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THERMAL PHYSICS
UNIT-II
Engineering Physics-I PH 6151
Prepared by
Dr.N.R.SHEELA (Asst.Prof)Dept. Of Applied Physics
SVCE
Unit- II Thermal Physics
Modes of heat transfer-thermal conductivity-Newton’s law of cooling-Linear heat flow-Lee’sdisc method- Radial heat flow-Rubber tubemethod-Conduction through compound media(series and parallel)
Heat Transfer
• Heat always moves from a warmer place to a cooler place.
• Hot objects in a cooler room will cool to room temperature.
• Cold objects in a warmer room will heat up to room temperature.
Methods
CONDUCTION CONVECTION RADIATION
Heat Transfer Methods
Conduction
When we heat a metal strip at one end, the heat
travels to the other end.
As we heat the metal, the particles vibrate, these
vibrations make the adjacent particles vibrate, and so on
a,thus the transmission of heat takes place by molecular
vibrations in the case of conduction.As it always requires
material medium(solid) it takes place in vacuum.
Metals are different
The outer electrons of metal atoms drift, and are free to move.
When the metal is heated, ‘electrons’ gain kinetic energy and transfer it throughout the metal.
Insulators, such as wood and plastic, do not have such electrons which is why they do not conduct heat as well as metals.
Why does metal feel colder than wood, if they
are both at the same temperature?
Metal is a conductor, wood is an insulator. Metal
conducts the heat away from our hands. Wood does
not conduct the heat away from our hands, so the
wood feels warmer than the metal.
Convection
What happens to the particles in a liquid or a
gas when we heat them?
The particles spread out and become less dense.
This effects fluid movement.
What is a fluid?
A liquid or gas.
Fluid movement
Cooler, more dense, fluids sink through warmer, less dense fluids.
In effect, warmer liquids and gases rise up.
Cooler liquids and gases sink.
Water movement
Hot water
risesCooler water
sinks
Convection
current
Cools at the
surface
Why is it windy at the seaside?
Cold air sinks
Where is the freezer
compartment put in a fridge?
Freezer compartment
It is put at the top, because cool air sinks, so it cools the food on the way down.
It is warmer at the
bottom, so this warmer air rises and a convection current is
set up.
The third method of heat transfer
How does heat energy get
from the Sun to the Earth?
There are no particles
between the Sun and the
Earth so it CANNOT
travel by conduction or
by convection. Hence no
material medium
needed.?
RADIATION
Emission experiment
Four containers were filled with warm water. Which
container would have the warmest water after ten minutes?
Shiny metal
Dull metal
Dull black
Shiny black
The shiny metal container would be the warmest after ten
minutes because its shiny surface reflects heat radiation
back into the container so less is lost. The dull black
container would be the coolest because it is the best at
emitting heat radiation.
Absorption experiment
Four containers were placed equidistant from a heater. Which
container would have the warmest water after ten minutes?
The dull black container would be the warmest after ten
minutes because its surface absorbs heat radiation the best.
The shiny metal container would be the coolest because it is
the poorest at absorbing heat radiation.
Shiny metal
Dull metal
Dull black
Shiny black
Convection questions
Why are boilers placed beneath hot water
tanks in people’s homes?
Hot water rises.
So when the boiler heats the water, and the hot
water rises, the water tank is filled with hot water.
Why does hot air rise and cold air sink?
Cool air is more dense than warm air, so
the cool air ‘falls through’ the warm air.
Radiation questions
Why are houses painted white in hot countries?
White reflects heat radiation and keeps the house cooler.
Why are shiny foil blankets wrapped around marathon
runners at the end of a race?
The shiny metal reflects the heat radiation from the
runner back in, this stops the runner getting cold.
Newton’s Law of CoolingIt states that the rate of fall of temperature of the body isproportional to the difference between the temperature of the bodyand that of the surrounding medium.
Cooling is fast when the temperature difference is high.
Cooling becomes slower when the temperature of the hot body falls
to a temperature closure to room temperature.
