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11th Physics Notes - Thermal Expansion for JEE Main

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Page 1: 11th Physics Notes -  Thermal Expansion for JEE Main

9011041155 / 9011031155

• Live Webinars (online lectures)

with

recordings.

• Online Query Solving

• Online MCQ tests with detailed

solutions

• Online Notes and Solved

Exercises

• Career Counseling

1

Page 2: 11th Physics Notes -  Thermal Expansion for JEE Main

9011041155 / 9011031155

Thermal Expansion

Increase in the Dimensions of a substance on heating

is called its Thermal Expansion.

Expansion in Solids

Almost all the solids (except bismuth, ice and cast iron)

expand on heating. In solids all the atoms and

molecules are closely bound to each other. So, they

exert strong electromagnetic forces on each other. Due

to this tight bonding between them, though they vibrate

about their mean positions, they cannot leave it

forever. When a solid is heated, the atoms and

mmolecules absorb thermal energy and start vibrating

with greater amplitude.

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So, they move away from each other or intermolecular

distance between them increases. This causes

changes in different dimensions of the solid.

1. Linear Expansion

Increase in the length of the solid on heating is

called linear expansion.

Let,

L0 = length of the solid at 0oC.

Lt = length of the solid at toC.

t = rise in temperature of the solid.

∴ Lt – L0 = increase in the length.

It is found that Lt – L0 α L0t

∴ Lt – L0 = α L0t .....(1) where α is coefficient of

linear expansion of the solid.

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If L0 = 1m or cm and t = 1℃, then

α = Lt – L0

Thus, temperature coefficient of linear expansion of

a solid is increase in its length per unit original

length at 0℃, per unit rise in its temperature. Its

unit is per ℃ (/℃) and values for solids are ranging

from 10-6 to 10-5 / ℃.

From equation (1)

Lt = L0 + L0 α t

∴ Lt = L0 (1 + α t) ..... (2)

Let, L1 = length of the solid at t1℃

L2 = length of the solid at t2℃

From equation (2),

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L1 = L0 (1 + α t1) and

L2 = L0 (1 + α t2)

∴ L2 / L1 = (1 + α t2) (1 + α t1)-1

Binomial expansion of (1 + α t1)-1 will be

1 – α t1 + terms containing higher powers of α.

As α is very small, its higher powers can be

neglected.

∴ (1 + α t1)-1 = 1 – α t1

∴ L2/L1 = (1 + α t2) (1 – α t1)

∴ L2/L1 = 1 – α t1 + α t2 – α2t1t2

∴ L2/L1 = 1 + α (t2 – t1)

∴ L2 = L1 + L1α (t2 – t1) ... (3)

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∴ L2 – L1 = L1 α (t2 – t1)

Thus, by knowing L1, L2 and rise in temperature

(t2 – t1), a can be calculated approximately.

2. Areal or Superficial Expansion

Let, A0 = area of the solid at 0℃.

At = area of the solid at t℃.

t = rise in temperature of the solid.

∴ At – A0 = increase in the area.

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It is found that At – A0 α A0t

∴ At – A0 = β A0t ..... (1) where β is coefficient of

areal expansion of the solid.

If A0 = 1m2 or cm2 and t = 1℃, then

β = At – A0

Thus, temperature coefficient of areal expansion of

a solid is increase in its area per unit original area

at 0℃, per unit rise in its temperature. Its unit is per

℃(/ ℃).

From equation (1)

At = A0 + A0 βt

∴ At = A0 (1 + βt) ..... (2)

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Every time it is not essential to measure area of the

solid at 0℃, to calculate β. It can be calculated at

any temperature with the formula :

3. Volumetric Expansion

Let,

V0 = volume of the solid at 0℃.

Vt = volume of the solid at t℃.

t = rise in temperature of the solid.

∴ Vt – V0 = increase in the volume.

It is found that Vt – V0 α V0t 8

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∴ Vt – V0 = γ V0t ..... (1) where b is coefficient of

areal expansion of the solid.

