Thermal Expansion

Thermal expansion refers to the tendency of matter to change in volume or shape when exposed to changes in temperature. When heated, most materials expand due to increased molecular movement, while cooling causes them to contract. This physical property is crucial in various fields such as engineering, construction, and manufacturing, where temperature fluctuations impact material performance. Understanding thermal expansion helps in designing structures and devices that can withstand temperature variations without damage.

1.0Potential Energy V/s Distance Curve

When matter is heated without a change in state, it usually expands.
According to atomic theory, this happens because the potential energy curve is asymmetric.

As temperature rises from $T_{1}$ to $T_{2}$

Atomic vibration amplitude increases.
Atomic energy rises from $E_{1}$ to $E_{2}$
Average distance between atoms increases from $r_{1}$ to $r_{2}$ (where $r_{2}$ > $r_{1}$ )
Increase in atomic spacing causes the material to expand (thermal expansion).
If the potential energy curve were symmetric, no expansion would occur despite heating.

2.0Linear Expansion

When the rod is heated, its increase in length $Δ L$ is proportional to its original length $L_{0}$ and change in temperature $Δ T$ where ΔT is in $^{\circ} C or K.$

$d L = α L_{0} d T ⟹ Δ L = α L_{0} Δ T$

$α L_{0} T << 1$

$α = \frac{Δ L}{L _{0} Δ T}$ where $α$ is called the coefficient of linear expansion whose unit is $^{\circ} C^{- 1} or K^{- 1} .$

$L = L_{0} (1 + α Δ T)$ where L is the length after heating the rod.

Variation of $α$ with temperature and distance

If varies with distance, for example $α = a x + b$

Then total expansion = $\int (a x + b) Δ T d x$

If varies with temperature, $α = f (T) Then Δ L = \int α L_{0} d T$

Note: Actually, thermal expansion is always 3-D expansion. When other two dimensions of an object are negligible with respect to one, then observations are significant only in one dimension and it is known as linear expansion.

3.0Superficial (Areal) Expansion

The increase in the surface area of a solid when it is heated without a change in state.

$A = A_{0} (1 + β Δ T)$

$A_{0} = l_{0}^{2} and A = l^{2}$

$(1 + β Δ T) = [l_{0} (1 + α Δ T)]^{2} ⟹ β \approx 2 α$

4.0Volume Expansion

The increase in the volume of a substance when it is heated without a change in state.

$V = V_{0} (1 + γ Δ T)$

$V = l^{3} and V_{0} = l_{0}^{3}$ so

$α : β : γ$ = 1 : 2 : 3

Unit of :: is $α, β is 1^{\circ} C or 1/ K or K^{- 1} .$

5.0Variation of Time Period of Pendulum Clocks

The clock’s time depends on the number of oscillations of its pendulum.
Each time the pendulum reaches an extreme position, the second hand advances by one second.
One full oscillation moves the second hand by two seconds.
A pendulum with a time period of 2 seconds is called a seconds pendulum.

Let $T = 2 π \frac{L _{0}}{g}$ at temperature $θ_{0}$ and T'=2lg $T^{'} = 2 π \frac{L}{g}$

$\frac{T ^{'}}{T} = \frac{L}{L _{0}} = \frac{L _{0} ( 1 + α Δ θ )}{L _{0}} = 1 + α Δ θ \approx 1 + \frac{1}{2} α Δ θ$

Therefore change (loss or gain) in time per unit time lapsed is $T = 12 \frac{T ^{'} - T}{T} = \frac{1}{2} α Δ θ$

Gain or loss in time in duration of 't' in

$Δ T = \frac{1}{2} α Δ θT$ if T is the correct time then

(a) $θ < θ_{0}, then T^{'} < T$ clock becomes fast and gain time

(b) $θ > θ_{0}, then T^{'} > T$ clock becomes slow and loose time

6.0Effect of Thermal Expansion on Measurements

Case (1): Only Object Expands

Scale does not expand.
Actual and measured length at temperature 2

$l_{2} = l_{1} [1 + α_{0} (θ_{2} - θ_{1})]$

$α_{0}$ = Linear expansion coefficient of object

Case (2): Only Scale (Measuring Instrument) Expands

Object length remains unchanged.
Measured value (MV) decreases due to scale expansion:

