Applying the first law of thermodynamics,
H = du + P.dV (1)
Where H is the quantity of heat supplied to a system, the amount of external work done be
P.dV and the increase in internal energy of the molecules be dV.
Applying thermodynamical relation,
VT T
PVH
(2)
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VT T
PTV
VPU
PTPT
VU
VT
(3)
3
PVU
or
TVU
Here is a function of temperature alone.
Substituting these values in equation (3),
33
dt
dT
dt
dT 33
4
TdTd 4
integrating log = 4logT + constant
or 4KT (4)
Where K is a constant.
Also the total rate of emission per unit area of a blackbody is proportional to the energy
dentsity.
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Steam
G
Rh
C
H B
T T
A W
E F
Fig.1.12a
R T4
R = T4 (5)
Where is Stefans constant.
Note: The value of Stefans constant = 5.672 x 10-8 in MKS units
3.2 DETERMINATION OF STEFANS CONSTANT (Laboratory Method):
The laboratory apparatus used to determine the Stefans constant is shown in Fig 1.12. A
hollow hemispherical metallic vessel A is enclosed in a wooden box W. The inner surface of A is
coated with lamp black and wooden box W is lined with tin plates. The whole apparatus is
placed on a wooden base having a small hole at its centre. The vessel A is heated by passing
steam inside the box and A acts as a black body radiator. The thermometers T , T record the
temperature of A.
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A
B C
Deflection
Fig1.12b
A small silver disc B whose upper surface is coated with lamp black is placed at the central hole.
The ebonite covering C is used to cover and uncover the disc B from the radiations of the
enclosure. It can be arranged from outside with the help of the handle H. The disc B is connected
to a thermocouple arrangement. One junction of the thermocouple is immersed in a tube
containing oil. The tube is surrounded by a beaker containing water. A sensitive galvanometer G
is used in the circuit. The leads connected to the terminals of the galvanometer are immersed in
cotton wool in the box F to avoid any distribution effect due to the difference of temperature in
the leads. A rheostat Rh can be used in the circuit to obtain the deflection within the range. The
actual experiment consists of two parts.
1. The thermocouple is first standardized. Before passing steam into the chamber the disc B is
at the room temperature.
The water bath E acts as a hot junction. It is heated and at various temperatures of the hot
junction, the corresponding deflections in the galvanometer are noted. A graph between the
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31
Time (sec)
D
G F
E
Deflection
Fig.1.12c
difference of temperature of the hot junction and the room temperature along the Y-axis and
galvanometer deflection along X-axis is plotted in fig 1.12b. From the graph,
BCAB
ddT
tan (i)
2. The disc is completely covered with C and steam is passed into the chamber. After some
time, the thermometers T. T show constant temperature. The bath E is kept at room temperature.
With the help of the handle H, the cover C is tilted so that the upper surface of the disc B
receives the radiations from the enclosure. The deflections in the galvanometer are observed after
equal intervals of time (say 10 seconds). A graph is plotted between time and deflection
(fig1.12c). A tangent is drawn on the curve at a point D.
GFEF
ddt
tan (ii)
Theory:
Let, at any instant, the temperature of the enclosure and the disc be T1 and T2 (degrees
Kelvin) respectively. The disc will absorb more heat from the surroundings and radiate less heat
to the surroundings. Its temperature will rise. From Stefans law,
R1 = T14 and R2 = T24
(R1 - R2) = ( T14 - T24) (iii)
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32
Here R1 is the amount of heat radiation absorbed per unit area per second by the disc and R2 is
the amount of heat radiation emitted per unit area per second by the disc.
Let the mass of the disc be m, specific heat S, rate of rise of temperature dT/dt, and area of the
upper surface A.
Then
dtdTmS
JARR
21
dtdTmS
JATT
244
1
dtdT
TTAJmS
2
441
(iv)
To find (dT/dt), equations (i) and (ii) are used
tantan
dtd
ddT
dtdT
To find T2, the deflection in the galvanometer corresponding to the point D on graph in fig
1.12c is noted and for this deflection, the temperature difference from the graph fig 1.12bis
noted. To this reading add the room temperature and find T2 in degrees Kelvin.
Substituting these values in equation (iv),
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33
tan
tan2
441 TTAJmS
(v)
Hence can be calculated.
3.3 SOLAR CONSTANT:
The sun is the source of heat radiations and it emits heat radiations in all directions. The
earth receives only a fraction of the energy emitted by the sun. The atmosphere also absorbs a
part of the heat radiations and air, clouds, dust particles etc. in the atmosphere scatter the heat
and light radiations falling on them. From the quantity of heat radiations received by the earth, it
is possible to estimate the temperature of the sun. Therefore, to determine the value of a constant,
called solar constant, certain ideal conditions are taken into consideration.
Solar constant:
It is the amount of the heat energy (radiation) absorbed per minute by one sq cm of a perfectly
black body surface placed at a mean distance of the earth from the sun, in the absence of the
atmosphere, the surface being held perpendicular to the suns rays.
