The tendency of matter to change its shape, area, volume, and density with a change in temperature is called Thermal Expansion. When most substances are heated, their molecules begin to vibrate and move creating more distance between themselves. With increase in energy the particles start moving faster and the inter-molecular forces between them weakens. Thermal expansion is expressed as the relative or fractional change in length or volume per unit change in temperature. The relative expansion or strain divided by the change in temperature is called the material’s coefficient of linear thermal expansion. Read on to learn more about Thermal Expansion and see solved example questions.

Thermal Expansion: Effect of Heat

Why does the average separation between the molecules increases with the supply of heat energy? To understand this, let us observe matter at the atomic level. From the figure given below, we can observe that the mean distance between the atoms increases with the increase in potential energy when heat is supplied. When the temperature increases, this causes an increase in the potential energy of atoms and an increase in mean distance.

What are the Three Types of Thermal Expansion?

Based on the geometry of the material, there are three types of expansion.

1. Linear Expansion:- When the expansion due to heating occurs only along one direction, the expansion is said to be one dimensional or linear expansion.

Suppose that the temperature of a thin rod of length $l$ changed from $T$ to $T+∆T.$ It is found experimentally that, for small $∆T,$ the change in length $∆l$ of the rod is directly proportional to $∆T$ and length $l.$

$\mathrm{\Delta }l\propto \mathrm{\Delta }T..\dots (1)$

$\mathrm{\Delta }l\propto l\dots ..(2)$

From $(1)$ and $(2),$   

$\mathrm{\Delta }l\propto l\mathrm{\Delta }T$

$\Rightarrow \mathrm{\Delta }l=\alpha l\mathrm{\Delta }T$

Here $\alpha$ is the proportionality constant. It is different for different materials and is called the coefficient of linear expansion. Its $\mathrm{S}\mathrm{I}$ unit is ${}^{\mathrm{o}}{\mathrm{C}}^{–1}$or ${\mathrm{K}}^{–1}.$

2. Superficial Expansion:- When there is any change in the area of a body due to heating, the expansion is called areal or superficial expansion. Change in area $∆A$ varies in the same manner as in the case of linear expansion and is given by,

$∆A=\beta A∆T$

Here $\beta$ is the coefficient of superficial expansion. 

Relation between the coefficient of linear expansion and superficial expansion:- When the temperature of the square plate of isotropic material of side length $l$ is increased by $dT.$ Then the area will be increased by an amount $dA$ is given by,

$\mathrm{d}A=\left(\frac{dA}{dl}\right)\times \mathrm{d}l=2l\times \mathrm{d}l\left(A=l^2\right)$

$\Rightarrow \mathrm{d}A=2l\times \mathrm{d}l=2l^2\alpha \mathrm{d}T(\mathrm{d}l=l\alpha \mathrm{d}T)$

$\Rightarrow \mathrm{d}A=2\alpha A\mathrm{d}T=\beta A\mathrm{d}T$

Thus $\beta =2\alpha$

3. Volumetric Expansion:- Just like linear expansion, the change in volume $∆V$ is proportional to volume $V$ and $∆T.$

$\mathrm{\Delta }V\propto \mathrm{\Delta }T$ and $\mathrm{\Delta }V\propto V$

Introducing a proportionality constant $\gamma ,$ we may write $∆V$ as,

$\mathrm{\Delta }V=\gamma V\mathrm{\Delta }T$

Here $\gamma$ is called the coefficient of volume expansion.   

