thermal expansion,heat transfer and radiation)

BM SAPB.M.SHARMA A CA DEMY O F PHY SIC S

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Sunday, April 9, 2023 1B.M.Sharma Academy of Physics

Consider a one –dimensional solid of length l at temperature T. If the temperature changes by T the change in length is directly proportional to

(1)the original length l,

(2)the change in temperature, i.e,

l l t Where is coefficient of thermal expansion, a proportionality constant.

At a particular temperature

0

1 1limT

dT dT

2


A two- dimensional object, such as a thin metallic plate, changes area when their temperature is raised or lowered. For an isotropic material the area expansion may be expressed in terms of the linear expansion coefficient .

Consider a rectangular area of dimensions l1 and l2. The area is

1 2A l l

2 11 2

dA dl dll l

dT dT dT

1 2 1 2( ) ( )l l l l

1 22 l l

3


If a is constant over the temperature range of under consideration, this equation can be integrated.

1 22ff

i i

A T

A TdA l l dT

(2 )A A T A T Where is coefficient of superficial expansion, twice the coefficient of linear expansion.

4


Object expand along all dimensions as shown in figure; their areas and volumes increase with temperature as well as their lengths. If we cur a hole in a metal plate, the remaining material will expand exactly as it would if the plug were still in place.

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Consider a rectangular parallelpiped of slides l1, l2, l3

1 2 3V l l l

For a temperature change

3 2 11 2 1 3 2 3

dldV dl dll l l l l l

dT dT dT dT

33 ,

dll

dT 2

2 ,dl

ldT

11

dll

dT

1 2 3 1 2 3 1 2 3( ) ( ) ( )dV

l l l l l l l l ldT

1 2 3(3 )l l l

6


If is constant over the temperature range under consideration.

1 2 33ff

i i

V T

V TdV l l l dT

(3 )V V T

( )V V T

Where is the coefficient of volume expansion, which is three times the coefficient of linear expansion.

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Problem 1.Problem 1.

The figure shows a rectangular plate of dimension (l × b) from which two circular holes of radii R1 and R2 (R1 > R2) has been cut. The distance between the two holes is d (< R2).(a) What happens to all these distances and

dimensions when the rod is heated up.

(b) If l’, b’, R’1 , R’2 and d’ are the respective lengths at a higher temperature, then determine the relations between the ratios

' ' '1 2

1 2

' ', , ,

l b R R dand

l b R R d

R1

l

db

R2

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Solution.Solution.

(a) All the dimensions increases as the temperature is increased.

(a) All these ratios are equal.

' '2 2

1 2

' ' '1 tan

R Rl b dT cons t

l b R R d

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The difference between lengths of a certain brass rod and of a steel rod is claimed to be constant at all temperature. Is this possible ?

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Solution.Solution.

If LB and LS are the lengths of brass and steel rods respectively at a given temperature, then the lengths of the rods when temperature is changed by C0 will become,

' (1 )B B BL L ' (1 )S S SL L And

So that ' ' ( ) ( )S S B S B B S SL L L L L L


So (L’B – L’S) will be equal to (LB – LS) at all temperature if,

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0B B S SL L [ 0]as

( / ) ( / )B S S BL L i.e, the difference in the lengths of the two rods will be independent of temperature if the lengths are in the inverse ratio of their coefficients of linear expansion.



There are two spheres of same radius and material at same temperature but one being solid while the other hollow. Which sphere will expand more if

(a) They are heated to the same temperature,

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(b) Same heat is given to them ?

Solution.Solution.

(b) As thermal expansion of isotropic solids is similar to true photographic enlargement, expansion of a cavity is same as if it had been a solid body of the same material, i.e.,

V V Sunday, April 9, 2023 12B.M.Sharma Academy of Physics

As here V, and are same for both solid and hollow spheres treated (cavity); so the expansion of both will be equal.

(b) If same heat is given to the two spheres due to lesser mass, rise in temperature of hollow sphere will be more [as = Q/mc] and hence its expansion will be more [as V = V ].

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V V



A steel ring of 3.00 inches inside diameter at 200C is to be heated and slipped over a brass shaft measuring 3.002 inches in diameter at 200C.

To what temperature should the ring be heated?

Solution.Solution.

Let be the temperature to which the ring must be heated.

Final diameter of ring should be 3.002 inches.

3.002 3[1 ( 20)]

3.002 320

3

75.6 C

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Density of a substance is mV

1 1(1 ) 1t

V VV V T T

1(1 )t T

(1 )t T

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1t

mand

V


The Buoyant force on a completely submerged body of volume V in a liquid of density is

B V g

With the increase in temperature,

' (1 )sV V T

and '(1 )L T

Hence new buoyant force is B’ = V’ ’ g

(1 )(1 )s

L

V T gT

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A solid cube of constant density s floats in a liquid of density , at 00C.(a) Determine the fraction f of the cube immersed

in the liquid.

(b) What happens to the factor f when the liquid is heated to a temperature T.