The instruments used to measure the solar constant are called pyrheliometers. The heat energy
absorbed by a known area in a fixed time is found with the help of the pyrheliometers. To
eliminate the effects of absorption by the atmosphere, the value of the solar constant is found at
various altitudes of the sun on the same day under similar sky conditions. If S is the observed
solar constant, S0 the true solar constant and Z the altitude (angular elevation) of the sun, then
S = S0 sec Z (i)
or log S = log S0 + sec Zlog (ii)
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34
M3
M3
M2
P
light source
Thermopile
Fig.8.14a
Infra-red spectrometer
Here is a constant.
A graph is platted between log S along the y-axis and sec Z along the x-axis. The graph is a
straight line. Produce the graph to meet the y-axis. The intercept on the y-axis gives log S0, the
value of S0 the solar constant can be calculated. The value obtained varies between 1.90 and 2.60
calories per sq cm per minute.
3.4 SOLAR SPECTRUM:
The radiation received from the sun is similar to that of a perfectly black body. In solar
spectrum the wavelength corresponding to maximum energy is about 5000 A for a surface
temperature of 5750 K. The solar spectrum consists of a large number of dark lines called
Franunhofer lines. These lines were first observed by Wollaston in 1802 and later studied by
Fraunhofer in 1814. These lines are observed in the complete range of the spectrum viz. ultra-
violet, visible and infrared. Fraunhofer measured the wavelengths of many of these lines
accurately and found that they occupied exactly the same positions as the bright lines emitted by
different gases and vapours.
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35
According to Kirchhoffs law of radiation, any substance at a lower temperature will absorb the
radiations of those wavelengths which it will emit when excited by electric discharge. This the
solar spectrum is an absorption spectrum and the Fraunhofer lines are the absorption lines of
relatively cooler gases and vapours present in the earths atmosphere and the suns outer
atmosphere. The central portion of the sun is called the photosphere which is at a very high
temperature. Surrounding the photosphere is the chromosphere which is at a much lower
temperature than the photosphere. The positions of the Fraunhofer lines in the solar spectrum
give the atmosphere. The presence of more than sixty elements in the suns atmosphere is found
this way. In fact, helium was first discovered by the study of the Fraunhofer lines in the solar
spectrum before it had been isolated in the laboratory. The most prominent of the Fraunhofer
lines are denoted by the letters of the alphabet. Some of the lines, with their wavelengths and the
elements responsible for their absorption are given below:
Line Element Wavelength in
A Atmospheric oxygen 7594
B Atmospheric oxygen 6867
C Hydrogen 6563
D1 Sodium 5896
D2 Sodium 5890 F Hydrogen 4861
G1 Hydrogen 4341
H Calcium 3969
K Calcium 3934
Infra-red-Spectrum:
Extending on either side of the visible spectrum, there are invisible radiations which do not
cause the sensation of sight. The radiations beyond the red end of the visible spectrum are called
infra-red radiations and their wavelengths extend up to 400,000 . Sun is a powerful source of
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36
infra-red radiation. Beyond the violet end of the visible spectrum, the radiations extending up to
a wavelength of 100 are called ultraviolet radiations.
The heating effect of the infra-red radiations is used in measuring the wavelength of the
radiation. An infra-red ray spectrometer is shown in Fig 1.14(a). Light from a strong source of
light such as an electric arc is rendered parallel by reflection from a concave stainless steel
mirror M1. This parallel beam is refracted through the rock salt prism P and the emergent
dispersed beam after reflection from the mirrors M1 and M3 is incident on a thermopile or a
bolometer. M2 is a plane mirror and M3 is a concave mirror. Wiens law of radiation is given by,
mT = 0.2892
where T is the absolute temperature and m is the wavelength of the radiation. The thermopile
readings help in the calculation of temperature and from the temperature, can be calculated.
As infra-red radiations are not absorbed by air or thick frog, infra-red ray photographs can be
taken over long distances of fog and mist where visible light cannot penetrate. For this, specially
designed photographic plates are used with suitable filters. A solution of iodine in alcohol is a
suitable filter, because this transmits the infra-red radiations and absorbs the visible light. In
World War II, infra-red photography played a very useful part in detecting objects in the dark
through mist, fog and clouds. Infra-red radiations have a wide application in the field of
medicine, industry etc. Infra-red radiations can penetrate deep into the human body and by their
property of heating can dilate the blood vessel at the portion exposed to the radiations. This
enables increased flow of blood.
Ultra-violet Spectrum:
The spectrum that covers the wavelengths from 4000 to 100 is called the ultra-violet
spectrum. An electric arc of carbon, iron or other materials, mercury vapour lamps, discharge of
electricity through hydrogen contained in quartz tubes are some of the artificial sources that give
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37
ultra-violet radiations. Ordinary glass absorbs the ultra-violet radiations. Hence quartz lenses
and prisms are used. But quartz is double refracting material. If a single prism of quartz is used,
due to the property of double refraction of the prism, two images of the prism, two images of the
slit (ordinary and extraordinary) corresponding to a single wavelength are observed. This will
reduce the sharpness of each image. To compensate for this, the collimating lens is
made from right-handed quartz and the telescope objective is made from left-handed quartz.
Similarly the prism used for the dispersion of the incident beam consists of two halves. The two
halves are held together by glycerine. One half is made from right-handed quartz and the other
half is made from left-handed quartz.