$M V = l_{1} [1 + α_{s} (θ_{2} - θ_{1})]$

$α_{s}$ = linear expansion coefficient of scale

Case (3): Both Object and Scale Expand

Measured value: $M V = l_{1} [1 + (α_{0} - α_{s}) (θ_{2} - θ_{1})]$
If $α_{0} > α_{s}, measured value > actual length.$
If $α_{0} < α_{s}, measured value < actual length.$

General Formula for Measured Value:

$Measured Value = Calibrated Value \times [1 + α (θ - θ_{c})]$

$α = (α_{0} - α_{s}), θ = measurement temperature, θ_{c} = calibration temperature$

$If θ > θ_{c}, true length$ < $scale reading$
$If θ < θ_{c}, true length$ > $scale reading$

7.0Thermal Stress And Thermal Strain

No stress occurs if a rod is free to expand or contract with temperature.
Thermal stress arises when expansion or contraction is restricted.
Tensile stress → when heated (expansion restricted)
Compressive stress → when cooled (contraction restricted)

Thermal Strain

$Strain = \frac{Δ L}{L _{0}} = α Δ T$

α = coefficient of linear expansion,ΔT = temperature change

L0 = original (natural) length at new temperature $L_{0} = original (natural) length at new temperature$

Note:Final and original lengths must be measured at the same temperature.

Example:Rod Fixed Between Two Rigid Walls

Initial length = $L_{0}$

Temperature increases byΔT , but ends can't move → strain develop

$St r ain = α Δ T$

Leads to thermal stress in the rod.

$Strain = \frac{Length of rod at new temperature - Natural length of the rod at new temperature}{Natural length of the rod at new temperature}$

$Strain = \frac{l _{0} - l _{0} ( 1 + α Δ θ )}{l _{0} ( 1 + α Δ θ )} = \frac{- l _{0} α Δ θ}{l _{0} ( 1 + α Δ θ )}$

$α$ is very small so strain = $- α δ θ$ (Negative sign represents that the length of the rod is less than the natural length that means is compressed by the ends)

8.0The Bimetallic Strip

A bimetallic strip is made by bonding two metals with different linear expansion coefficients.
Common metals used: $Brass: α = 19 \times 1 0^{- 6},^{\circ} C, Steel: α = 12 \times 1 0^{- 6},^{\circ} C$

When heated:

Brass expands more than steel.
The strip bends towards the steel side (brass on the outside of the curve).

When cooled:

Brass contracts more than steel.
The strip bends towards the brass side.

Note:Used in thermostats and temperature-sensing devices.

Mathematical Solution:

$(R + \frac{d}{2}) ϕ = L_{0} (1 + α_{1} Δ θ)$

$(R - \frac{d}{2}) ϕ = L_{0} (1 + α_{2} Δ θ)$

$\frac{R + \frac{d}{2}}{R - \frac{d}{2}} = \frac{1 + α _{1} Δ θ}{1 + α _{2} Δ θ}$

9.0Thermal Expansion in Liquids

Liquids do not exhibit linear or superficial expansion, only volume expansion.
Liquids are heated along with the container holding them.

On heating:

Initially, liquid level falls → container expands first.
Then, liquid level rises → liquid expands faster than the container.

In a volume vs. temperature graph:

PQ: Expansion of the vessel.
QR: Real expansion of the liquid.

Real volume expansion of liquid=Apparent expansion+Expansion of vessel Real Volume Expansion of Liquid = Apparent Expansion + Expansion of Vessel

Co–Efficient of Real Expansion $r (γ_{r})$

It is due to the actual increase in volume of liquid due to heating.

