Unit – II Thermal Physics

Introduction- Modes of Heat Transfer

Normally there are three modes of transfer of heat from one

place to another viz., conduction, convection and radiation.

Conduction : Conduction refers to the heat transfer that occurs across the medium. Medium

can be solid or a fluid.

Convection: It is the process in which heat is transferred from hotter end to colder end by

the actual movement of heated particles.

.Radiation : In radiation, in the absence of intervening medium, there is net heat transfer between two surfaces at different temperatures in the form of electromagnetic waves.

Specific Heat Capacity

The specific heat capacity of a material is the amount of energy (in Joules) needed to increase the temperature of one kilogram of mass of the material by one Kelvin.

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.

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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

dTTemperature gradient at section P =

dx

d 2TdT

KA Gx

AGxUS EUGxTdx2 dt

KAGx KAGx KAGx

d 2T US dT Ep Tdx2 K dt KA

Excess temperature at section Q = T dT

Gxdx

d § dT ·Temperature gradient at Q = ¨T Gx¸

dxdx © ¹§ dT

d 2T ·

= ¨ Gx¸dx

2

© dx ¹

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Heat flowing (entering) through P in one second

Q1 = KAdT

…………………………(1)dx

Heat flowing (leaving) through Q in one second

§ dT d 2T ·

Q2 = KA¨

Gx¸dx

dx 2

© ¹

Q2 = KA dT K d 2T Gx …………(2)2

dx dx

?Net heat gain by the element Gxin one second

Q = Q1- Q2 ………………………. …… (3)

dT § dT

d 2T ·

= - KA ¨ KA

KA Gx¸

dx

dxdx 2

© ¹

= - KA dT

+ KA dT

KA d

2T

Gxdxdx2dx

d 2T G

Q = KA dx2 x ………………………..(4)

Before the steady state is reached

Before the steady state is reached, the amount of heat Q is used in two ways. Apart of the heat is used in raising the temperature of the rod and the remaining heat islost by radiation from the surface.

Heat absorbed per second to raise the temperature of the rod

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dT

dt

= (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

Heat lost per second due to radiation = E p δx T …………..(6)

WhereE – Emissive power of the surfacep - Perimeter of the barδx – Surface area of the elementT - Average excess of temperature of the element over that of the

surroundingsAmount of heat (Q) = Amount of heat absorbed + Amount of heat lost

Q = (A x δx)ρ x S x + E p δx T …….(7)On Comparing the eqns (4) and (7)

KA

d 2TGx = (A x δx)ρ x S x + E p δx T………(8)

2

dxDividing both LHS and RHS of the above equation by KA, we have,

KA

d 2T Gx AGxUS

dTEUGxTdx2 dt

KAGx KAGx KAGx

d 2TUS

dT

Ep T ………………………(9)

dx2 K dt KA

The above equation is standard differential equation for the flow of heat through the rod.

Special cases:-

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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 EpδxT = 0

d 2TUS

dT 1

dT …………….(10)

dx2 K dt h dt

where,

K

h , thermal diffusivity of the rod.US

\ Case – 2: After the steady state is reached.

After the steady state is reached, there is no raise of temperature

Hence, dT

= 0

dt? equation (9) becomes

d 2T

Ep T

dx2 KASubstituting,

Ep

P 2 , we haveKA

d 2T P 2T ……….(11)

dx2

d 2T 2

P T 0 (This represent second order differential equation).dx2

The general solution of this equation is

T AePx BePx ……..(12)

Where A and B are two unknown constants which can be determined from the

boundary conditions of the problem.

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Suppose the bar is of infinite length,

Excess temperature above the surrounding of the rod of the hot end = T0

Excess temperature above the surrounding of the rod at the cold end = 0

Boundary condition

(i)When x = 0, T = T0 (ii) when x = f, T = 0

T0 = A+B 0 = Aef + Be-f

0 = Aef

As we know ef cannot be zero, therefore A should be zero i.e., A=0then, T0 = B

Substituting A and B in equation (12), we have

T = T0e-μx

………(13)The above equation represents the excess temperature of a point at a distance x from

the hot end after the steady state is reached and it exponentially falls from hot end.

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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 θ1 & θ2 and temperature at the contact surface

be θ, which is unknown. Heat will flow from A to B through the surface of contact only

if θ1 > θ2. After steady state is reached heat flowing per second (Q) through every layer

is same. A is the area of cross section of both layers

Amount of heat flowing per sec through A

Q =K

1 A(T 1

T )…………(1)

x1

Amount of heat flowing per sec through B

Q=K

2A(T

1 T ) …………(2)

x1

The amount of heat flowing through the materials A and B is equal in steady

conditions

Hence (1) and (2) are equal

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K1

A(T1

T

) =

K 2

A(T1

T

)……….(3)

x1 x1

Rearranging the (3), we have

K1A(T1-T)x2 = K2A(T-T2)x1

K1T1x2 - K1Tx2 = K2Tx1- K2T2x1

K1T1x2 +K2T2x1 = K2Tx1+ K1Tx2

K1T1x2 +K2T2x1 = T( K2x1+ K1x2)

T = K

1T

1x

2 K

2T

2 x

1 ………..(4)K2 x1 K1 x2

This is the expression for interface temperature of two composite slabs in

series.