(c) If is the coefficient of volume expansion of the liquid, then determine the temperature at which the solid will be completely immersed.(

(d) If f1 and f2 are the fractions of the volume of the solid immersed in the liquid at temperatures T1 and T2 (T2 > T1) respectively, then obtain an expression for calculating in terms of f1, f2, T1 and T2.

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Solution.Solution.

(a) Let V be the total volume of the solid and V be the volume of solid immersed in the liquid. Then,

s

l

Vf

V

(b) Density of the liquid decreases as the temperature is increased to T.

Now '

1l

l T

'' (1 ) (1 )s s

l l

f T f T

The solid gets more immersed in the liquid.

V - V

V

s

l

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(c) The solid gets completely immersed in the liquid when f’ becomes unity.

1 fT

f

Thus, for f’ = 1

(d) 1 1(1 );ff T 2 2(1 )ff T

On dividing, we get 1 1

2 2

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f Tf T

1 2

2 1 1 2

fff T fT

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A solid cube with coefficient of linear expansion floats in a liquid whose coefficient of volume expansion is . When both the solid and the liquid are heated, discuss the conditions when

(a) The solid sinks more

(b) The solid does not sink or lift.

(c) The solid lifts up.

Solution.Solution.

We know that

s

l

Vf

V

V - V

V

s

l

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Thus1' ' '1

s

L

TB V gB V g T

The apparent weight of a body in a liquid is

appW W B W Vg

Apparent weight when temperature increases,

' ' 'appW W V g

Generally s L

'B B 'app appand W W

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Thus with the rise in temperature buoyant force decreases and apparent weight will increase.


'

'

1 1'

1 3 1 3s s

l l

T Tff

T T

On heating,

(a) The solid sinks more, i.e., f’ > f when > 3

(b) The solid does not sink or lift when = 3, and

(c) The solid lifts up, i.e, f’ < f when < 3.

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Consider a scale designed to give correct reading at temperature T, When the temperature rises, due to linear expansion of scale the scale reading will be less than the actual reading . Consider a unit reading on scale, it will become (1+ T) . Thus

Actual reading= Scale reading(1+ T)

Similarly when temperature is decreased, the scale will contract and the actual reading will be less than the scale reading.

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A surveyor’s 30m steel tape is correct at a temperature of 200C. The distance between two points, as measured by this tape on a day when the temperature is 350C, is 26m.

What is the true distance between the point? (steel = 1.2 × 10-5/0C)

Solution.Solution.

Let temperature rise above the correct temperature be = 35 - 20 = 150C

Using the Relation : Correct length = measured length (1 + )

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true distance between the points 26 (1 + 1.2 × 10-5 × 15)

true distance = 26.00468m.Sunday, April 9, 2023 24B.M.Sharma Academy of Physics


A steel –scale is to be prepared such that the millimeter intervals are to be accurate with in 5 × 10-4 mm at a certain temperature.

Determine the maximum permissible temperature variation during the ruling of the millimeter marks if steel = 13.22 × 10-6/C0.

Solution.Solution.

We know that in case of linear expansion L = L

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Here L < 5 × 10-4 mm; L = 1 mm and = 13.22 × 10-6/C0. 6

6

( 5 10 )1 13.22 10

LL

037.83 C 0

max( ) 37.83 C Sunday, April 9, 2023 25B.M.Sharma Academy of Physics


The brass scale of a barometer gives correct reading at 00C. Coefficient of linear expansion of brass is 2.0 × 10-5/C0. The barometer reads 75 cm at 270 C.

What is the atmospheric pressure at 270C ?

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Solution.Solution.

We know that in case of scale reading.

True value = scale reading [1 + ]Here SR = 75 cm; = 2 × 10-5/C0

And = (’ – ) = 27 - 0 = 27 C0

So True value h = 75 (1 + 2 × 10-5 × 27) = 75.04 cm


Note :Note :

The pressure at 00C will be given by

H0d0 = hd = P, i.e, h0 = h(d/d0)

But as d = d0/(1 + Hg ),

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0 (1 )Hg

hh

With h = hs( 1 + ) = 75.04 cm


When a liquid is heated in a vessel, both the liquid and the vessel expand. The change in volume of liquid relative to the vessel is called its apparent expansion. The apparent change in volume of the liquid is given by

( 3 )appV V T

Where V is the volume of the liquid

is the coefficient of volume expansion of the liquid.

is the coefficient of linear expansion of the vessel

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(i) If < 3, level of liquid in the vessel rises on heating.

(ii) If > 3, level of liquid in the vessel falls on heating.

(ii) If = 3, level of liquid in the container will remain stationary with respect to the vessel.



A one litre glass flask contains some mercury. It is found that at different temperatures the volume of air inside the flask remains the same.

What is the volume of mercury in this flask if coefficient of linear expansion of glass is 9 × 10-6/C0 while volume expansion of mercury is 1.8 × 10-4/C0 ?

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Solution.Solution.