$γ_{r} = \frac{Real increase in volume}{Initial volume \times Δ θ}$

$= \frac{Δ V}{V \times Δ θ}$

$γ_{real} = γ_{apparent} + γ_{vessel}$
Change (apparent change) in volume in liquid relative to vessel is

$Δ V_{app} = V (γ_{real} - γ_{vessel}) Δ θ$

$α =$ Coefficient of linear expansion of the vessel

Different level of liquid in vessel

\gamma	V\Delta V	Level
$γ_{real} > γ_{vessel} (= 3 α), γ_{app} > 0$	$V_{a} pp$ is Positive	Level of liquid in Vessel will rise on heating
$γ_{real} < γ_{vessel} (= 3 α), γ_{app} < 0$	$V_{a} pp$ is Negative	Level of liquid in Vessel will fall on heating
$γ_{real} = γ_{vessel} (= 3 α)$ $V_{a} pp = 0$	$V_{a} pp = 0$	Level of liquid in Vessel will remain same

10.0Solved Example

Q-1.If volume of metal increases by 0.12% on increasing temperature by 20°C. Find ?

Solution:

$γ = \frac{Δ V}{V Δ T}$

$\frac{Δ V}{V} = \frac{0.12}{100} = 12 \times 1 0^{- 4}$

$3 α = \frac{12 \times 1 0 ^{- 4}}{20}$

$α = \frac{1}{5} \times 1 0^{- 4} = 0.2 \times 1 0^{- 4}$

Q-2.Iron rod of 50 cm is heated from20°C to 100°C. Find length of rod at 100°C if

$α_{iron} = 12 \times 1 0^{- 6},^{\circ} C^{- 1}$

Solution:

$l_{2} = l_{1} [1 + α (T_{2} - T_{1})]$

$l_{2} = 50 [1 + 12 \times 1 0^{- 6} \times 80]$

$l_{2} = 50 [1 + 96 \times 1 0^{- 5}] = 50 + 48 \times 1 0^{- 3}$

$l_{2} = 50.048 cm$

Q-3.The difference between lengths of a certain brass rod and of a steel rod is claimed to be constant at all temperatures. Is this possible ?

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 °C.

$L_{B}^{'} = L_{B} (1 + α_{B} Δ θ)$

and

$L_{S}^{'} = L_{S} (1 + α_{S} Δ θ)$

So that

$L_{B}^{'} - L_{S}^{'} = (L_{B} - L_{S}) + (L_{B} α_{B} - L_{S} α_{S}) Δ θ$

So,

$For (L_{B}^{'} - L_{S}^{'}) = (L_{B} - L_{S}) at all temperatures,$

if, $L_{B} α_{B} - L_{S} α_{S} = 0$ as ≠0

Or

$\frac{L _{B}}{L _{S}} = \frac{α _{S}}{α _{B}}$ .

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.

Q-4.A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at 20°C, how fast or slow will it go in 24 hours at40°C? Coefficient of linear expansion of $α_{iron} = 1.2 \times 1 0^{- 6}$ , $^{\circ} C^{- 1}$

Solution:The time difference occurred in 24 hours (86400 seconds) is given by

$Δ t = \frac{1}{2} \times 1.2 \times 1 0^{- 6} \times 20 \times 86400$

$Δ t = 1.04 sec$

This is loss of time as is greater than 0 . As the temperature increases, the time period also increases. Thus, the clock goes slow.

Q-5.A rectangular plate has a circular cavity as shown in the figure. If we increase its temperature then which dimension will increase in the following figure.

Solution:Distance between any two points on an object increases with increase in temperature. So, all dimensions a, b, c and d will increase.

Thermal Expansion

Thermal expansion refers to the tendency of matter to change in volume or shape when exposed to changes in temperature. When heated, most materials expand due to increased molecular movement, while cooling causes them to contract. This physical property is crucial in various fields such as engineering, construction, and manufacturing, where temperature fluctuations impact material performance. Understanding thermal expansion helps in designing structures and devices that can withstand temperature variations without damage.

1.0Potential Energy V/s Distance Curve

When matter is heated without a change in state, it usually expands.
According to atomic theory, this happens because the potential energy curve is asymmetric.

As temperature rises from $T_{1}$ to $T_{2}$

Atomic vibration amplitude increases.
Atomic energy rises from $E_{1}$ to $E_{2}$
Average distance between atoms increases from $r_{1}$ to $r_{2}$ (where $r_{2}$ > $r_{1}$ )
Increase in atomic spacing causes the material to expand (thermal expansion).
If the potential energy curve were symmetric, no expansion would occur despite heating.