Substituting T from equation (4) in equation (1), we get

Q =K1 Aª § K1T1x2 K2T2 x1·º

«T1¨ ¸

x K x K x¨ ¸»

1 ¬ © 2 11 2 ¹¼

ª§ K2T1x1 K1T1x2 K1T1 x2·º

= K1 A «¨ K2T2 x1 ¸»

x1

¨

K2 x1 K1x2

¸

¬© ¹¼

=K1 A ª K2T1x1 K2T2 x1 º

« »

K2 x1 K1 x2

x1 ¬

¼

=K1K2 A ª T1x1 T2 x1 º

« »

x1 ¬

K2 x

1 K

1x

2¼

=K

1K

2 A(T

1 T

2 )

K2 x1 K1 x2

=A(T

1 T

2 )

K2 x

1 K

1 x

2

K1K2 K1K2

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Q =A(T

1 T2 )

……………..(5) x1

x2

K1 K2

‘Q’ is the amount of heat flowing through the compound wall of two

materials.

This method can also be extended to comp[osite slab with more than two

slabs.

Generally, the amount of heat conducted per sec for any number of slabs is

given by ,

Q =A(T

1 T2 )

§ x ·¦¨ ¸

©K ¹

BODIES IN PARALLEL

Let us consider a compound wall of two different materials A and B of thermal

conductivities K1 and K2 and of thickness d1 and d2 respectively. These two materiallayers are arranged in parallel.

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The temperatures θ1 is maintained at one faces of the material A and B and opposite

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.

Q1 =K

1 A1 (T

1 T2 ) …………….(1)

x1

Amount of heat flowing through the second material (B) in one second.

Q2 = K

2 A2 (T1 T2 ) …………….(2)

x2

The total heat flowing through these materials per second is equal to the sum of Q1 and

Q2

Q = Q1+Q2 ……………………….(3)

Substituting equations (1) and (2) in (3), we get,

Q=

K1

A1

(T1

T2

) +

K2

A2

(T1

T2

)

x1

x2

? Amount of heat flowing per second

Q = T1

§ K1 A1 K2 A2 ·T2

¨ ¸¨

x1 x2

¸

© ¹

In general, the net amount of heat flowing per second parallel to the composite slabs

is given by

¦Q (T1 T2 ) ¦

KA

x

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RADIAL FLOW OF HEAT

In this method heat flows from the inner side towards the other side along theradius of the cylindrical shell. This method is useful in determining the thermalconductivity of bad conductors taken in the powder form.

CYLINDRICAL SHELL METHOD (or) RUBBER TUBE METHOD

Consider a cylindrical tube of length l, innerradius r1 andouter 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 thetube. Consider an element of thickness dr and length l at a distance r fromthe axis.

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 T1 and on the outer surface is noted as T2.

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Calculation:

The cylinder may be considered to consists of a large number of co-axial cylinders of increasing radii. Consider such an elemental cylindricalshell of thr thickness dr at a distance ‘r’ from the axis. Let the temperaturesof inner and outer surfaces of the elemental shell be T and T+dT. Then,

The Amount ofheat conducted per second Q KA dT

drHere Area of cross section A = 2πrl

?Q

2SrlK dT

drRearranging the above equation we have

dr 2 SlK dTr Q

…………(1)

? The Thermal conductivity of the whole cylinder can be got by, integrating equation (1) within the limits r1 to r2 and T1 to T2,

r2 dr 2S lK

T2

³ ³ dTr Qr T

1 1

§ r2 · 2SlK Tlog ¨ ¸1

Te ¨ ¸

Q2

© r

1 ¹Rearranging we get,

§ r2Q.log ·¨ ¸¨ ¸

K =

e© r1 ¹

2SlT1 T2

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Qu 2.3026 u log10§ r2 ·¨ ¸

K =

© r1 ¹

W m-1K-12Sl T1 T2

By knowing the values in RHS, the thermal conductivity of the given material can be found.

DETERMINATION OF THERMAL CONDUCTIVITY OF RUBBER

It is based on the principle of radial flow of heat through a cylindrical shell.

Description: It consists of a calorimeter, stirrer with a thermometer. The setup is kept

inside the woodenbox. The space between the calorimeter and the box is filled with insulating

materials such as cotton, wool, etc. to avoid radiation loss, as shown in fig.

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 10qC is noted, let it be ‘t’ seconds.