If V0 = 1m3 or cm3 and t = 1℃, then

γ = Vt – V0

Thus, temperature coefficient of volume expansion

of a solid is increase in its volume per unit original

volume at 0℃, per unit rise in its temperature. Its

unit is per ℃ (/℃).

From equation (1)

Vt = V0 + V0 β t

∴ Vt = V0 (1 + γ t)..... (2)

Every time it is not essential to measure volume of

the solid at 0℃, to calculate γ. It can be calculated

at any temperature with the formula :9

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Relation between α, β and γ

Consider a thin square plate of a solid having side L0

at 0℃.

∴ Surface area at 0℃ = A0 = L02

Let the plate be heated to t℃. Let,

Lt = length of the plate at t℃

∴ At = area of the plate at t℃ = Lt2

For linear expansion, Lt = L0 (1 + α t)

∴ Lt2 = L02 (1 + α t)2 i.e. At = A0 (1 + α t)210

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∴ At = A0 (1 + 2αt + α2t2)

As α is very small α2t2 can be neglected.

∴ At = A0 (1 + 2αt)

But, At = A0 (1 + β t) ∴ 2α = β

Consider a uniform of a solid having side L0 at 0℃.

∴ Volume of the cube at 0℃ = V0 = L03

Let the plate be heated to t℃. Let,

Lt = length of the plate at t℃

∴ Vt = volume of the plate at t℃ = Lt3

For linear expansion, Lt = L0 (1 + α t)

∴ Lt3 = L03 (1 + α t)3 i.e. Vt = V0 (1 + α t)3

∴ Vt = V0 (1 + 3αt + 3α2t2 + α3t3)

As α is very small 3α2t2, a3t3 can be neglected.

∴ Vt = V0 (1 + 3αt)

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But, Vt = V0 (1 + γ t)

∴ 3α = γ

∴ 6α = 3β = 2γ

This is the relation between the three expansion

coefficients.

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Boiling point / Steam point

The temperature at which pure water freezes at

standard atmospheric pressure is called as ice point or

freezing point. The temperature at which pure water

boils at standard atmospheric pressure is called as

boiling point or steam point.

1. Celsius scale

On this scale the ice point (melting point of pure

ice) is marked as 0ºC and the steam point (boiling

point of water) is marked as 100ºC, both taken at

normal atmospheric pressure. The interval between

these points is divided into 100 equal parts. Each of

these division is called as one degree Celsius and

is written as 1ºC.

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2. Fahrenheit scale

On this scale the ice point (melting point of pure

ice) is marked as 32ºF (boiling point of water)

steam point is marked as 212ºF. The interval

between these two reference points is divided into

180 equal parts. Each division is called as degree

Fahrenheit and is written as ºF.

The graph of Fahrenheit temperature (tf) versus

Celsius temperature (tc). It is a straight line whose

equation is,

.............(1 (a))

.........(1(b))

Ideal gas equation

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Boyle’s law states that at constant temperature, the

volume of a given mass of a gas is inversely

proportional to its pressure.

Mathematically this law may be expressed as,

at constant temperature ......(2)

Charle’s law states that at constant pressure, volume

of a given mass of a gas is directly proportional to

its absolute temperature. Mathematically this law

may be expressed as,

V ∝ T, at constant pressure .... (3)

Combining (2) and (3) we have,

= constant .....(4)

For one mole of a gas the constant of proportionality is

R

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.............(5)

PV = RT---This relation is called as ideal gas equation.

R is universal gas constant.= 8.31 JK-1 mol-1

PV = nRT ......... (6)

From equation (6) PV ∝ T.

This relationship allows a gas to be used to measure

temperature in a constant volume gas thermometer

keeping the volume of a gas constant, it gives P ∝ T.

Specific heat Capacity

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Specific heat of a substance is defined as the quantity

of heat required to rise the temperature of unit mass of

a substance through 1ºC (or 1 K).