2.0Linear Expansion

When the rod is heated, its increase in length $Δ L$ is proportional to its original length $L_{0}$ and change in temperature $Δ T$ where ΔT is in $^{\circ} C or K.$

$d L = α L_{0} d T ⟹ Δ L = α L_{0} Δ T$

$α L_{0} T << 1$

$α = \frac{Δ L}{L _{0} Δ T}$ where $α$ is called the coefficient of linear expansion whose unit is $^{\circ} C^{- 1} or K^{- 1} .$

$L = L_{0} (1 + α Δ T)$ where L is the length after heating the rod.

Variation of $α$ with temperature and distance

If varies with distance, for example $α = a x + b$

Then total expansion = $\int (a x + b) Δ T d x$

If varies with temperature, $α = f (T) Then Δ L = \int α L_{0} d T$

Note: Actually, thermal expansion is always 3-D expansion. When other two dimensions of an object are negligible with respect to one, then observations are significant only in one dimension and it is known as linear expansion.

3.0Superficial (Areal) Expansion

The increase in the surface area of a solid when it is heated without a change in state.

$A = A_{0} (1 + β Δ T)$

$A_{0} = l_{0}^{2} and A = l^{2}$

$(1 + β Δ T) = [l_{0} (1 + α Δ T)]^{2} ⟹ β \approx 2 α$

4.0Volume Expansion

The increase in the volume of a substance when it is heated without a change in state.

$V = V_{0} (1 + γ Δ T)$

$V = l^{3} and V_{0} = l_{0}^{3}$ so

$α : β : γ$ = 1 : 2 : 3

Unit of :: is $α, β is 1^{\circ} C or 1/ K or K^{- 1} .$

5.0Variation of Time Period of Pendulum Clocks

The clock’s time depends on the number of oscillations of its pendulum.
Each time the pendulum reaches an extreme position, the second hand advances by one second.
One full oscillation moves the second hand by two seconds.
A pendulum with a time period of 2 seconds is called a seconds pendulum.

Let $T = 2 π \frac{L _{0}}{g}$ at temperature $θ_{0}$ and T'=2lg $T^{'} = 2 π \frac{L}{g}$

$\frac{T ^{'}}{T} = \frac{L}{L _{0}} = \frac{L _{0} ( 1 + α Δ θ )}{L _{0}} = 1 + α Δ θ \approx 1 + \frac{1}{2} α Δ θ$

Therefore change (loss or gain) in time per unit time lapsed is $T = 12 \frac{T ^{'} - T}{T} = \frac{1}{2} α Δ θ$

Gain or loss in time in duration of 't' in

$Δ T = \frac{1}{2} α Δ θT$ if T is the correct time then

(a) $θ < θ_{0}, then T^{'} < T$ clock becomes fast and gain time

(b) $θ > θ_{0}, then T^{'} > T$ clock becomes slow and loose time

6.0Effect of Thermal Expansion on Measurements

Case (1): Only Object Expands

Scale does not expand.
Actual and measured length at temperature 2

$l_{2} = l_{1} [1 + α_{0} (θ_{2} - θ_{1})]$

$α_{0}$ = Linear expansion coefficient of object

Case (2): Only Scale (Measuring Instrument) Expands

Object length remains unchanged.
Measured value (MV) decreases due to scale expansion:

$M V = l_{1} [1 + α_{s} (θ_{2} - θ_{1})]$

$α_{s}$ = linear expansion coefficient of scale

Case (3): Both Object and Scale Expand

Measured value: $M V = l_{1} [1 + (α_{0} - α_{s}) (θ_{2} - θ_{1})]$
If $α_{0} > α_{s}, measured value > actual length.$
If $α_{0} < α_{s}, measured value < actual length.$

General Formula for Measured Value:

$Measured Value = Calibrated Value \times [1 + α (θ - θ_{c})]$

$α = (α_{0} - α_{s}), θ = measurement temperature, θ_{c} = calibration temperature$