In fig 1.14(b) ABD is a right-angled prism of right-handed quartz and ADC is of left-handed
quartz. For recording an ultra-violet spectrum for wavelengths shorter than 1,200 , a concave
reflection grating is used. The diffracted beam is photographed. The grating and the
photographic plate are enclosed in a metal chamber which is evacuated.
Ultra-violet radiations have a variety of applications. Sterilisation of rooms in which blood
plasma, drugs, vaccines etc. are prepared and sealed in the containers is done by ultra-violet
radiations. Drugs, poisons, dyes etc. fluoresce under the action of ultra-violet rays. The resolving
A
B C R L
Fig1.14
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38
power of a microscope is increased when ultra-violet light is used for illumination. Fluorescent
tubes depend on the principle of fluorescence effected by ultra-violet radiations.
Electromagnetic Spectrum:
Visible spectrum includes those wavelengths which can stimulate the sense of sight. But there
is no basic difference between light waves and electromagnetic waves produced by electrical
oscillating circuits. The term electromagnetic spectrum is used for the range of wavelengths from
104 meters to 1 (10-8cm). There is in fact no limit to the production of electromagnetic waves
of very long wavelengths. The frequency of an alternating current generator can be made as low
as possible by decreasing the speed of the generator. The wavelength of waves transmitted by a
50 cycle transmission line is 5x108 cm. Waves of shorter wavelength can be produced by
electrical oscillators. X-rays and gamma rays represent the waves of very short lengths.
It is interesting to note that the visible range of the spectrum comprises only a small range of
the electromagnetic spectrum extending approximately from 4000 in the extreme violet region
to 8000 in the extreme red. Beyond the violet region of the visible spectrum is the ultraviolet,
the X-rays and -rays.
3.5 TEMPERATURE OF THE SUN:
The sun consists of a central hot portion surrounded by the photosphere. The central portion
has a temperature of the order of 107 K. The photosphere has a temperature of about 6000 K.
This temperature is also called the effective temperature of the sun. Considering the sun as a
perfect black body radiator, the temperature of the sun can be calculated.
Let the mean distance of the sun from the earth be R and S the solar constant. Then, the total
amount of heat energy received by the sphere of radius R in one minute = 4R2S.
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39
If r is the radius of the sun, then the amount of the heat energy radiated by 1 sq cm surface of
the sun in one minute,
E = (4R2S / 4r2) = (R / r)2 x S
Taking, R = 148.48x107 km
r = 6.928x105km
The mean value of S = 1.94 cals per cm3 per minute.
E = (148.48x107 / 6.928x105)2 x (1.94 / 60) cals per sec -----(i)
Also, E = T4
But = 5.75x10-5 ergs per cm2 per second
= (5.75x10-5 / 4.2x107) cal per cm2 per second
E = (5.75x10-5 / 4.2x107) . T4 -----(ii)
Equating (i) and (ii),
(5.75x10-5 / 4.2x107) . T4 = (148.48x107 / 6.928x105)2 x (1.94 / 60)
T4 = 5730 K -----(iii)
This temperature gives the effective temperature of the sun acting as a black body radiator.
The actual temperature of the sun is higher than this value. The temperature of the sun is usually
taken as 6000 K.
Temperature of the sun can also be calculated from Wiens displacement law,
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40
max T = 0.2892
The wavelength of the radiations for the radiations for which the energy is maximum in the
spectrum is 4900x10-8 cm.
Substituting the value of m, the value of T comes out to be 5902 K. This value is in
agreement with the accepted value. Hence, the effective temperature of the sun (photosphere) is
about 6000 K.
3.6 LET US SUM UP
In this lesson an important law viz., Stefan Boltzman law is discussed and its derivation and
determination is explained. Also you learned about solar constant and details of various spectrum
such as solar spectrum, infrared, ultraviolet spectrum. The topic of solar constant, and a method
to obtain temperature of the sun is also described in this lesson.
3.7 CHECK YOUR PROGRESS
1. State Stefan Boltzman law
2. Define Solar Constant
3. Write a note on Solar spectrum
3.8 LESSON END ACTIVITIES
1. Calculate the radiant emittance of a black body at a temperature of i) 400 K and ii) 4000 K
using Stefans law.
[Hint: R = T4, = 5.67 x 10-8 MKS units]
2. If a solar constant at the surface of the earth is 1400 W m-2, compute its value at Jupiter which
is 5.2 AU away from the sun.
[Hint: S2/S1 = (R1/R2)2, Here R1 = 1 AU ]
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41
3. The total luminosity of the sun is 3.9 x 1026 watts. The mean distance of the sun from the earth
is 1.496 x 1011 m. Calculate the value of solar constant.
[Hint: S = (E/4R2) ]
3.9 POINTS FOR DISCUSSIONS
(1) (i) State Stefans law of heat radiation and derive the law mathematically.
(ii) Describe an experiment for the verification of Stefans constant. What is its
numerical value in M.K.S units?
(2) Define solar constant. How is it experimentally determined? Comment on the source
of energy in the sun.