The rate of cooling
Where k is proportionality
constant, which depends on
the area and nature of the
surface
t
d
dt
Q
tk
d
dt
Q
Experimental verification
A beaker shaped copper
calorimeter blackened outside is
taken and filled with hot water of
about 90C for (3/4)th of its
capacity. The mouth of the
calorimeter is closed with
insulating cap and is suspended
inside a double walled vessel
with water in between two walls
at room temperature as
measured by a thermometer.
A stirrer inserted inside the calorimeter is kept stirred.
A thermometer inserted inside the calorimeter measures the temperarure tFor
every 30 seconds until temperature t
falls to (+5) C
When loge(t-) is plotted against time t,
we get a straight line as shown.
verifying Newton’s law of cooling.
Applications of Newton’s Law of Cooling
1. Helps to design radiators/cooling system in thermal machine.
2. To calculate the time taken for a hot object to cool down to a lower temperature.
3. To determine the specific heat capacity of a substance.
4. It helps to estimate the time of death by measuring the temperature of dead body.
RECTILINEAR FLOW OF HEAT THROUGH
A ROD
• Consider a long rod AB of uniform cross section heated at one end A as shown in figure.
• Then there is flow of heat along the length of the bar and heat is also radiated from its surface. B is the cold end.
• Consider the flow of heat between the sections P and Q at distance x and x+δx from the hot end.
• Excess temperature above the surroundings at section P = 0
• Temperature gradient at section P =dx
d
Excess temperature at section Q = xdx
d
x
dx
d
dx
d
x
dx
d
dx
d
2
2
=
dx
dKA
Q1 = …………………………(1)
Heat flowing (entering) through P in
one second
Temperature gradient at Q =
Heat flowing (leaving) through Q in one second
x
dx
d
dx
dKA
2
2
Q2 =
xdx
dK
dx
dKA
2
2
Q2 =
Net heat gain by the element x in one second
Q = Q1- Q2 ………………………. …… (3)
…………(2)
dx
dKA
x
dx
dKA
dx
dKA
2
2
= -
dx
dKA
x
dx
dKA
dx
dKA
2
2
-
xdx
dKA
2
2
Q = ………………………..(4)
Before the steady state is reached
Before the steady state is reached, the amount of heat Q is used in two ways.
(I) A part of the heat is used in raising the temperature of the rod
and
(II) the remaining heat is lost by radiation from the surface.
Heat absorbed per second to raise the temperature of the rod
= mass x specific heat capacity x
= (A x δx)ρ x S x …………(5)
where A – Area of the cross-section of the rodρ – Density of the rod S – Specific heat capacity of the rod
- Rate of rise in temperature
dt
d
dt
d
Heat lost per second due to radiation = E p δx …………..(6)
Where
E – Emissive power of the surface
p - Perimeter of the bar
δx – Surface area of the element
- Average excess of temperature of the element over that
of the surroundings
Amount of heat (Q) = Amount of heat absorbed + Amount of heat lost
Q = (A x δx)ρ x S + E p δx …….(7)
On Comparing the eqns (4) and (7)
= (A x δx)ρ x S xxdx
dKA
2
2
dt
d
KA
Ep
dt
d
K
S
dx
d
2
2
xKA
xE
xKA
dt
dSxA
xKA
xdx
dKA
2
2
………………………(9)
The above equation is standard differential equation for
the flow of heat through the rod.
Special cases:-
Case – 1: when heat lost by radiation is negligible.If the rod is completely covered by insulating materials, then there is no
loss of heat due to radiation.Hence Epδx = 0
where, , thermal diffusivity of the rod.
Case – 2: After the steady state is reached.
After the steady state is reached, there is no raise of temperature
dt
d
hdt
d
K
S
dx
d 12
2
…………….(10)
hS
K
Hence, = 0
equation (9) becomes
Substituting,
dt
d
KA
Ep
dx
d
2
2
2KA
Ep
2
2
2
dx
d
, we have
………. ..... (11)
02
2
2
dx
d(This represent second order differential equation).
The general solution of this equation is
……..(12)
Where A and B are two unknown constants which can be determined from the boundary conditions of the problem.