If V is the volume of flask, VL of mercury and VA of air in it, L AV V V

Now as with change in temperature volume of air remains constant, the expansion of mercury will be equal to that of the whole flask i.e., LV V


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G L LV V [ ]as V V

Here 1 1000V litre cc

and 6 03 27 10 /G G C

So VL = (1000 × 27 × 10-6/1.8 × 10-4) =150 cc



A glass flask whose volume is exactly 1000 cm3 at 00C is filled level full of mercury at this temperature. When the flask and mercury are heated to 1000C, 15.2 cm3 of mercury overflow.

The coefficient of cubical expansion of Hg is 1.82 × 10-4/0C. Compute the coefficient of linear expansion of glass.

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15.21000g l

0.000182 0.000152 0 5 0 10.00003 / 3 10 ( )g C C

5 01 10 /3g C

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Solution.Solution.

As 15.2 cm3 of Hg overflow at 1000C,

(final volume of Hg) - (final volume of glass flask) = 15.2 cm3.

1000(1 + l) - 1000(1 + g) = 15.2

where = rise in temperature = 100 - 0 = 1000C



A 250 cm3 glass bottle is completely filled with water at 500C. The bottle and water are heated to 600C. How much water runs over if :

(a) the expansion of the bottle is neglected;

(b) the expansion of the bottle is included? Given the coefficient of areal expansion of glass = 1.2 × 10-5/0C and water = 60 × 10-5/0C.

Solution.Solution.

Water overflow = (final volume of water) - (final volume of bottle)

(a) If the expansion of bottle is neglected :

water overflow 250(1 ) 250l

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5250 60 10 10

Water overflow =1.5 cm3.Sunday, April 9, 2023 34B.M.Sharma Academy of Physics

(b) If the bottle (glass) expands :

Water overflow= (final volume of water) - (final volume of glass)

Water overflow 250 (1 + l) - 250 (1 + g)

Water overflow =250 (l g) where g = 3/2 = 1.8 × 105/0C

Water overflow 250 (58.2 × 10-5) × (60 - 50)

water overflow = 1.455 cm3.

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Fractional charge in time period,

1 12 2

dT dl dgT l g

If acceleration due to gravity is uniform,

0dgg

12

dT dlT l

dll

Time period of a pendulum is given by

2l

Tg

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Thus we have1 12 2

dT dlT l


If temperature is increased, the length of pendulum will increase, consequently time period will increase. The pendulum clock will run slow.

The time lost in time t, is

Similarly time gained when temperature is decreased is also given by

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12

t t

12

t t



A pendulum clock with a pendulum made of Invar ( = 0.7 × 10-6/C0) has a period of 0.5 s and is accurate at 250C. If the clock is used in a country where the temperature averages 350 C, what correction is necessary at the end of a month (30 days) to the time given by the clock ?

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Solution.Solution.

As explained in earlier, in time interval t the clock will become slow (or will lose time) by

12

t t

71(7 10 ) (30 86400) (35 25) 9.1

2t s



A clock with a brass pendulum shaft keeps correct time at a certain temperature.

(a) How closely must the temperature be controlled if the clock is not to gain or lose more than 1 sec. a day? Does the answer depend on the period of the pendulum?

(b) Will an increase of temperature cause the clock to gain or lose?

(bras = 2 × 10-5/0C)

Solution.Solution.

(a) No. of seconds lost or gained per day

186400

2

39


where; = rise or drop in temperature

= coeff. of linear expansion of shaft.

We want that

5

2| |

2 10 86400

186400 1

2

| | 1.1574 C Hence temperature should not increase or decrease by more than 1.15740C

(b) An increase in temperature makes the pendulum slow and hence clock loses time.

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A pendulum clock loses 12 sec. a day if the temperature is 400C and goes fast by 4 sec. a day if the temperature is 200C.

Find the temperature at which the clock will show correct time and the coefficient of linear expansion of the metal of the pendulum shaft.

Solution.Solution.

Let T be the temperature at which the clock is correct.

time lost per day = (½)(rise in temperature) × 86400

12 =(1/2) (40 - T) × 86400 …(i)

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4 = (½)(T - 20) × 86400

on adding (1) and (2), we get

64 = 86400(40 - 20)

= 1.85 × 10-5/0C.

on dividing (1) and (2), we get

12 (T - 20) = 4(40 - T)

T = 250C

clock shows correct time at 250C.

…(ii)

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time gained per day = (½)(drop in temperature) × 86400


When a metal rod is heated or cooled, it expands or contracts. If it is prevented from the expansion or contraction, then stresses are produced in it corresponding to the thermal strain.

Thermal strain :

t

LT

L

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Thermal stress : tY Y T

Force : F A YA T


The following points must be noted about this force.

(i) When the rod is heated, the force is compressive; when the rod is cooled, the force is tensile.

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(ii) The magnitude of the force produced is independent of the length of the rod.



The metal rods of same length l and area of cross- section A are fixed end to end between two rigid supports, as shown in figure. The materials of the rods have Young’s Moduli Y1 and Y2 and the coefficients of linear expansion 1 and 2. When the system is heated up,(a) How many types of strains are produced in each

rod ?

(a) Determine the condition when the junction between the rods does not shift.