3.10 REFERENCES
(1) Heat and Thermodynamics by Brijlal and Subramanyam
(2) Treatise on Heat by Srivastava
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42
************************************************************************
Annexure-I
************************************************************************
The given cardboard is placed between the disc Lees and the steam chamber. Two
thermometers are inserted into the radial holes drilled on the side of the metallic discs. Steam is
then passed through the steam chamber from a boiler until a steady state is reached. The steady
temperatures of the disc 1 and steam chamber 2 recorded by the thermometers are noted.
The cardboard is then removed and the disc heated directly until the temperature rises 5C above
1. Then the steam chamber is removed, and the disc is allowed to cool. A stop clock is started
and the time for every 1C fall in temperature of the disc is noted until the temperature of the disc
falls 5C below 1. The values are tabulated.
A cooling curve is drawn with time along x-axis and the temperature along the y-axis as shown
in figure below,
rom these readings, thermal conductivity of the bad conductor is calculated by usin equation.
Observation:
Mass of the disc M=kg
Specific heat of the material of the disc, s=Jkg-1K-1
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43
Radius of the disc, r=m
Thickness of the disc, l=m
Thickness of the bad conductor, d=m
The steady temperatures of the disc, 1=m
The steady temperatures of the steam chamber, 2=m
Rate of cooling at 1 [from the cooling curve], (d/dt) =Ks-1
Results:
The coefficient of the given thermal conductivity of the bad conductor
KWm-1K-1
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44
QUESTIONS
(1) (i) Give the theory of cylindrical flow of heat.
(ii) Describe Lees disc method to find the coefficient of thermal conductivity of a
. bad conductor.
(2) (i) Explain coefficient of thermal conductivity. What is temperature gradiant?
(ii) Describe with a neat diagram explain how you would determine the thermal
conductivity of rubber.
(3) (i) State Stefans law of heat radiation and derive the law mathematically.
(ii) Describe an experiment for the verification of Stefans constant. What is its
numerical value in M.K.S units?
(4) Define solar constant. How is it experimentally determined? Comment on the source
of energy in the sun.
(5) Write short notes on,
(iii) Black body radiation
(iv) Solar spectrum.
(v) Temperature of the sun.
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45
PROBLEMS
(1) Two sides of a square copper plates of side 15cm and of thickness 0.5cm are kept at 0o C
and 100o C. Calculate the quantity of heat conducted through the plate per hour. The
thermal conductivity of copper is 0.9.
Solution:
td
KAQ
21
41863600105.0
100100159.0 2
2
= 6.1103
(2) Calculate the value of the surface temperature of the moon using Weins displacement
law. (Value of m = 14.46 micron).
Weins displacement law m = constant
The value of constant = 289210-6M.K
= temperature to be calculated
14.4610-6 = 289210-6
= 200K
(3) Calculate the radiant emittance of a black body at a temperature (i) 400 K (ii) 4000 K.
( = 5.67210-8 M.K.S units)
Solution:
R = T4
i) R = 5.762 10-8[400]4 = 1452 watts/m2
ii) R = 5.762 10-8[4000]4 = 14520 Kilo.watts/m2
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46
(4) Calculate the energy radiated per minute from the filament of an incandescent lamp at
2000 K, if the surface area is 5.010-5m2
and its relative emittance is 0.85.[ = 5.67210-8]
Solution:
E = A e t (T4)
Here A = Area; e= relative emittance; t=time
E = 5 10-5 0.85 5.672 10-8 60 (2000)4
E = 2315 Joules
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47
UNIT II
LESSON 4 CONTENTS
4.0 Aims and Objectives
4.1 First Law of Thermodynamics.
4.1 (a) First Law of Thermodynamics for a Change in State of a Closed Systems
4.2 Isothermal Process
4.2(a) Adiabatic Process.
4.2(b) Isochoric Process
4.3 Gas Equation During an Adiabatic Process
4.3 (a) Slopes of Adiabatic and Isothermals
4.3 (b) Work done During an Isothermal Process
4.3 (c) Work done during and Adiabatic Process.
4.4 Let us Sum up
4.5 Check your progress
4.6 Lesson end activities
4.7 Points for discussion
4.8 References
4.0 AIMS AND OBJECTIVES In this lesson you will learn the first law of thermo dynamics, adiabatic and isothermal processes and also will learn the work done during adiabatic and isothermal processes.
4.1 First Law of Thermodynamics.
Joules law gives the relation between the work done and the heat produced. It is true
when the whole of the work done is used in producing heat or vice varsa. Here, W= JH where J
is the Joules mechanical equivalent of heat. But in practice, when a certain quantity of heat is
supplied to a system the whole of the heat energy may not be converted into work. Part of the
heat may be used in doing external work and the rest of the heat might be used in increasing the
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48
internal energy of the molecules. Let the quantity of heat supplied to a system be H, the amount
of external work done be W and the increase in internal energy of the molecules be dU. The
term U represents the internal energy of a gas due to molecular agitation as well as due to the
forces of inter-molecular attraction. Mathematically.
H = dU + W
.(i) Equation (i) represents the first law of thermodynamics. All the quantities are measured
in heat units. The first law of thermodynamics states that the amount of heat given to a systems is
equal to the sum of the increase in the internal energy of the system and the external work done.