Suppose the bar is of infinite length,
Excess temperature above the surrounding of the rod of
the hot end = 0
Excess temperature above the surrounding of the rod at
the cold end = 0
xx BeAe
Boundary conditions(I) When x = 0, = 0 (ii) when x = , = 0
0 = A+B 0 = Ae + Be-
0 = Ae
i.e., A=0then, 0 = BSubstituting A and B in equation (12), we have
= 0e-μx ………(13)
The above equation represents the excess temperature of a pointat a distance x from the hot end after the steady state is reachedand it exponentially falls from hot end.
RADIAL FLOW OF HEAT
In this method heat flows from the inner side towards the other side along
the radius of the cylindrical shell.
This method is useful in determining the thermal conductivity of bad
conductors taken in the powder form.
Consider a cylindrical tube of length l, inner radius r1 and outer radius r2.
The tube carries steam or some hot liquid.
After the steady state is reached, the temperature on the innersurface is θ1 and
on the outer surface is θ2 in such a way θ1 > θ2.
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.
CYLINDRICAL SHELL METHOD (or) RUBBER TUBE METHOD
Working:
• Steam is allowed to pass through the axis of the cylindricalshell.
• The heat flows from the inner surface to the other surfaceradially.
• After the steady state is reached, the temperature at the innersurface is noted as 1 and on the outer surface is noted as 2.
Calculation:
• The cylinder may be considered to consists of a large number of co-axial cylinders of increasing radii.
• Consider such an elemental cylindrical shell of thr thickness drat a distance ‘r’ from the axis.
• Let the temperatures of inner and outer surfaces of the elemental shell be and +d. Then,
The Amount of heat conducted per second
Here Area of cross section A = 2πrl
Rearranging the above equation we have
…………(1)
The Thermal conductivity of the whole cylinder can be obtained by
dr
dKAQ
dr
drlKQ
2
dQ
lK
r
dr 2
2
1
22
1
dQ
lKr
rr
dr
21
1
2 2log
Q
lK
r
re
21
1
2
2
log.
l
r
r
eQ
K =
21
1
210
2
log3026.2
l
rr
QW m-1K-1K =
DETERMINATION OF THERMAL CONDUCTIVITY OF
RUBBER
It is based on the principle of radial flow of heat through a cylindrical shell.
Working:
• The empty calorimeter is weighed, let it be (w1).
• It is filled with two third of water and is again weighed, let it be (w2)
• A known length of rubber tube is immersed inside the water contained in the calorimeter.
• Steam is passed through one end of the rubber tube and let out through the other end of the tube.
• The heat flows from the inner layer of the rubber tube to the outer layer and is radiated.
• The radiated heat is gained by the water in the calorimeter.
• The time taken for the steam flow to raise the temperature of the water about 10C is noted, let it be ‘t’ seconds.
Observation and calculation:
Let w1 Weight of calorimeter
w2 Weight of calorimeter and water
w2 – w1 Weight of the water alone
1 Initial temperature of the water
2 Final temperature of the water
2 -1 Rise in temperature of the water
S Temperature of the steam
l Length of the rubber tube (immersed)
r1 Inner radius of the rubber tube
r2 Outer radius of the rubber tube
s1 Specific heat capacity of the calorimeter
s2 Specific heat capacity of the water
3 Average temperature of the rubber tube.
2
21 3 =
From the theory of cylindrical shell method the amount of heat conducted by the rubber tube per second is given by
The amount of heat gained by
water per second = ……(3)
1
2
3
log
2
r
r
lK
e
S Q = ……………….(1)
The amount of heat gained by
calorimeter per second =
t
sw1211
……………(2)
t
sww 12212
The amount of heat gained by the water and calorimeter per second is obtained by
(2) +(3)
t
swsww )()()( 121112212 Q =
t
swwsw 2121112 )( Q = ..............(4)
Under steady state
The amount of heat conducted by = The amount of heat gained by the water
the rubber tube per second and the calorimeter per second
Hence, Equation (1) = Equation (4)
Wm-1K-1
1
2
3
log
2
r
r
lK
e
S t
swwsw 2121112 )( =
2
21
2
)(2
log)(
21
212111
212
s
e
lt
swwswr
r
K =
3 =
The thermal conductivity of a
material is determined by
various methods
Searle’s method
Forbe’s method
Lee’s disc method
Radial flow method
LEE’S DISC METHOD FOR DETERMINATION OF THERMAL CONDUCTIVITY OF BAD CONDUCTOR
The thermal conductivity of bad conductor like ebonite or card board is determined by this method.