$If θ > θ_{c}, true length$ < $scale reading$
$If θ < θ_{c}, true length$ > $scale reading$

7.0Thermal Stress And Thermal Strain

No stress occurs if a rod is free to expand or contract with temperature.
Thermal stress arises when expansion or contraction is restricted.
Tensile stress → when heated (expansion restricted)
Compressive stress → when cooled (contraction restricted)

Thermal Strain

$Strain = \frac{Δ L}{L _{0}} = α Δ T$

α = coefficient of linear expansion,ΔT = temperature change

L0 = original (natural) length at new temperature $L_{0} = original (natural) length at new temperature$

Note:Final and original lengths must be measured at the same temperature.

Example:Rod Fixed Between Two Rigid Walls

Initial length = $L_{0}$

Temperature increases byΔT , but ends can't move → strain develop

$St r ain = α Δ T$

Leads to thermal stress in the rod.

$Strain = \frac{Length of rod at new temperature - Natural length of the rod at new temperature}{Natural length of the rod at new temperature}$

$Strain = \frac{l _{0} - l _{0} ( 1 + α Δ θ )}{l _{0} ( 1 + α Δ θ )} = \frac{- l _{0} α Δ θ}{l _{0} ( 1 + α Δ θ )}$

$α$ is very small so strain = $- α δ θ$ (Negative sign represents that the length of the rod is less than the natural length that means is compressed by the ends)

8.0The Bimetallic Strip

A bimetallic strip is made by bonding two metals with different linear expansion coefficients.
Common metals used: $Brass: α = 19 \times 1 0^{- 6},^{\circ} C, Steel: α = 12 \times 1 0^{- 6},^{\circ} C$

When heated:

Brass expands more than steel.
The strip bends towards the steel side (brass on the outside of the curve).

When cooled:

Brass contracts more than steel.
The strip bends towards the brass side.

Note:Used in thermostats and temperature-sensing devices.

Mathematical Solution:

$(R + \frac{d}{2}) ϕ = L_{0} (1 + α_{1} Δ θ)$

$(R - \frac{d}{2}) ϕ = L_{0} (1 + α_{2} Δ θ)$

$\frac{R + \frac{d}{2}}{R - \frac{d}{2}} = \frac{1 + α _{1} Δ θ}{1 + α _{2} Δ θ}$

9.0Thermal Expansion in Liquids

Liquids do not exhibit linear or superficial expansion, only volume expansion.
Liquids are heated along with the container holding them.

On heating:

Initially, liquid level falls → container expands first.
Then, liquid level rises → liquid expands faster than the container.

In a volume vs. temperature graph:

PQ: Expansion of the vessel.
QR: Real expansion of the liquid.

Real volume expansion of liquid=Apparent expansion+Expansion of vessel Real Volume Expansion of Liquid = Apparent Expansion + Expansion of Vessel

Co–Efficient of Real Expansion $r (γ_{r})$

It is due to the actual increase in volume of liquid due to heating.

$γ_{r} = \frac{Real increase in volume}{Initial volume \times Δ θ}$

$= \frac{Δ V}{V \times Δ θ}$

$γ_{real} = γ_{apparent} + γ_{vessel}$
Change (apparent change) in volume in liquid relative to vessel is

$Δ V_{app} = V (γ_{real} - γ_{vessel}) Δ θ$

$α =$ Coefficient of linear expansion of the vessel

Different level of liquid in vessel

\gamma	V\Delta V	Level
$γ_{real} > γ_{vessel} (= 3 α), γ_{app} > 0$	$V_{a} pp$ is Positive	Level of liquid in Vessel will rise on heating
$γ_{real} < γ_{vessel} (= 3 α), γ_{app} < 0$	$V_{a} pp$ is Negative	Level of liquid in Vessel will fall on heating
$γ_{real} = γ_{vessel} (= 3 α)$ $V_{a} pp = 0$	$V_{a} pp = 0$	Level of liquid in Vessel will remain same

10.0Solved Example

Q-1.If volume of metal increases by 0.12% on increasing temperature by 20°C. Find ?

