kvpy thermal expansion.pdf

7/23/2019 KVPY THERMAL EXPANSION.pdf

1/7

96

1. INTRODUCTION

Solids are made of atoms and molecules. At a given

temperature, the atoms and molecules are placed at

some equilibrium distance. When heat is supplied to

solid, the interatomic separation increases by which

there is an expansion of solids. This expansion can

be in terms of length/area/volume.

From the above tree, it is clear that the thermal

expansion of solid is classified into three categories.

(1) Linear expansion of solids

(2) Superficial expansion of solids

(3) Cubical expansion of solids

Now we shall discuss these topics one by one.

2. LINEAR EXPANSION OF SOLID

(i) Almost all solids expand on heating. On

increasing the temperature of a solid, its lengthincreases. This change in length of a solid on

heating is called linear expansion.

(ii) Coefficient of linear expansion is defined as

fractional increase in length per C rise in

temperature. If is the length of the rod at

T K and as the temperature is changed to

T +T its length becomes +, so coefficientof linear expansion is given by

=T

=dT

d

(iii) Unit ofis Per Kelvin or Per C. It is positivefor metals except carbon. The value of isnegative for plastic because in plastic when the

temperature increases, length decreases.

(iv)The numerical value of is same in both theunits i.e. in Per Kelvin or Per C

(v) Ifis coefficient of linear expansion at t1C.1 = length of the rod at t1C

2 = length of the rod at t2C

2 =1[1 +(t2 t1)]t may be in any unit C or K because in the

formula there is a difference of temperature which

remains same for C or K.

(vi) If length of the rod is0at 0C andtat tC than

t =0 [1 + (t 0)] t =0(1 +t)

where is temperature coefficient of linearexpansion at 0C. Here t should be in C only

because initial temperature is taken as 0C.

3. APPLICATION OF LINEAR EXPANSION

3.1 Differential expansion of two solid rod

(1) Suppose there are two rods of length1and2.The first rod is kept on the other such that the

initial separation between the free ends of the

rod is

S =2 1

Both the rods are initially at a temperature of

t1C.

On heating the entire system, the temperature

increases to t2 such that the length of both the

rod increases. So, if the new length of the rod be

1 and2 then

1=1[1 +1 (t2 t1)]

2=2 [1 +2 (t2 t1)]

Here 1 and 2 are the coefficients of linearexpansion at t1 and t2 C

Now the separation between the free ends of the

rod also changes such that

S =2 1

=2 [1 +2 (t2 t1)] 1 [1 +1(t2 t1)]

= (2 1) + (22 11) (t2 t1)


2/7

97

If the new separation Sof the composite rod isequal to the original separation then

S = S

(2 1) + (22 11) (t2 t1) =2 1(2211)(t2 t1) = 0

22 11 = 0

2

2 =1

1

3.2 Radius of Bimetallic strip

If two strips of different metals are welded together

to form a bimetallic strip, when heated uniformly it

bends in the form of an arc, the metal with greater

coefficient of linear expansion lies on convex side.

The radius of arc thus formed by bimetal is :

)tt)((

dR

1212

or

t)(

dR

12

at Ct01

Ct02

Where t = temp. difference between the two ends.

d = thickness of each strip

1&2coefficients of linear expansion

Differential expansion of two solid rods

Ex. 1 The length of the steel rod which would have

the same difference in length with a copper rod

of length 24cm at all temperatures.

(copper = 18 106k1,steel= 12 10

6k1)

is

(A) 20 cm (B) 18 cm

(C) 24 cm (D) 36 cm

Sol. (D) By linear expansion of solids, we have

=..Tsosteel . steel .T =copper . copper . T

steel =steel

coppercopper .

=6

6

1012

101824

= 36 cm

Ex. 2 A metal sheet with a circular hole is heated. The

hole

(A) Gets larger

(B) Gets smaller

(C) Remains of the same size

(D) Gets deformed

Sol. (A) When a body is heated, the distance between

any two points on it increases. So due to thermal

expansion of solids, the hole gets larger.

Ex. 3 Two identical rectangular strips one of copper

and other of steel are riveted together to form a

bimetallic strip (copper >steel). On heating thisstrip will

(A) Remains straight

(B) Bend with copper on convex side

(C) Bend with steel on convex side

(D) Get twisted

Sol. (B) When a bimetallic strip of different metals is

heated, the strip bends due to unequal linear

expansion of the two metals. The strip will bend

with metal of greater on the convex side i.e.outer side.

3.3 Thermal stress

(1) When a rod is heated or cooled, it expands or

contracts. It is turned as free expansion of the

rod

t1C

t2C strain = 0

Actually no strain is being developed because

on increasing the temperature the length of the

rod increases so at t2C, 2 because natural

lengthof rod.

(2) Now if the ends of the rods are rigidly fixed so

as to prevent it from expansion or contractionthan stress is produced in the rod. By virtue of

this thermal stress the rod exerts a large force on

the supports.