For a cyclic process, the change in the internal energy of the system is zero because the
system is brought back to the original condition. Therefore for a cyclic process dU = 0
and H = w (ii)
This equation represents Joules law. (Both are expressed in heat units).
For a system carried through a cyclic process, its initial and final internal energies are
equal. From the first law of thermodynamics, or a system undergoing any number of complete
cycles.
U2 U1 = 0
H = W
H = W (both are in the heat units)
4.1 (a) First Law of Thermodynamics for a Change in State of a Closed Systems
For a closed system during a complete cycle, the first law of the Thermodynamics in
H = W
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49
In Practice, however, we are also concerned with a process rather than a cycle. Let the system
undergo a cycle, changing its state from 1 to 2 along the path A and from 2 to 1 along the path B.
This cyclic process is represented in the P-V diagram (Fig. 2.1a).
H = w
Fig. 2.1 (a)
For the complete cyclic process
2A 1B 2A 1B H + H = W + W (i) 1A 2B 1A 2B Now, consider the second cycle in which the system changes from state 1 to state 2 along the
path A and returns from state 2 to state 1 along the path C. For this cyclic process.
2A 1C 2A 1C H + H = W + W (ii) 1A 2C 1A 2C Subtracting (ii) from (i)
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50
1B 1C 2B 1C H - H = W + W 2B 2C 2B 2C (or) 1B 1C (H - W) = (H - W) (iii) 2B 2C
Here B and C represent arbitrary process between the states 1 and 2. Therefore, it can be
concluded that the quantity (H - W ) is the same for all processes between the states 1 and 2.
The quantity (H-W) depends only on the initial and the final states of the system and is
independent of the path followed between the two states.
Let dE = (H W)
From the above logic, it can be seen that
dE = constant and is independent of the path
This naturally suggests that E is a point function and dE is an exact differential.
The point function E is a property of the system.
Here dE is the derivative of E and it is an exact differential.
H - W dE --- (iv)
or H = dE + W --- (v)
Integrating equation (v), from the initial state 1 to the final state 2
1H2 = (E2 E1) + 1W2
(Note. 1H2 cannot be written as (H2 H1), because it depends upon the path).
Similarly, 1W2 cannot be written as (W2 W1), because it also depends upon the path.
Here 1H2 represents the heat transferred,
1W2 represents the work done,
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51
E2 represents the total energy of the system in state 2,
E1 represents the total energy of the system in state 1,
At this point, it is worthwhile discussing what this E can possibly mean. With reference
to the system, the energies crossing the boundaries are all taken care of in the form of H and W.
For dimensional stability of Eq. (v), this E must be energy and this must belong to the system.
Therefore, E2 represents the energy of the system in state 2
E1 represents the total energy of the system in state 1
This energy E acquires a value at any given equilibrium condition by virtue of its
thermodynamic state. The working substance for example a gas, has molecules moving in all
random fashion. The molecules have energy associated by virtue of mutual attraction and this
part is similar to the potential energy of a body in macroscopic terms. They also have velocities
and hence kinetic energy. This energy E therefore can be visualized as comprising of molecular
potential and kinetic energies in addition to macroscopic potential and kinetic energies. The first
part, which owes its existence to the thermodynamic nature is often called the internal energy
which is completely dependent on the thermodynamic state and the other two depend on
mechanical or physical state of the system.
E = U + KE + PE + others which depend upon chemical nature etc.
For a closed system (non chemical) the changes in all others except U are insignificant
and dE = dU
From equation (v)
H = dU + W Here all the quantities are in consistent units.
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52
4.2 Isothermal Process
If a system is perfectly conducting to the surroundings and the temperature remains
constant throughout the process, it is called an isothermal process. Consider a working substance
at a certain pressure and temperature and having a volume represented by the point A (Fig.2.2)
Fig. 2.2
Pressure is decreased and work is done by the working substance at the cost of its internal
energy and there should be fall in temperature. But, the system is perfectly conducting to the
surroundings. It absorbs heat from the surroundings and maintains a constant temperature. Thus
from A to B the temperature remains constant. The curve AB is called the isothermal curve or
isothermal.
Consider the working substance at the point B and let the pressure be increased. External
work is done on the working substance and there should be rise in temperature. But the system is
perfectly conducting to the surroundings. It gives extra heat to the surroundings and its
temperature remains constant from B to A.
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53
Thus during the isothermal process, the temperature of the working substance remains
constant. It can absorb heat or give heat to the surroundings. The equation for an isothermal
process is
PV = RT = constant (For one gram molecule of a gas)
For n gram molecules of a gas PV = nRT
4.2(a) Adiabatic Process.
During an adiabatic process, the working substance is perfectly insulated from the
surroundings. It can neither given heat nor take heat from the surroundings. When work is done
on the working substance, there is rise in temperature because the external work done on the
working substance increases its internal energy. When work is done by the working substance, it
is done at the cost of its internal energy. As the system is perfectly insulated from the
surroundings, there is fall in temperature.
Thus, during an adiabatic process, the working substance is perfectly insulated from the
surroundings. All along the process there is change in temperature. A curve between pressure
and volume during the adiabatic process is called an adiabatic curve or an adiabatic.
Examples.
1. The compression of the mixture of oil vapour and air during compression stroke of an internal
combustion is an adiabatic process and there is rise in temperature.