Relation between $\gamma$ and $\alpha$:- When the temperature of a cube of isotropic material of side length $l$ is increased by $dT.$ Then the volume will be increased by an amount $dV$ is given by,

$\text{d}V=\left(\frac{\text{d}V}{\text{d}l}\right)\times \text{d}l=3l\times \text{d}l\left(V=l^{\mathbf{3}}\right)$

$\Rightarrow \mathrm{d}V=3l^2\times \mathrm{d}l=3l^3\alpha \mathrm{d}T(\mathrm{d}l=l\alpha \mathrm{d}T)$

$\Rightarrow \mathrm{d}V=3\alpha V\mathrm{d}T=\gamma V\mathrm{d}T$

Thus, $\gamma =3\alpha$

The Anomalous Expansion of Water:- Almost all the material expands when they are heated. On the other hand, some substances contract when they are heated over a specific temperature range. The most common example is water. From the below graph, we can observe that the density of water is maximum at $4^{\mathrm{o}}\mathrm{C}.$ So, for a fixed mass of water at $0^{\mathrm{o}}\mathrm{C},$ the density will increase with the increase in temperature up to $4^{\mathrm{o}}\mathrm{C}.$ For fixed mass, the density of water will be inversely proportional to volume. Hence the volume of water will decrease with an increase in temperature from $0^{\mathrm{o}}\mathrm{C}$ to $4^{\mathrm{o}}\mathrm{C}.$

Thermal Stress and Strain

When the temperature of a rod of length $l$ will change by $∆T,$ then the change in length $∆l$ of the rod will be given by,

$∆l=\alpha l∆T$

Now the thermal Strain $\left({\epsilon }_T\right)=\frac{\mathrm{C}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}}{\mathrm{O}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}}=\frac{\alpha l\mathrm{\Delta }T}{l}$

$\Rightarrow {\epsilon }_T=\alpha \mathrm{\Delta }T$

When we prevent the thermal expansion of the rod by fixing both ends, the rod acquires a compressive strain due to external forces provided by the rigid support at the ends. The corresponding stress developed in the rod due to the external force is called thermal stress. It is given by,

$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{S}\mathrm{t}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\left({\sigma }_T\right)=\mathrm{Y}\mathrm{o}\mathrm{u}\mathrm{n}{\mathrm{g}}^{\prime }\mathrm{s}\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{l}\mathrm{u}\mathrm{s}(\mathrm{E})\times \mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{S}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{i}\mathrm{n}$

$\Rightarrow {\sigma }_T=E{\epsilon }_T=E\alpha \mathrm{\Delta }T$

When the temperature increases, the stress will be compressive. When the temperature decreases, the stress will be tensile.

Effect of Rise in Temperature on Thermal Expansion Coefficient

The thermal expansion coefficient depends slightly on the temperature, but its variation is small enough to be negligible, even over a temperature range of ${100}^{\mathrm{o}}\mathrm{C}.$ So, we treat $\alpha$ as constant.

Effects of Thermal Expansion on Density

As most of the material expands when the temperature of the material increased, but the mass remains constant. So, the density of most materials decreased during thermal expansion.
Final density,

${\rho }_{\mathrm{F}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}}={\rho }_{\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}(1–\gamma \mathrm{\Delta }T)$

Application of Thermal Expansion

We see many applications of thermal expansion in our daily life. Some of them are given below:

1. Hot balloons – You may have observed hot balloons on beaches. When the air inside the balloons is heated, it expands. Due to this, it displaces more air, and more upthrust acts on it.
2. Rail Joint – Small space is left when two rails are joined together. When the train passes through the track, it gets heated due to friction between the wheel of the train and the rail track. So, due to the increase in temperature of the track, the track expands, and its length increase. This increase in length can distort the rail track if a gap is not provided.
3. Thermostats – These are used in the electrical circuit to avoid overload. Thermostats consist of a bimetallic strip of materials having different thermal expansion coefficients. When the temperature rise, both the materials expand in different amounts. This causes bending of the strip and helps in breaking the circuit.