3/7

98

The first figure indicates the rod kept at a

temperature t1C having length 1. The rod is

between two rigid supports. If the supports were

not there then on increasing the temperature the

free expansion occurs by which length becomes

2 at higher temperature t2C. The third figure

indicates that on increasing the temperature to

t2C, the length remains same but a compressivestrain is induced in the rod.

If1 is the length of the rod at t1C and on

increasing the temperature to t2C its length

becomes 2 then

2=1 [1 + (t2 t1)]

2 =1 +1 (t2 t1)

1

12

= (t2 t1)

The above relationship indicates the thermal

strain developed in the rod. i.e.

Thermal strain =2

12

= (t2 t1)

In elasticity, Young's modulus =strain

stress

Stress = Y (t2 t1)

As force =Area

stress

Force = YA (t2 t1)

(3) If the rod is in its natural length at t1C while at

t2C it is in compressed state, then

Strain =lengthOriginal

lengthinChange

In this case first figure represents the rod oflengthat t1C when the temperature is lowered

to t2C the length of the rod remains same but a

tensile strain is developed in the rod.

=t

)t(

=

t

t

=t

t

~ t

Stress = t

(4) When the temperature of the rod is increased,

the compressor stress is developed while on

decreasing the temperature of the rod the tensile

stress is developed.

Thermal stress

Ex. 4 A steel rod of length 1m rests on a smoothhorizontal base. If it is heated from 0C to 100C,what is the longitudinal strain developed ?

Sol. In this case rod rests on a horizontal base whichis the free expansion on heating. Hence no strainis developed in the rod i.e.

Strain = 0

Ex. 5 A steel rod of length 50 cm has a crosssectionalarea of 0.4cm2. What force would be required tostretch this rod by the same amount as theexpansion produced by heating it through 10C.

(= 105k1and Y = 2 1011N/m2)Sol. We have

Force = YA. = 2 1011 0.4 104 105 10

= 0.8 103

= 800 N

3.4 Effect of temperature on Pendulum clock

A pendulum clock consists of a metal rod or wire withthe bob at one end.

Let1be the length of the simple pendulum at 1Cthan time period T1 is given by

T1 = 2 g1

...(1)

Now when the temperature increases to 2C theeffective length becomes2 so that

T2 = 2 g2

...(2)

Dividing eqn. (2) by (1), we get

1

2

T

T=

1

2

but 2 =1 [1 + (2 1)]

so 1

2

T

T

= 1

121 )](1[

1

2

T

T= [1 + (2 1)]

1

2

T

T= 1 + 1/2 (2 1)

1

12

T

TT= 1/2 (2 1)


4/7

99

Change in time periodT = T2 T1= 1/2..T1The above expression represents the time lost per

oscillation. Thus a pendulum clock loses time in

summer and gains time in winter.

Note: If a pendulum clock is giving correct time when

time period is T than

If T increases, clock becomes slow.

If T decreases, clock becomes fast.

Pendulum clock

Ex. 6 A pendulum clock with a pendulum made of invar

(= 0.7 106 C1) has a period of 0.5 s andis accurate at 25C. If the clock is used in a

country where the temperature averages 35C,what correction is necessary at the end of a

month (30 days) to the time given by the clock

Sol. In time interval t, the clock will become slow by

t = 2

1

.t.

=2

1 7 107 30 86400 (35 25)

= 9.1 s

4. SUPERFICIAL EXPANSION OF SOLID

(i) On increasing the temperature of solid, its area

increases. This change in area is referred as

superficial expansion of solids.

(ii) If A0is the area of solid at 0C. On heating the

rod to t1

C, the area becomes At

so that

At = A0 [1 +t]

Whereis coefficient of superficial expansion at0C and t should be in C only.

(iii) If the area of solid at temperature t1C is A1and

on heating the rod, the area becomes A2at t2C

than

A2 = A1 [1 + (t2 t1)]

Whereis coefficient of superficial expansion att1C.

(iv) Coefficient of superficial expansionis defined as

fractional increase in area per C rise in temperature.

=T

AA

=dT.A

dA

(v) Unit of is per C or per Kelvin

5. VOLUME EXPANSION OF SOLID

(i) On increasing the temperature of rod, its volume

changes.

(ii) If V0 is the volume of solid at 0C and on

increasing the temperature, volume becomes V tthan,

Vt = V0 [1 +t]

Where is coefficient of volume expansion at0C. Here also t should be in C only.

(iii) If V1 is the volume of solid at t

1C and on

increasing the temperature to t2C the volume

becomes V2 then,

V2 = V1 [1 + (t2 t1)]

Where is coefficient of volume expansion att1C.

(iv) Coefficient of volume expansion is defined as

the fractional increase in volume per C rise in

temperature.

=T

VV

=dT.V

dV

(v) Unit of is per C or per Kelvin.