Principle:
In the steady state the quantity of heat conducted across any cross section is equal to the quantity of heat radiated from its surface to the surroundings.
Description:
The given bad conductor (B) is shaped with the diameter
as that of the circular slab (or) disc ‘D’.
The bad conductor is placed inbetween the steam chamber (S) andthe disc (D), provided the bad conductor, steam chamber and the slabshould be of same diameter.
Holes are provided in the steam chamber (S) and the disc (D) in whichthermometer are inserted to measure the temperatures.
The total arrangement is hanged over the stand as shown in fig.
Working:
Steam is passed through the steam chamber till the steady
state is reached.
Let the temperature of the steam chamber (hot end) and
the disc (cold end) be 1 and 2 respectively.
Observation and Calculation:
• Let ‘x’ be the thickness of the bad conductor (B), ‘m’ is the mass of the slab, ‘s’ be the specific heat capacity of the slab.
• ‘r’ is the radius of the slab and ‘h’ be the height of the slab, then
Amount of heat conducted by the
Bad conductor per second = …….(1)
Area of the cross section is = πr2 ……………… (2)
Amount of heat conducted per second = ……(3)
x
KA )( 21
x
rK )( 21
2
The amount of heat lost by
slab per second = m x s x Rate of cooling
= msRc…………..(4)
Hence, we can write equation (3) = equation (4)
Under steady state
The amount of heat conducted by the = Amount of heat lost by the slab
Bad conductor (B) per second (D) Per second
x
rK )( 21
2 = msRc
)( 21
2 r
msxRcK = …….(5)
To find the rate of cooling Rc
To find the rate of cooling for the disc alone, the bad conductor is removed
and the steam chamber is directly placed over the disc and heated.
When the temperature of the slab attains 5C higher than 2, the steam
chamber is removed. The slab is allowed to cool, simultaneously a stop
watch is switched ON.
A graph is plotted taking time along ‘x’ axis and temperature along
‘y’ axis, the rate of cooling for the disc alone (i.e) is found from
the graph as shown in fig.
dt
d
The rate of cooling is directly proportional to the surface area exposed.
Case(i)
Steam chamber and bad conductor are placed over slab, in which radiation takes place from the bottom surface of area (πr2) of the slab and the sides of the of area (2πrh).
Rc = 2 πr2 + 2πrh
Rc = πr(r+2h)………(6)
Case(ii)
The heat is radiated by the slab alone, (i.e) from the bottom of area(πr2), top surface of the slab of area (πr2) and also through the sides of the slab of area 2πrh.
= πr2 + πr2 + 2πrh
= 2 πr2 + 2πrh
2
dt
d
2
dt
d
2
dt
d= 2πr(r+h)
From (6) and (7)
)2
)2
2
hrr
hrr
dt
d
Rc
2)(2
)2(
dt
d
hr
hrRc =
Substituting (8) in (5) we have
K =
Hence, thermal conductivity of the given bad conductor can be determined from the above relation.
)(2)(
)2(
21
2
2
hrr
hrdt
dmsx
Wm-1K-1
HEAT CONDUCTION THROUGH A COMPOUND MEDIA (SERIES AND PARALLEL)
Consider a composite slab of two different materials, A & B of thermal conductivity K1 & K2 respectively. Let the thickness of these two layers A & B be d1 and d2
respectively.
• Let the temperature of the end faces be 1 & 2 andtemperature at the contact surface be θ, which isunknown.
• Heat will flow from A to B through the surface ofcontact only if 1 > 2.
• After steady state is reached heat flowing per second(Q) through every layer is same.
• A is the area of cross section of both layersAmount of heat flowing per sec through A
Q = …………(1)
Amount of heat flowing per sec through BQ = …………(2)
1
11 )(
x
AK
1
12 )(
x
AK
Hence (1) and (2) are equal
=
Rearranging the (3), we have
K1A(1-)x2 = K2A(2)x1
K11x2 - K1x2 = K2x1 K22x1
K11x2 +K22x1 = K2x1+ K1x2
K11x2 +K22x1 = ( K2x1+ K1x2)
= ………..(4)
This is the expression for interface temperature of two composite slabs in series.