2. The expansion of the combustion products during the working stroke of an engine is an
adiabatic process and there is fall in temperature.
3. The sudden bursting of a cycle tube is an adiabatic process.
Apply the first law of thermodynamics to an adiabatic process, H = 0,
H = dU + W
or 0 = dU + W
The process that take place suddenly or quickly are adiabatic processes.
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54
4.2(b) Isochoric Process
If the working substance is taken in a non-expanding chamber, the heat supplied will
increase the pressure and temperature. The volume of the substance will remain constant. Such a
process is called an Isochoric Process. The work done is zero because there is no change in
volume. The whole of the heat supplied increases the internal energy. Therefore, during the
isochoric process W = 0.
H = dU
The heat transferred in such a process
H = CvdT
C2dT = dU
Hence C2 is the specific heat for one gram-molecule of a gas at constant volume.
4.3 Gas Equation During an Adiabatic Process
Consider 1 gram of the working substance (ideal gas) perfectly insulated from the
surroundings. Let the external work done by the gas be W.
Applying the first law of thermodynamics
H = dU + W
But H = 0
and W = P.dV
Where P is the pressure of the gas and dV is the change in Volume.
P.dV 0 = dV + .(1) J
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55
As the external work is done by the gas at the cost of its internal energy, there is fall in temperature by dT. dU = 1 x Cv x dT
P.dV Cv dT + = 0 .(2) J For an ideal gas PV = rT Differentianting, P.dV + V.dP = r.dT Substituting the value of dT in equation (ii), P.dV + V.dP P.dV CV + = 0 r J
CV (P.dV + V.dP) + r. P.dV = 0 . . . . (3)
J But, r = Cp - Cv J Cv P.dV + Cv.V.dP + Cp.PdV - Cv.PdV = 0
Cp P.dV + Cv.V.dP = 0
Dividing by Cv.PV Cp dV dP + = 0 Cv V P Cp
But = Cv
dP dV + = 0 P V
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56
Integrating, log P + log V = const.
log PV = const..
or PV = const. (4)
This is the equation connecting pressure and volume during an adiabatic process.
Taking PV = rT
rT or P = V rT V = const. V But r is const rTV -1 = const.
TV -1 = const.
Also rT V = P rT
P = const. P r T = const. or T -1 = const.
P - ------ = const
or T Thus, during an adiabatic process (i) PV = const.
(ii) TV -1 = const. and
(iii) P -1 = const. T = const.
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57
4.3 (a) Slopes of Adiabatics and Isothermal
In an isothermal process PV = const. Differentiating, PdV + VdP = 0 dP P
= - ...(1) or dV V
In an adiabatic process
PV = Const.
Differentiating,
P V -1 dV + VdP = 0
dP P = - ...(2)
dV V Therefore, the slope of an adiabatic is times the slope of the isothermal
.
Fig. 4.3 (a)
Hence, the adiabatic curve is steeper than the isothermal curve at a point where the two
curves intersect each other. (Fig:4.3(a).
4.3 (b) Work done During an Isothermal Process
When a gas is allowed to expand isothermally, work is done by it.
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Let the initial and final volumes be V1 and V2 respectively. In Fig.2.3 (b), the area of the
shaded strip represents the work done for a small change in volume dV. When the volume
changes from V1 to V2.
V2 Work done = P. dV = area ABba ...(i)
V1
Fig. 4.3 (b)
Fig.4.3(b) represents the indicator diagram. Considering one gram molecule of the gas
PV = RT
RT P =
or V
V2 dV W = RT V1 V
V2 = RT loge . (2) V1
Also P1V1 = P2V2
or V2 = P1 . . . (3) V1 P2 W = RT x 2.3026 x log 10 P1 (4)
P2
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Here, the change in the internal energy of the system is zero (because the temperature
remains constant). So the heat transferred is equal to the work done.
4.3 (c) Work done during and Adiabatic Process.
During an adiabatic process, the gas expands from volume V1 to V2. As shown by the
indicator diagram the work done for an increase in
Fig. 2.3 (c)
Fig:4.3(c) Volume dV = P.dV. Work done when the gas expands from V1 to V2 is given by,
v2 W= PdV = Area ABba V1 During an adiabatic process,
PV = const = K
K P = V V2 dV W= K V1 V = 1 1 1 - 1- V2 -1 V1 r-1
Since A and B lie on the same adiabatic P1V1 = P2V2 = K
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W = 1 K K - 1- V2 -1 V1 -1 W = 1 P2V2 P1V1 - 1- V2 -1 V1 -1 W = 1 P2V2 - P1V1 (2) 1- Taking T1 and T2 as the temperatures at the points A and B respectively and considering
one gram molecule of gas.
P1V1 = RT1
and P2V2 = RT2
Substituting these values in equation (ii)
W = 1 RT2 - RT1 (3) 1-r Here heat transferred is zero because the system is thermally insulated from the surroundings. 4.4 LET US SUM UP
From this lesson you have understood about the first law of thermodynamics and also learned
about the difference between the adiabatic and isothermal processes. The work done during the
adiabatic and isothermal processes is also discussed in this lesson.