1 2Y1 Y2

l l

45


Solution.Solution.

(a) When the system is heated up both the rods try to expand, thus thermal strain is produced in each rod.

Since each rod is prevented from expansion, therefore, they are under compression. Thus, each rod also experiences a mechanical strain.

The net change in length of each rod is

1 11

Fll T

AY

And 2 22

Fll T

AY

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(b) The junction between the rods will not change if the net strain in each rod is zero.

11

0Fl

l TAY

1 2 0

22

Fll T

AY

or 1 1 2 2 tanF

Y Y cons tA T

Hence, the junction does not shift if 1Y1 = 2Y2

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BM SAPB.M.SHARMA A CA DEMY O F PHY SIC S

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Methods of Heat Transfer

• Need to know the rate at which energy is transferred

• Need to know the mechanisms responsible for the transfer

• Methods include• Conduction• Convection• Radiation

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Conduction • The molecules vibrate about their equilibrium

positions• Particles near the flame vibrate with larger

amplitudes• These collide with adjacent molecules and transfer

some energy• Eventually, the energy travels entirely through the

rodConduction can occur only if there is a Conduction can occur only if there is a difference in temperature between two difference in temperature between two parts of the conducting mediumparts of the conducting medium

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HEAT CONDUCTIONHEAT CONDUCTION

Consider a solid slab of cross- sectional area A and thickness x, whose faces are kept at different temperature T1 and T2, respectively as shown in figure.

52

x T1 2 T = T

T2 T1H

t

H

t


53

•Experiments show that the rate of heat flow H/t is proportional to -----------------

• cross-sectional area A and

• the temperature gradient

Tx

x T1 2 T = T

T2 T1H

t

H

t


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That isH T

At x

H TKA

t x

Where k is the proportionality constant called the thermal conductivity.

Good conductors have high k values and good insulators have low k values


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If a slab of infinitesimal thickness dx, across which there is a temperature difference dT, we can write the fundamental law of heat conduction as :

dH dTKA

dt dx


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Note that we choose the direction of heat flow to be the direction in which x increases;

since heat flows in the direction of decreasing T,

therefore, we introduce a minus sign in the equation so as to get a positive value of

dHdt

dTdx

When is negative

dH dTQ KA

dt dx


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Electrical Analogy for Thermal condition

It is an important fact to appreciate that there exists an exact similarity between thermal and electrical conductivities of a conductor.

Electrical Conduction (i) Electric charge flows from higher potential to

lower potential.

(ii) The rate of flow of charge is called the electric current, i.e.

dqI

dt


(iii) The relation between the electric current and the potential difference is given by Ohm’s law, that is

1 2V VI

R

where R is the electric resistance of the conductor.

(iv) The electrical resistance is defined as l l

RA A

58


where is the resistivity of the conductor and is the electrical conductivity of the conductor.1

l

V2 V1dq

dt

dq

dtA

1 21 2( )

dq V V AI V V

dt R l

59


Thermal ConductionThermal Conduction

(i) heat flows from higher temperature to lower temperature.

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(ii) The rate of flow of heat may be called as heat current

(iii) Similarly, the heat current may be related with the temperature difference as :

dHI

dt

1 2

T

T TI

R

where RT is the thermal resistance of the conductor.


61

(iv) The thermal resistance may be defined as

T

lR

kA

where k is the thermal conductivity of the conductor.

2 11 2( )

T

dH KA T TT T

dt l R

T2 T1dH

dt

dH

dtA


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ImportantImportant

• Using the electrical analogy, a problem of heat conduction may be transformed into a problem of electrical conduction and can be solved by using the elementary concepts of electric circuits.

• Good conductors of heat are also the good conductors of electricity.


Problem .Problem .

Explain why

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(a) The glass bulb of a mercury thermometer is long (cylindrical and not spherical) and thin;

(a) The rate of flow of heat through conduction is given by

Solution.Solution.

(b) Clinical thermometer cannot be sterilized by boiling

( )H LdQKA

dt d


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(b) Clinical thermometer is usually calibrated to read from 950F to 1100 F while the boiling point of water is 2120 F. So on sterilization by heating due to thermal expansion of mercury in the capillary, the capillary of the thermometer will burst.

In order that a mercury thermometer inspite of having low thermal conductivity (K) of glass may quickly conduct heat from body to mercury, the glas bulb containing mercury is made long ( so that its area is increased) and thin (so that d is

decreased).


Problem 10.Problem 10.Two solid copper spheres of radii r1 = 15 cm and r2 = 20 cm are both at a temperature of 600C. If the temperature of surrounding is 500C, then find:

(a) the ratio of the heat loss per second from their surfaces initially

(b) the ratio of rates of cooling initially.

Solution.Solution.

(a) Ratio of heat loss

1 1

2 2

/

/

dQ dtHH dQ dt

11

2 2

60 50

60 50

KAHH KA

65


by using Newton’s Law of cooling

221 1 1

22 2 2

15 920 16

H A rH A r

2

1 1

2 2

H rH r

(b) The ratio of initial rates of cooling

1

2

/

/

d dt

d dt

We have

2

1 11

2 22

/

/

M s d dt rM s d dt r

66


As the spheres have the same densities, the ratio of their masses is equal to the ratio of their volumes.