Sample Problems on Thermal Expansion

Q.1. A steel ruler of length $20\mathrm{c}\mathrm{m}$ is calibrated to give a correct reading at ${20}^{\mathrm{o}}\mathrm{C}.$
(a) Will it give a too short or too long reading at a lower temperature?
(b) Find the actual length of the rod at ${40}^{\circ }\text{C}$. $\alpha =1.2\times {{10}^{–5}}^{\circ }{\text{C}}^{–1}.$
Ans: (a) If the temperature of the rod will decrease, the length of the rod will also decrease due to thermal contraction. So, at the lower temperature, the centimeter division will be shorter than $1\mathrm{c}\mathrm{m}.$ So, the ruler will give too long readings.
(b) The final temperature of the ruler is ${40}^{\mathrm{o}}\mathrm{C}.$ So, the change in temperature $∆T$ will be,
 $∆T=$Final temperature $–$ Initial temperature
$\Rightarrow \mathrm{\Delta }T={40}^{\mathrm{o}}\mathrm{C}–{20}^{\mathrm{o}}\mathrm{C}={20}^{\mathrm{o}}\mathrm{C}$
Now the final length will be,
The final length $=$ Initial length $+$ Change in length
$\Rightarrow$ The final length$=l+\mathrm{\Delta }l=l(1+\alpha \mathrm{\Delta }T)$
$\Rightarrow l_f=20\left(1+1.2\times {10}^{–5}\times {20}^{\mathrm{o}}\mathrm{C}\right)$
$\Rightarrow l_f=20.0048\mathrm{c}\mathrm{m}$

Summary

When the temperature of the material increase, the randomness of the atoms or molecules of the material increases. After gaining the heat energy, the potential energy of the atoms increases. It causes an increase in the mean distance between the atoms of the material. As a result, the expansion occurs in the material. However, water shrinks when it is heated from $0^{\mathrm{o}}\mathrm{C}$ to $4^{\mathrm{o}}\mathrm{C}.$ It happens because of the highest density of water at $4^{\mathrm{o}}\mathrm{C}.$
There are three types of expansion,

Linear Expansion:- The expansion will be in only one direction.
Change in length, $\mathrm{\Delta }l=\alpha l\mathrm{\Delta }T.$

Superficial Expansion:- The expansion will be in two directions.
Change in area, $\mathrm{\Delta }A=\beta A\mathrm{\Delta }T=2\alpha A\mathrm{\Delta }T$

Volumetric Expansion:- When the expansion occurs in all three directions, it is called volumetric expansion.
Change in volume, $\mathrm{\Delta }V=\gamma V\mathrm{\Delta }T=3\alpha A\mathrm{\Delta }T$  

When the rod is fixed at both ends, thermal stress is developed in the rod. It is given by,

Thermal stress, ${\sigma }_T=E{\epsilon }_T=E\alpha \mathrm{\Delta }T$

FAQs on Thermal Expansion

The most frequently asked questions on thermal expansion and their answers are given below:

Q.1. What will be the effect of heat supply on the volume of one liter of water at zero degree celcious? Ans: Unlike most materials, the water will contract when heated between zero degree celcious and four degree celcious. The density of water is highest at four degree celcious. So, the volume of water will decrease up to four degree celcious and then it will start increasing.

Q.2. What will be the value of thermal stress when a rod of length l is heated, having one end of the rod to expand? Ans: Since one end of the rod is to expand; the resistive force will not develop. Thus, the value of thermal stress will be zero in the rod.

Q.3. A pendulum watch is made in winter. Will it be faster or slower in summer? Ans: We know that the Time period of a pendulum is directly proportional to the square root of the length of the rod of the pendulum. So, when the temperature increases in the summer, the length of the rod will also increase. It will cause an increase in the time period of the pendulum. Thus, the pendulum watch will lose time and will become slower in summer.

Q.4. What are the disadvantages of thermal expansion? Ans: 1. Due to thermal expansion, the shape, and dimensions of the measuring instrument change. This can cause errors in measurement. 2. It causes thermal stress. This may lead to crack in fixed components. 3. Bursting of a tire occurs in summer due to thermal expansion of air inside the tube.

Q.5. What is caused by the thermal expansion of a substance, when heated? Ans: When a substance is heated, the molecules of substance start oscillating with greater amplitudes. As a result, the average separation between them increases. When the separation between the molecules increases, the material expands.

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