6. RELATION BETWEEN COEFFICIENT OF LINEAR

EXPANSION ( ), COEFFICIENT OF SUPER FICIAL

EXPANSION ( ) & COFFICIENT OF CUBICAL

EXPANSION ()

= 2

= 3

=

3

2

Superficial and volume expansion of

solid

Ex.7 The coefficient of linear expansion of a crystal in

one direction is 1 and that in every directionperpendicular to it is2. The coefficient of cubicalexpansion is

(A) 1+2 (B) 21+ 2(C)1+ 22 (D) None of above

Sol. (C) V = V0(1 +T)

or L3= L0(1 + 1T) 20L (1 +2T)2

or = 30L (1 +1T) (1 +2T)2

or V = V0(1 +1T)(1 +2T)2

Hence 1 +T = (1 +1T)(1 +2T)2

or 1 +T = (1 +1T)(1 + 22T)

= 1 + (1+ 22)T

Hence = 1+ 22


5/7

100

Ex. 8 Consider the following statements

(1) The coefficient of linear expansion has

dimension K1.

(2) The coefficient of volume expansion has

dimension K1.

(A) Both 1 and 2 are correct

(B) 1 is correct but 2 is wrong

(C) 2 is correct but 1 is wrong(D) 1 and 2 are both wrong

Sol. (A) Coefficient of linear expansion is given by

=dt.

d

and coefficient of volume expansion is

given by=dt.V

dV. So from above formula it is

clear that both have units of per Kelvin.

7. CHANGE IN DENSITY OF SOLID WITH

TEMPERATURE

Suppose m is the mass of a solid which at a given

temperature occupies a volume V so that density at

0C is d0= V

m

Now if the temperature is increased by tC, mass will

remain unchanged but due to thermal expansionvolume increases so that,

V= V (1 + t)

Now density dt= Vm

=)t1(V

m

dt = )t1(

d0

Here is coefficient of cubical expansion at 0C.


6/7

101

SOLVED EXAMPLES

Ex.1 A clock which keeps correct time at 25C has a

pendulum made of brass whose coefficient of

linear expansion is 0.000019. How many seconds

a day will it gain if the temperature fall to 0C.

Sol. Let L0and L25be the length of pendulum at 0C

and 25C respectively.

We know that

L25= L0(1 + T)

= L0(1 + 0.000019 25) = 1.000475 L0

If T25and T0be the time periods at 25C and 0C

respectively, then

T25 = 2

g

L25 and T0 = 2

g

L0

0

25

T

T=

0

25

L

L=

0

0

L

L000475.1

= )000475.1( = 1.000237.

Now0

025

T

TT= 0.000237.

Gain in time for one vibration

= 2 0.000237 sec.

Number of vibration in one day

=2

606024 sec. (T = 2 sec.)

Hence, the gain in time in one day

= 2 0.0002372

606024 =20.52 sec.

Ex. 2 A steel tape gives correct reading at 20C. A

piece of wood is being measured with the steel

tape at 0C. The reading is 25 cm on the tape.State whether the real length of the wood is more

than or less than or equal to 25 cm.

Sol. The steel expands on heating and contracts on

cooling. Given that tape gives correct reading at

20C. At 0C. the tape contracts. The situations

at 0C and 20C are shown in figure. Obviously

real length of the wood is less than 25 cm.

Ex.3 A metal disc has a hole in it. What happens to

size of hole when disc is heated?

Sol. On heating the size of hole increases.

Ex.4 A circular hole in an aluminium plate is 2.54 cm in

diameter at 0C. What is the diameter when the

temperature of the plate is raised to 100C ?

Given

Al = 2.3 105 (C)1

Sol. Let Doand Dtbe diameters of hole at 0C andtC respectively.

Circumference of hole at 0C

= 2r0 =D0Circumference of hole at t = 100 C

t = 2t = DtFrom relationt=0 (1 +.t), we get

Dt= D0(1 + 2.3 105 100)

Dt= 2.54 ( 1 + 0.0023)

= 2.5458 cm.

Ex.5 A pendulum clock keeps correct time at 0C. Mean

coefficient of linear expansion is a per C. If the

temperature of the room increases by tC, then

show that the clock loses per day by

86400t2

1

Sol. Let L0 be length at 0C and L1 that at tC. If T0and T1are time periods at 0C and tC respectively,

then

T0= 2 gL0 and

gL2T tt

Dividing

2/1

0

0

0

t

0

t )t1(L

)t1(L

L

L

T

T

.....(1)


7/7

102

[For section pendulum time period is 2 sec i.e.

T0/2 = 1 sec.]

If n0 and nt are number of seconds in one day,

then

t

0

0

t00

tt

n

n

T

T

2

Tn

2

Tn

From (1)

t

0

n

n= 2/1t )1( =

t

2

11

ort

0

n

n 1 = t

2

1

or t2

1

n

nn

0

0t

or nt n0 = 0nt2

1

But n0= 24 60 60 = 86400

nt = t2

1 86400

Negative sign shows that the clock loses time.

Ex.6 If the volume of a block of a metal changes by

0.12% when it is heated through 20C, what is

the coefficient of linear expansion of metal?

Sol. Coefficient of cubical expansion of metal is given

by

Vt

V

Here100

12.0

V

V

, t = 20C

201000.12

= 6.0 105per C

Coefficient of linear expansion

3

106.0

3

5

= 2.0 105Per C

kvpy thermal expansion.pdf

Documents