2

1 21

2 12

/

/

d dt r Md dt r M

2 3

1 2

2 1

r rr r

2

1

20 415 3

rr

67



Two plates each of area A, thickness L1 and L2 and thermal conductivities K1 and K2 respectively are joined to form a single plate of thickness (L1 + L2). If the temperature of the free surfaces are T1 and T2, calculate

(a) rate of flow of heat

(b) temperature of interface and

(c) equivalent thermal conductivity.

68


Las R

KA

1 21 2

1 2S

L LR R R

AK AK

1 2

1 2

( )dQ T TH

dt R R R

1 2

1 2

1 2

( )A T TL LK K

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L1

K1

T

K2

L2T2T1

Solution.Solution.

(a) If the thermal resistances of the two plates are R1 and R2 respectively, then as the plates are in series,


(b) IF T is the common temperature of interface, then, as in series the rate of flow of heat remains same, i.e., H = H (=H2), so

1 2 1

1 2 1

,T T T TR R R

1 2 1 1

1 2( )T R T R

TR R

2 11 2

2 1

1 2

1 2

L LT T

K KT

L LK K

Las R

KA

L1

K1

T

K2

L2T2T1

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2 11 2

2 1

1 2

1 2

L LT T

K KT

L LK K

Las R

KA

L1

K1

T

K2

L2T2T1

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(c) If K is the equivalent conductivity of composite slab, i.e., slab of thickness L1 + L2 and cross- sectional area A, then as in series

1 2SR R R 1 21 2

( )

eq

L LR R

AK

1 2 1 2

1 2 1 2

1 2

( )eq

L L L LK

A R R L LK K

Las R

KA

72

Note: If L1 = L2 = L, 1 2

1 2

2K KK HM

K K



Two rods A and B are of equal length. Each rod has its end at temperatures T1 and T2. What is the condition that will ensure equal rates of flow of heat through the rods A and B ?

Solution.Solution.We know that in case of heat conduction (dQ/dt) = KA(1 – 2)/L, so for equal rates of flow,

1 2 ,dQ dQdt dt

1 2 1 2( ) ( )A A B B

A B

K A K AL L

( / ) ( / )A B B AA A K K [as LA = LB = L given]

i.e., the ratio of cross-sectional area of the two rods must be in the inverse ratio of their thermal conductivities.

73



Cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of a material of thermal conductivity K2. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. What is the effective thermal conductivity of the system ?

74


1 2dQ dQ dQdt dt dt

Now if the equivalent conductivity is K,

2 21 1 2 2 1 2( ) 3 ( )K R K R

L L

21 24 ( )dQ R

Kdt L

[ as A = (2R)2]

So from equations (i) and (ii) we have ,

1 24 3K K K 1 2. , ( 3 ) / 4i e K K K

75

In this situation a rod of length L and area of cross-section R2 and another of same length L and cross-section [(2R)2 – R2] = 3R2 will conduct heat simultaneously; so total heat flowing per sec will be

Solution.Solution.


Alternative answer

Lwith R

KA

This problem is equivalent to two thermal resistances [of equal length L and cross-sections R2and [(2R)2 – R2] = 3R2 in parallel so that ,

1 2

1 1 1

eqR R R

Now if K is equivalent conductivity of rod of length L and radius 2R, i.e., cross-section 4R2.

2 2 21 24 3K R K R K R

L L L

1 23

4K K

K

76



Four square pieces of insulation, of two different materials all with same thickness and area A, are available to cover an opening of area 2A. This can be done in either of the two ways shown in fig. Which arrangement (A) or (B) would ensure lesser heat flow if K1 # K2?

K 1

K 2

K 1

K 2

K 1

K 2K 1

( a) ( b)

K 2

77


Solution.Solution.

If the thermal resistance of piece of conductivity K1 while that of K2 is R2, for case (A), 2R1 and 2R2 will be in parallel.

So,1 2

1 1 12 2AR R R

1 2

1 2

2( )A

R RR HM

R R

So,

78

K 1

K 2

K 1

K 2

K 1

K 2K 1

( a) ( b)

K 2


i.e.,

1 2

2B

R RR AM

Now as arithmetic mean is greater than harmonic mean, if R1 ≠ R*2, thermal resistance

in case (B) is higher than in case (A). So case (B) would ensure lesser heat flow as compared to (A).

79

While for case (B) (R1 + R2) and (R1 + R2) will be parallel.

1 2 1 2

1 1 1( ) ( )BR R R R R


Problem .Problem .Two identical rectangular rods of metal are welded end to end as shown in figure (A) and 10 J of heat flows through the rods in 2 min. How long would it take for 30 J of heat to flow through the rods if they are welded as shown in figure (B).

80

100 C0

0 C0

(A)

100 C0 0 C

0

(B)Sunday, April 9, 2023 80B.M.Sharma Academy of Physics

81

If the thermal resistance of each rod is R, in case (A) the rods are in series; so 2SR R R R

So rate of flow of heat in steady state will be

,S

dQdt R

10 (100 0)2 2R

Solution.Solution.