Solution:

$γ = \frac{Δ V}{V Δ T}$

$\frac{Δ V}{V} = \frac{0.12}{100} = 12 \times 1 0^{- 4}$

$3 α = \frac{12 \times 1 0 ^{- 4}}{20}$

$α = \frac{1}{5} \times 1 0^{- 4} = 0.2 \times 1 0^{- 4}$

Q-2.Iron rod of 50 cm is heated from20°C to 100°C. Find length of rod at 100°C if

$α_{iron} = 12 \times 1 0^{- 6},^{\circ} C^{- 1}$

Solution:

$l_{2} = l_{1} [1 + α (T_{2} - T_{1})]$

$l_{2} = 50 [1 + 12 \times 1 0^{- 6} \times 80]$

$l_{2} = 50 [1 + 96 \times 1 0^{- 5}] = 50 + 48 \times 1 0^{- 3}$

$l_{2} = 50.048 cm$

Q-3.The difference between lengths of a certain brass rod and of a steel rod is claimed to be constant at all temperatures. Is this possible ?

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 °C.

$L_{B}^{'} = L_{B} (1 + α_{B} Δ θ)$

and

$L_{S}^{'} = L_{S} (1 + α_{S} Δ θ)$

So that

$L_{B}^{'} - L_{S}^{'} = (L_{B} - L_{S}) + (L_{B} α_{B} - L_{S} α_{S}) Δ θ$

So,

$For (L_{B}^{'} - L_{S}^{'}) = (L_{B} - L_{S}) at all temperatures,$

if, $L_{B} α_{B} - L_{S} α_{S} = 0$ as ≠0

Or

$\frac{L _{B}}{L _{S}} = \frac{α _{S}}{α _{B}}$ .

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.

Q-4.A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at 20°C, how fast or slow will it go in 24 hours at40°C? Coefficient of linear expansion of $α_{iron} = 1.2 \times 1 0^{- 6}$ , $^{\circ} C^{- 1}$

Solution:The time difference occurred in 24 hours (86400 seconds) is given by

$Δ t = \frac{1}{2} \times 1.2 \times 1 0^{- 6} \times 20 \times 86400$

$Δ t = 1.04 sec$

This is loss of time as is greater than 0 . As the temperature increases, the time period also increases. Thus, the clock goes slow.

Q-5.A rectangular plate has a circular cavity as shown in the figure. If we increase its temperature then which dimension will increase in the following figure.

Solution:Distance between any two points on an object increases with increase in temperature. So, all dimensions a, b, c and d will increase.

Frequently Asked Questions

What is the fundamental atomic reason why most matter expands when heated?

How are the different types of expansion (length, area, and volume) related for a solid?

How does thermal expansion affect a pendulum clock, and will it run fast or slow when heated?

What causes thermal stress in a solid rod, and when is this stress created?

When a liquid is heated in a vessel, why do we need to distinguish between real and apparent volume expansion?

Join ALLEN!

Thermal Expansion

1.0Potential Energy V/s Distance Curve

2.0Linear Expansion

3.0Superficial (Areal) Expansion

4.0Volume Expansion

5.0Variation of Time Period of Pendulum Clocks

6.0Effect of Thermal Expansion on Measurements

7.0Thermal Stress And Thermal Strain

8.0The Bimetallic Strip

9.0Thermal Expansion in Liquids

10.0Solved Example

Table of Contents

Frequently Asked Questions

What is the fundamental atomic reason why most matter expands when heated?

How are the different types of expansion (length, area, and volume) related for a solid?

How does thermal expansion affect a pendulum clock, and will it run fast or slow when heated?

What causes thermal stress in a solid rod, and when is this stress created?

When a liquid is heated in a vessel, why do we need to distinguish between real and apparent volume expansion?

Join ALLEN!

Thermal Expansion

1.0Potential Energy V/s Distance Curve

2.0Linear Expansion

3.0Superficial (Areal) Expansion

4.0Volume Expansion

5.0Variation of Time Period of Pendulum Clocks

6.0Effect of Thermal Expansion on Measurements

7.0Thermal Stress And Thermal Strain

8.0The Bimetallic Strip

9.0Thermal Expansion in Liquids

10.0Solved Example

Table of Contents