4.5 CHECK YOUR PROGRESS
1. State first law of thermodynamics and mention its importance in heat
2. What is isothermal and adiabatic processes ?
3. What are the differences in the slope of isothermal and adiabatic processes ?
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4.6 LESSON END ACTIVITIES
1. A water fall is 500 meters high. Assuming that the entire kinetic energy gained during fall is
converted into heat. Calculate the rise in temperature of water in the base of the fall.
(g = 9.8 m/s2)
[Hint: i) H = (mgh)/J, ii) H= ms
Using first law of thermodynamics calculate the change in the internal energy of the system if
the latent heat of vaporization is 5.4 x 105 cal/Kg. Given dw=4.027x104 cal.
[Hint: dH = dv+dw]
4.7 POINTS FOR DISCUSSION
1. (i) Explain Isothermal and Adiabatic Process.
(ii) Derive an expression for the work done during an adiabatic process.
2. (i) Deduce the equation for the work done during Isothermal Process.
4.8 SOURCES
1. Heat and thermodynamics by R. Murugesan
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LESSON - 5 CONTENTS
5.0 Aims and Objectives
5.1 Clement and Desormes Method determination of
5.2 Irreversible Process
5.3 Reversible Process
5.4 Let us Sum up
5.5 Check your progress
5.6 Lesson end Activities
5.7 Points for Discussion
5.8 References
5.0 AIMS AND OBJECTIVES In this lesson you will learn the ratio of the specific heat of gat at constant volume and at constant pressure and its experimental determination. Also you will learn about the reversible and irreversible processes.
2.9 Clement and Desormes Method determination of
Clement and Desormes in 1819 designed an experiment to find , the ratio between the
two specific heats of a gas.
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Fig. 5.1
The vessel A has a capacity of 20 to 30 litres and is fitted in a box containing cotton and
wool. At the top end, three tubes are fitted as shown in figure.5.1. Through S1, dry air is forced
into the vessel A. The stop cock S1, is closed when the pressure inside A is slightly greater than
the atmospheric pressure. Let the difference in level on the two sides of the manometer M be H
and the atmosphere pressure be P0. The pressure of air inside the vessel is P1.
The stop-cock S is suddenly opened and closed just at the moment when the level of the
liquid on the two sides of the manometer are the same. Some quantity of air escapes to the
atmosphere. The air inside the vessel expands adiabaticaliy. The temperature of air inside the
vessel falls due to adiabatic expansion. The air inside the vessel is allowed to gain heat from the
surroundings and it finally attains the temperature of the surroundings. Let the pressure at the end
be P2 and the difference in levels on the two sides of the manometer be h.
Theory. Consider a fixed mass of air left in the vessel in the end. This mass of air has
expanded from volume V1 (Less than the volume of the vessel) at pressure P1 to volume V2 at
pressure P0. The process is adiabatic as shown by the curve AB (fig 2.4(a) .
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P2V1 = P0V2
P1 V2 = P0 V1 (1)
Finally the point C is reached. The points A and C are at the room temperature. Therefore
AC can be considered as an isothermal.
P1 V1 = P2V2 V2 P1 = V1 P2 (2)
Substituting the value of V2 in equation (i). V1
P1 P1 r = P0 P2
Fig. 5.1(a)
Taking Logarithms, log P1 log Po = (log P1 log P2)
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log P1 - log Po
=
log P1 - log P2 (3)
But P1 = Po + H and P2 + Po + h = log (Po + H)- log Po log (Po + H) log (Po + h) log (Po + H) Po =
log (Po + H) Po+h log 1 + H Po
log 1 + H-h Po+h H Po H Approximately, = = H-h H-h
H
Hence = H-h ...(4)
Similarly, for any gas can be determined by this method.
Drawbacks. When the stop-cock is opened, a series of oscillations are set up. This is
shown by the up and down movement of the liquid in the manometer. Therefore, the exact
moment when the stopcock should be closed is not known. The pressure any not be equal to the
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atmospheric pressure when the stop-cock is closed. It may be higher or less then the atmospheric
pressure. Thus the result obtained will not be accurate.
5.2 Ireversible Process
The thermo dynamical state of a system can be defined with the help of the thermo
dynamical coordinates of the system. The state of a system can be changed by altering the
thermo dynamical coordinates. Changing from one state to the other by changing the thermo
dynamical coordinates is called a process.
Consider two states i.e, state A and State B. Changes of state from A to B or vice versa is
a process and the direction of the process will depend upon a new thermo dynamical coordinate
called entropy. All process are not possible in the universe.
Consider the following process:
(1) Let two blocks A and B at different temperatures T1 and T2 (T1>T2) be kept in
contact but the system as a whole is insulated from the surroundings. Conduction
of heat takes place between the blocks, the temperature of A falls and the
temperature of B rises and thermo dynamical equilibrium will be reached.
(2) Consider a flywheel rotating with an angular velocity its initial kinetic energy is
I2. After some time the wheel comes to rest and kinetic energy is utilized in
overcoming friction at the bearings. The temperature of the wheel and the
bearings rises and the increase in their internal energy is equal to the original
kinetic energy of the fly wheel.
(3) Consider two flasks A and B connected by a glass tube provided with a stop cock.