,P

dQdt R

30 (100 0)

( / 2)t R

Substituting the value of R (= 10) from equation (i) and (ii), we get t = 1.5 minute.

82

In case (B) the rods are in parallel; so

1 1 1

PR R R . ,

2P

Ri e R

So rate of flow of heat in this situation will be



A body which has a surface area 5.00 cm2 and a temperature of 7270C radiates 300 joule of energy each minute.

What is its emissivity ? (Stefan’s Boltzmann constt. = 5.67 × 10-8 W/m2 K4)

83


So 8 4 3 4

55.67 10 (5 10 ) (10 )

e

10.18

5.67

84


Solution.Solution.

According to Stefan’s Boltzmann law, energy radiated per sec by a body

4P e AT

Here

273 727 1000T K

(300 / 60) / 5 /P J s J s


The rate at which the radiant energy reaches the surface of earth from the sun is about 1.4 kW/m2. The distance from earth to the sun is about 1.5 × 1011m, and the radius of sun is about 0.7 × 109m.

(a) What is the rate of radiation of energy, per unit area, from the sun’s surface?

(b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?

85

Solution.Solution.Let D = distance from SUN to the earth

= 1.5 × 1011m.

Let R = radius of SUN - 0.7 × 109m.

Let power of SUN = P = energy radiated from the surface of SUN per second.

86


Sun

big sphere

of radius D

D

earth

(a) Hence in every one second, P joules of energy are radiated from the surface of SUN and this energy passes through a big sphere of radius D centred at SUN.

Hence at the circumference of this big sphere (i.e., near the surface of the earth), the energy crossing through a unit area per sec.

2area of big sphere 4 DP P

3 22 1.4 10

4 DP

Wim

11 2 34 (1.5 10 ) 1.4 10P W 263.96 10P W

87

Sun

big sphere

of radius D

D

earthSunday, April 9, 2023 87B.M.Sharma Academy of Physics

Rate of radiation of energy per sec. per unit area of SUN’s surface is given by :

2area of big sphere 4P P

R

7 26.43 10 Wim

88

Sun

big sphere

of radius D

D

earth


(b) If SUN is an ideal black body, e = 1.

4E T1/ 4E

T

1/ 47

8

6.43 105803

5.67 10K

89



The emissivity of tungsten is approximately 0.35. A tungsten sphere 1 cm in radius is suspended within a large evacuated enclosure whose walls are at 300K.

What power input is required to maintain the sphere at a temperature of 3000K if heat conduction along the supports is neglected? = 5.67 × 10-8 S.I. Units.

90


To maintain constant temperature, Power input required = net heat loss per sec from the surface 4 4

0( )inputP e A T T 8 2 4 40.35 5.67 10 4 (0.01) (3000 300 )inputP

2019.8 watts

91

Solution.Solution.

Net heat lost by sphere per second Hnet = eA (T4 - T0

4)

where T = temperature of sphere = 3000 K

T0 = temperature of surrounding = 300K

A = 4 r2 = 4 (0.01)2


Problem .Problem .A double pane window used for insulating a room thermally from outside, consists of two glass sheets each of area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state, the room- glass interface and the glass-outdoor interface are at constant temperature of 270C and 00C respectively.

Calculate the rate of heat flow through the window pane. Also find the temperature of other interfaces. Given, thermal conductivities of glass and air as 0.8 and 0.08 Wm-1 K-1 respectively.

92

Air

K = .08

.05m.01 .01

G

K=.8

G

K=.8

Room

27 C0

0 C0

Out side

1Sunday, April 9, 2023 92B.M.Sharma Academy of Physics

with LR

KA

1 0.01 0.052

0.80 0.08eq

LR

KA A

Air

K = .08

.05m.01 .01

G

K=.8

G

K=.8

Room

27 C0

0 C0

Out side

1

93

Solution.Solution.

In case of thermal conduction as

dQKA

dt L R


And as here

21 ,A m

1 5 2640 8 40eqR

And hence

(27 0) 4041.5

26dQ

Wdt R

94

Air

K = .08

.05m.01 .01

G

K=.8

G

K=.8

Room

27 C0

0 C0

Out side

1


Now if 1 and 2 are the temperatures of air in contact with glass in the room and outside as shown in figure.

1(27 )41.5 0.80 1

0.01

2( 0)41.5 0.80 1

0.01

Solving these for 1 and 2 we get

1 26.48 C 2 0.52 C and

95



A uniform copper bar 100 cm long is insulated on sides, and has its ends exposed to the ice and steam respectively. If there is a layer of water 0.1mm thick at each end, calculate the temperature gradient in the bar. KCu = 1.04 and Kwater = 0.0014 in C.G.S. units.

Solution.Solution.

Let 1 and 2 be the temperatures at the ends of the copper bar.Heat transfer per sec. through the system is

(100 0)

0.01 100 0.01

w cu w

dH Adt

K K K

96


Heat transfer per sec. through copper bar

1 2

100CuK A

As the rods are in series, heat transfer per second must be same through each part.