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Observation and calculation:Weight of calorimeterLet w1

w2 Weight of calorimeter and waterw2 – w1 Weight of the water aloneT1 Initial temperature of the waterT2 Final temperature of the waterT2 -T1 Rise in temperature of the waterTS Temperature of the steaml Length of the rubber tube (immersed)r1 Inner radius of the rubber tuber2 Outer radius of the rubber tubes1 Specific heat capacity of the calorimeters2 Specific heat capacity of the water

T3 Average temperature of the rubber tube.

T3 =T1 T2

2

We know from the theory of cylindrical shell method the amount of heat

conducted by the rubber tube per second is given by

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The amount of heat gained by

calorimeter per second

The amount of heat gained

by water per second

K 2S lTS T3 ……………….(1)§ r2 ·¨ ¸

loge ¨ ¸

© r

1 ¹

= w1 s1 T2 T1 ……………(2)

t

=w 2

w 1

s 2

T 2

T 1 ……(3)

t? The amount of heat gained by the water and calorimeter per second is obtained

by(2) +(3)

Q = (w2 w1 )s2 (T2 T1 )w1s1 (T2 T1 )t

Q = (T2 T1 )>w1s1 w2 w1 s2 @ ..............(4)

Under steady statet

The amount of heat conducted by The amount of heat gained by therubber tube per second = the water and the calorimeter

per escond

Hence, equation (1) = Equation (4)

K 2S lTS T3 = (T2 T1 )>w1s1 w2 w1 s2 @§ r2 · t¨ ¸

loge ¨ ¸

© r1 ¹

Substituting T3 =T1 T2

2

§r2 · w2 w1 s2 @

? K =(T2 T1 )loge ©¨ r1 ¹¸>

w1s

1 Wm-1K-1

2S ltªTs (T1 T2 ) º

«

2»

¬ ¼By substituting the values in RHS, the thermal conductivity of the rubbercan be determined.

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Methods to determine thermal conductivity

The thermal conductivity of a material is determined by various methods

1. Searle’s method – for good conductors like metallic rods2. Forbe’s method - for determining the absolute conductivity of metals3. Lee’s disc method – for bad conductors4. Radial flow method – for bad conductors

LEE’S DISC METHOD FOR DETERMINATION OF THERMAL CONDUCTIVITY OF BAD CONDUCTORThe thermal conductivity of bad conductor like ebonite or card board is determined

by this method.Description:

The given bad conductor (B) is shaped with the diameter as that of the circularslab (or) disc ‘D’. The bad conductor is placed inbetween the steam chamber (S) and thedisc (D), provided the bad conductor, steam chamber and the slab should be of samediameter. Holes are provided in the steam chamber (S) and the disc (D) in which

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thermometer 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 isreached . Let the temperature of the steam chamber (hot end) and the disc(cold end) be T1 and T2 respectively.Observation and Calculation:

Let ‘x’ be the thickness of the bad conductor (B), ‘m’ is the mass of the slab, ‘s’ bethe specific heat capacity of the slab. ‘r’ is the radius of the slab and ‘h’ be the height ofthe slab, then

Amount of heat conducted by theKA(T1 T2 )Bad conductor per second = …….(1)

x

Area of the cross section is = πr2 ……………… (2)

Amount of heat conducted per second = KSr2 (T1 T2 ) ……(3)x

The amount of heat lost byslab per second =m x s x Rate of cooling

= msRc…………..(4)

Under steady state

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The amount of heat conducted by the = Amount of heat lost by the slab

Bad conductyor (B) per second (D) Per second

Hence, we can write equation (3) = equation (4)

KSr2 (T1 T2 ) = msRcx

K =

msxRc

…….(5)Sr2 (T1 T2 )

To find the rate of cooling Rc

Rc in equation (3) represents the rate of cooling of the disc along with the steamchamber. To find the rate of cooling for the disc alone, the bad conductor is removed and thesteam chamber is directly placed over the disc and heated.

When the temperature of the slab attains 5qC higher than T2, the steam chamber

is removed. The slab is allowed to cool, simulataneously a stop watch is switched ON.

A graph is plotted taking time along ‘x’ axis and temperature along ‘y’ axis, the§ dT ·

rate of cooling for the disc alone (i.e) ¨ ¸ is found from the graph as shown in fig.

© dt ¹

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 + 2πrhRc = πr(r+2h)………(6)

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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.

§ dT · 2 2? ¨ ¸

= πr

+ πr + 2πrh

© dt ¹T2

§ dT · = 2 πr 2 + 2πrh¨ ¸

© dt ¹T2

§ dT · = 2πr(r+h)……..(7)¨ ¸

© dt ¹T2

From (6) and (7)Rc Srr 2h)

§ dT · 2Srr h)¨ ¸

© dt ¹T2

Rc =(r 2h) § dT · ………(8)

¨ ¸

2(r h) © dt ¹T2

Substituting (8) in (5) we have

§ dT ·msx¨ ¸ (r 2h)

© dt ¹T2 -1 -1

K = Wm KSr2 (T1 T2 )2(r h)Hence, thermal conductivity of the given bad conductor can be determined from

the above relation.

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