1
11 )(
x
AK
1
2)( 2
x
AK ……….(3)
2112
122211
xKxK
xKxK
• Substituting from equation (4) in equation (1), we get
=
=
=
=
This is the expression for interface temperature of two composite slabs in series.
2112
1222111
1
1
xKxK
xKxK
x
AK
2112
122211211112
1
1
xKxK
xKxKxKxK
x
AK
2112
122112
1
1
xKxK
xKxK
x
AK
2112
1211
1
21
xKxK
xx
x
AKK
=
=
= ………..(5)
Q’ is the amount of heat flowing through the compound wall of two materials.
This method can also be extended to composite slab with more than two slabs.
Generally, the amount of heat conducted per sec for any number of slabs is given by
2112
2121 )(
xKxK
AKK
21
21
21
12
21 )(
KK
xK
KK
xK
A
2
2
1
1
21 )(
K
x
K
x
A
Generally, the amount of heat conducted
per sec for any number of slabs is given
by Q =
BODIES IN PARALLEL
Let us consider a compound wall of two different materials A and B ofthermal conductivities K1 and K2 and of thickness d1 and d2respectively.
These two material layers are arranged in parallel.
•
K
x
A )(21
The temperatures θ1 is maintained at one faces of the material A and B andopposite faces of the material A and B are at temperature θ2 .
A1 & A2 be the areas of cross-section of the materials.
Amount of heat flowing through the first material (A) in one second.
………..(1)
Amount of heat flowing through the second material (B) in one second.
………..(2)
The total heat flowing through these materials per second is equal to the
sum of Q1 and Q2
Q = Q1+Q2 ……………………….(3)In general, the net amount of heat flowing per second parallel to the composite slabs is given by
1
2111 )(
x
AK
2
2122 )(
x
AK Q2 =
Q1 =
x
KAQ )( 21
Practical Applications of Conduction of Heat• Metals are good conductors of heat and wood, felt, brick, glass, glass,
granite, cotton, wool, cork, ebonite, rubber are bad conductors of heat.
• Sauce pans, hot water buckets, kettles and other utensils are made of metal. They are provided with wooden or ebonite handles so that heat from the utensils is not conducted to the hand.
• Ice box has a double wall made of tin or iron. The space in between the two wall is packed with cork or felt. This is done because cork and felt are poor conductors of heat and prevent the flow of the outside heat into the box.
• 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.
• A steel blade appears colder than a wooden handle in winter.Steel is a good conductor of heat. As soon as a person touchesthe blade, heat flows from the hand to the blade. Therefore itappears colder. Since wood is a bad conductor of heat ,heat doesnot flow from the hand to the handle.
• Two shirts keep the body warmer than a single shirt of the samematerial and double the thickness. Between the two shirts a finelayer of air acts as a bad conductor and does not allow the heatfrom the body to flow out to the surroundings.
•
Davy’s Safety Lamp
• It is also based on the principle of conduction of heat.
• A wire gauze is placed over a Bunsen burner and the gas is lit above the wire gauze. The gas comes out of the wire gauze.
• A flame appears at the top surface of the wire gauze. The gas below the wire gauze does not get sufficient heat from ignition. The wire gauze does not get sufficient heat for ignition.
• The wire gauze conducts away the heat of the flame above it and the temperature at the lower surface does not reach the ignition temperature.
In Davy’s safety lamp, the cylindrical metal gauze of high
thermal conductivity surrounds the flame.
When this lamp is take inside a mine, even if explosive gases
are present, they do not get ignited because the wire gauge
conducts away the heat of the flame.
• The temperature outside the wire gauze remains lowerthan the ignition temperature of the gases. In theabsence of the wire gauze the gases outside can explode.
• As soon as miner notices the presence of the explosive gases from the colour of the flame, he gives a warning signal.
• The work is stopped and the mine is thoroughly ventilated.