1 2100 0

1000.01 100 0.01Cu

w Cu w

K AA

K K K

Putting Kcu = 1.04 and Kw = 0.0014 we get :

Temperature gradient

0 11 2 0.87100

C cm

97


Problem 15.Problem 15.A closed cubical box made of perfectly insulating material has walls of thickness 8 cm and the only way for the heat to enter or leave the box is through the solid, cylindrical, metallic plugs each of cross-sectional area 12 cm2 and length 8 cm fixed in the opposite walls of the box as shown in the figure. The outer surface A is kept at 1000C while the outer surface B of other plug is kept at 40C. K of the material of the plugs is 0.5 cal/s/C/cm. A source of energy generating 36 cal/s is enclosed inside the box. Find the equilibrium temperature of the inner surface of the box assuming that it is same at all points on the inner surface..

98

A

100 C0

4 C0

B

SourceSunday, April 9, 2023 98B.M.Sharma Academy of Physics

Solution.Solution.

Let be the temperature of inner surface of box.

Heat transfer per sec. through A + Heat produced by source per sec. = Heat transfer per sec. through B

36 /A B

dH dHcal s

dt dt

100 436

KA KAd d

99

A

100 C0

4 C0

B

SourceSunday, April 9, 2023 99B.M.Sharma Academy of Physics

( 4 100 ) 36KA d

Now d = 8 cm, A = 12 cm2, K = 0.5 cal/s/0C/cm.

36 82 104

12 0.5

076 C

100


Radiation• Radiation does not require

physical contact• All objects radiate energy

continuously in the form of electromagnetic waves due to thermal vibrations of the molecules


102

HEAT RADIATIONHEAT RADIATION

(i) All bodies emit radiation, the intensity and wavelength distribution of which depend on the nature and temperature of its surface.

(ii) Heat radiations are electromagnetic in nature with wavelength longer than that of the visible light.

• In the electromagnetic spectrum, The region of heat radiations extends from about 750 nm to 4×105nm.


Properties of heat radiations

The properties of heat radiations are similar to that of light waves.

103

* Radiant heat travels with the speed of light. * It travels along a straight line .* It can be reflected as well as

refracted like light rays.

* Like light intensity, the intensity of heat radiations decreases in inverse square proportion with distance.

• It also show the phenomena of interferance, differciation or polarization. Sunday, April 9, 2023 103B.M.Sharma Academy of Physics

104

Basic termsPerfectly black body

is a body, which absorbs all the radiations incident on it .


Blackbody• Blackbody is an idealized

system that absorbs incident radiation of all wavelengths

• If it is heated to a certain temperature, it starts radiate electromagnetic waves of all wavelengths

• Cavity is a good real-life approximation to a blackbody


106

Basic termsAbsorptive power (a)

of a surface is defined as the ratio of the radiant energy absorbed to the total energy incident on it.

a ≤ 1 (for any body)

a = 1 (For perfectly black body)

Spectral absorptive power (a)

is defined as the absorptive power of the surface in the wavelength range to + d.

By defination,

0

a a d


Emissivity (e)

of a surface is a measure of its relative remmitance with respect to a perfectly black body.

For example ,If the emmissity of a surface is 0.6,It means that it will radiate 60% of the energy radiated by a perfectly black body under similar conditions .

Spectral emmissive power (e)

is the emissive power of the body between the range to + d.

By defination,

0

e e d


Stefan’s Law

The radiant energy emitted by a body per unit area per second is directly proportional to the fourth power of its absolute temperature T.

That is, I T4

where e is the emissivity of the surface and

is the Stefan’s constant. Its numerical value is 5.68 × 108 Wm-2K4.

If A is the surface area of the body, then the rate of heat emitted by the body is

dHE e AT

dt 4

108

I eT4


Application of Stefan’s LawApplication of Stefan’s Law

If the temperature of the body is more than the temperature T0 of the surrounding, then the body loses heat at a faster rate than it absorbs the same from the surroundings. The net rate of loss of heat is given by

dHE e A T T

dt 4 4

0( )

If m be the mass of body and c be its specific heat, then we may write

dH dTdt dt

109

dH dTmc

dt dt


110

When the temperature difference between the body and its surrounding is not very large, i.e.

T T T T T T 0 0

dTT

dt

dHWe have E e A T T

dt 4 4

0( )

dHE e A T T T

dt 4 4

0 0[( ) )]

dH TE e AT

dt T

4 40

0

[(1 ) 1)]T T

ButT T

4

0 0

(1 ) 1 4

dH TE e AT

dt T

40

0

[(1 4 ) 1)]dH T

E e ATdt T

4

00

[4 ]

dHE e AT T

dt 3

04 dHE T

dt


Thus, when the temperature difference between the body and its surrounding is not very large,

then the rate of cooling is directly proportional to the temperature difference.

This is known as Newton’s Law of Cooling.

dTT

dt



A body cools in 7 minute from 600C to 400C. What will be the temperature after the next 7 minute ? The temperature of surroundings is 100C.