Q ∝ M ∆T

Q ∝ cM ∆T ............. (7)

where c is called as specific heat (or specific heat

capacity) of a substance. From equation (8) we get

.............. (9)

If m = 1 kg and ∆T = 10C then c = Q.

SI unit of specific heat is Jkg-10C-1 or Jkg-1 K-1, CGS Unit

is cal g-1 K-1 or Jkg-1 K-1. The specific heat of water is

4.2 Jkg-10C-1; it means that 4.2 Jkg-10C-1 of energy must

be added to 1g of water to rise its temperature by 1ºC.

In case of gas slight change in temperature is

accompanied with considerable changes in both i.e. in

volume and pressure. 17

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If gas is heated at constant pressure, volume changes

and hence some work is done on surrounding in

expansion. Hence more heat is required. Therefore

specific heat at constant pressure is greater than

specific heat at constant volume.

1. Specific heat of a gas at constant volume (cv) is

defined as the quantity of heat required to rise the

temperature of unit mass of a gas through 1K (or

10C) when its volume is kept constant.

2. Specific heat of a gas at constant pressure (cp) is

defined as the quantity of heat required to rise the

temperature of unit mass of a gas through 1K (or

10C) when its pressure is kept constant.

For unit mass of a gas, the specific heats are called

principal specific heats.

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i. Molar specific heat of a gas at constant volume

(Cv) is defined as the quantity of heat required

to rise the temperature of one mole of the gas

through 1 K (or 10C), when is volume is kept

constant.

ii. Molar specific heat of gas at constant pressure

(CP) is defined as the quantity of heat required

to rise the temperature of one mole of the gas

through 1 K (or 10C) when its pressure is kept

constant. The number of molecules in one mole

of a gas is given by Avogadro’s number

N = 6.025 × 1023 molecules per mole

= 6.025 × 1026 molecules per kilomole. SI unit

of molar specific heat is J/K and mole K.

The molar specific heat = molecular weight × principle specific heat.

Cp = M × cp and Cv = M cv .............. (10)19

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At standard pressure the temperature at which a

substance changes its state from solid to liquid is

called as its melting point. Melting point of water is 0ºC.

At standard pressure the temperature at which a

substance changes it state from liquid to gas is called a

boiling point. The boiling point of water is 100ºC.

The triple point of water is that point where water in a

solid, liquid and gas states coexists in equilibrium and

this occurs only at a unique temperature and a

pressure.

Latent heat of a substance is the quantity of heat

required to change the state of unit mass of the

substance without changing its temperature.

Conduction

It is mode of transfer of heat through a medium, without

actual migration of the particles of the medium.

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Temperature gradient

It is the rate of change of temperature with distance for

a metal rod in steady state.

The figure shows, a section of a metal rod of length ∆x.

In steady state, if the temperature of the left end is θ +

∆θ and that of the right and is θ, the temperature

gradient is ∆θ / ∆x.

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In steady state, the rod doesn’t absorb any heat. So,

the amount of heat entering the rod per second (Q) is

same as the amount of heat leaving the rod per

second.

Experimentally it is found that, the heat flowing through

a conductor or the amount of heat conducted by a

conductor is directly proportional to area (A) of the

conductor, temperature gradient (∆θ / ∆x) and time t

for which the heat is flowing.

i.e. Q ∝ A (∆θ / ∆x) t

∴ Q = kA (∆θ / ∆x)t where k is coefficient of

thermal conductivity.

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k = Q

So, coefficient of thermal conductivity is the amount of

heat flowing through a conductor, of unit area, in unit

time, when unit temperature gradient is maintained

along its length.

Its units are cal / cm ℃ s in C.G. S., kcal / m ℃ s in

M.K.S. and J / m ºK s in S.I.

Its dimensions are [M1L1T-3K-1]

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ConvectionTransfer of heat energy because of actual migration of particles from 1 point to another point is known as Convection

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Radiation

Transfer of heat energy through vacuum is known as

radiation

Newton’s law of cooling

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The rate of fall of temperature of a body is directly proportional to excess temperature of a body over the surrounding (provided that the excess is small)

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