Let A contain air at high pressure and B is evacuated. The System is isolated from
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the surroundings. If the stop cock is opened, air rushes from A to B, the pressure
in A decrease and the volume of air increases.
All the above three examples though different, are thermo dynamical processes involving
change in thermo dynamical coordinates. Also, in accordance with the first law of
thermodynamics, the principle of conservations of energy is not violated because the total energy
of the system is conserved. It is also clear that, with the initial conditions described above, the
three processes will takes place.
Let us consider the possibility of the above three processes taking place in the reverse
direction. In the first case, if the reverse process is possible, the block B should transfer heat to A
and initial conditions should be restored. In the second case, if the reverse process is possible, the
heat energy must again change to kinetic energy and the fly wheel start rotating with the initial
angular velocity . In the third case, if the reverse process is possible the air in B must flow back
to A and the initial condition should be obtained.
But, it is a matter of common experience, that none of the above conditions for the
reverse process are reached. It means that the direction of the process cannot be determined by
knowing the thermo dynamical coordinates in the two end states. To determine the direction of
the process a new thermo dynamical coordinate has been devised by Clausius and this is called
the entropy of the system. Similar to internal energy, entropy is also a function of the State of the
system. For any possible process, the entropy of an isolated system should increase or remain
constant. The process in which there is a possibility of decrease in entropy cannot take place.
If the entropy of an isolated system is maximum, any change of state will mean decrease
in entropy and hence that change of state will not take place.
To conclude, process in which the entropy of an isolated system decreases do not take
place or for all processes taking place in an isolated system the entropy of the system should
increase or remain constant. It means a process is irreversible if the entropy decreases when the
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direction of the process is reversed. A process is said to be irreversible if it cannot be retracted
back exactly in the opposite direction. During an irreversible process heat energy is always used
to overcome friction. Energy is also dissipated in the form of conduction and radiation. This loss
of energy always takes place whether the engine works in one direction or the reverse direction.
Such energy cannot be regained. In actual practice all the engines are irreversible. If electric
current is passed through a wire, heat is produced. If the direction of the current is reversed, heat
is again produced. This is also an example of an irreversible process. All chemical reaction are
irreversible. In general, all natural processes are irreversible.
5.3 Reversible Process
From the thermo dynamical point of view, a reversible process is one in which an
infinitesimally small change in the external conditions will result in all the changes taking place
in the direct process but exactly repeated in the reverse order and in the opposite sense. The
process should take place at an extremely slow rate. In a reversible cycle, there should not be any
loss of heat due to friction or radiation. In this process, the initial conditions of the working
substance can be obtained.
Consider a cylinder, containing a gas at a certain pressure and temperature. The cylinder
is fitted with a frictionless piston. If the pressure is decreased, the gas expands slowly and
maintain a constant temperature (isothermal process). The energy required for this expansion is
continuously drawn from the source (surroundings). If the pressure on the Piston is increased, the
gas contracts slowly and maintains constant temperature (isothermal process). The energy
liberated during compression is given to the sink (surroundings). This is also true for an adiabatic
process provided the process takes place infinitely slowly.
The process will not be reversible if there is any loss of heat due to friction, radiation or
conduction. If the changes take place rapidly, the process will not be reversible. The energy used
in overcoming friction cannot be retraced.
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5.4 LET US SUM UP
In this lesson you have learned about the relation between the two specific heats of gases namely
Cp and Cv. Also you learned about the reversible and irreversible processes.
5.5 CHECK YOUR PROGRESS 1. State the difference between reversible and irreversible processes.
2. Why gas has got two specific heat ? Give reasons.
5.6 LESSON END ACTIVITIES
1. Explain reversible and irreversible processes with suitable examples
2. 14 POINTS FOR DISCUSSION
1. How do you determine by Clement and Desormes method
5.8 REFERENCES
1. Heat by Anantha Krishnan
2. Heat and thermodynamics by Brijlal and Subramaniam
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LESSON 6 CONTENTS
6.0 Aims and Objectives
6.1 Second law of Thermodynamics
6.2 Carnots Reversible Engine.
6.3 Carnot's Engine and Refrigerator
6.4 Carnot's Theorem
6.5 Let us Sum up
6.6 Check your progress
6.7 Lesson end Activities
6.8 Points for Discussion
6.9 References
6.0 AIMS AND OBJECTIVES In this lesson you will learn the second law of thermo dynamics, and its applications to Carnot engine also you will derive Carnots theorem
6.1 Second law of Thermodynamics
A heat engine is chiefly concerned with the conservation of heat energy into mechanical
work. A refrigerator is a device to cool a certain space below the temperature of its
surroundings. The first law of thermodynamics is a qualitative statement which does not preclude
the possibility of the existence of either a heat engine or a refrigerator. The first law does not
contradict the existence of a 100% efficient or a self-acting refrigerator.
In practice, these two are not attainable. These phenomena are recognized and this led to
the formulation of a law governing these two devices. It is called second law of thermodynamics.
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A new term reservoir is used to explain the second law. A reservoir is a device having
infinite thermal capacity and which can absorb, retain or reject unlimited quantity of heat without
any change in its temperature.
Kelvin-Plank s