Solution.Solution.

According to Newton’s law of cooling,

1 2 1 202

Kt

60 40 60 40

107 2

K

114

K So that …(i)

112


Now if after cooling from 400C to 7 minute the temperature of the body becomes , according to Newton’s law of cooling,

40 4010

7 2K

Which in the light of equation (i), i.e., K = (1/14) gives 40 1 20

7 14 2

28 C 160 4 20

113


Alternative solution : According to Newton’s law of cooling,

0( )d

Kdt

or2

100

1( )

t ddt

K

or 1 0

2 0

1loget

K

So here1 60 10

7 log40 10eK

and also

1 40 107 log

10eK

114

5 303 ( 10)

50 30log log

30 10

5 – 50 = 90 or = 280C



A liquid takes 5 minute to cool from 800C to 500C. How much time will it take to cool from 600C to 300C ? The temperature of the surrounding is 200C.

Solution.Solution.

According to Newton’s law of cooling ;

1 2 1 202

Kt

80 50 80 50

205 2

K

So that

and60 30 60 30

202

Kt

Solving these for t, we get t = 9 minute.

115



A body cools down from 600C to 550C in 30 seconds. Using Newton’s Law of cooling, calculate the time taken by same body to cool down from 550 C to 500C. Assume that the temperature of surrounding is 450C.

Solution.Solution.

Assume that a body cools down from temperature i to f in t seconds, and c is the temperature of surroundings.Applying Newton’s Law of cooling,

According to Newton’ Law of cooling

0

dK

dt 045 C

116


45d

dt

from t = 0s to t = 30s, changes from 600C to 550C.

(K is constant)

55 30

60 045d

K dt

55 45ln (30 0)

60 45K

50

55 3045

tdK dt

50 45

ln ( 30)55 45

K t

…(i)

…(ii)

117


Divide II by I to set : 50 45

ln30 55 4555 4530 ln60 45

t

30 ln230 ln3 / 2

t

230 1

ln3 / 2ln

t

81.28 s

time from = 550C to = 500C is (t - 30)

= (81.28-30) = 51.28s

118


Distribution of Energy in the Spectrum of a Black Body

•The figure shows a graph plotted between spectral radiant intensity E and wavelength of the emitted radiations at a fixed temperature.

• The graph shows that at a given temperature, the spectral intensity is not uniformly distributed among different wavelengths.

119

Em

E

T = constant

m


120

It first increases with wavelength, reaches a maximum at a particular wavelength m and then decreases with increase in wavelength.

Em

E

T = constant

m


Blackbody Radiation Graph• Experimental data for

distribution of energy in blackbody radiation

• As the temperature increases, the total amount of energy increases• As the temperature

increases, the peak of the distribution shifts to shorter wavelengths

121


Wien’s Displacement Law• The wavelength of the peak of the

blackbody distribution was found to follow Wein’s Displacement Law

λmax T = 0.2898 x 10-2 m • K

•λmax is the wavelength at the curve’s peak

•T is the absolute temperature of the object emitting the radiation

122


•As the temperature of the body increases, the wavelength at which the spectral intensity is maximum shifts toward left.

•Therefore, it is also called. Wien’s Displacement law.

123


Problem 9.Problem 9.A blackened platinum wire, when gradually heated, appears first dull red, then blue and finally white. Why ?Solution.Solution.According to Wien’s law,

mT = constt, i.e, m = (constt.)/T

So when a black body is heated, lm decreases, i.e., with rise in temperature the maximum intensity of radiation emitted gets shifted towards the shorter wavelengths. So the colour of the heated object will change from that of longer wavelength (red) to that of shorter (blue) and when the temperature is sufficiently high and all wavelengths are emitted, the colour will become white (incandescent).

124



On a winter night you feel warmer when clouds cover the sky than when the sky is clear. Why ?

Solution.Solution.

In the day heat radiations coming from sun are absorbed by the earth while in the night the earth radiates this heat. Now as according to Wien’s law m T = constant, the temperature of hot earth is much lesser than that of sun ( 6000 K); the wavelength of radiations emitted by the earth will be of much longer wavelength than that incident on it.

125


Now as clouds (and glass) have the property of passing radiations of shorter wavelength and reflecting longer ones, the heat emitted from the earth will not pass through the clouds and, thus, heat energy emitted by the earth is trapped between the earth and the cloud causing warmer nights.

(Had there been no clouds, the energy emitted by the earth will be radiated in the sky causing cool nights. )

126


Note: Similarly if sunlight enters a glass room the glass acts as a one way gate, i.e., allows the heat of shorter wavelengths (coming from sun) to enter the room and does not allows the longer wavelength radiations emitted by the heated objects in the room to escape from it (by reflecting them).

In this way energy is trapped in the glass house and the room will be gradually warmed up. This effect is called ‘green house effect’ as it is used to keep the plants in warm atmosphere in winters by placing them in a glass house exposed to sun.

127


thermal expansion,heat transfer and radiation)

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