You may have observed that sometimes sealed bottles with metallic lids are so tightly screwed that one has to put the lid in hot water for some time to open it. This would allow the metallic lid to expand, thereby loosening it to unscrew easily. In the case of liquids, you may have observed that mercury in a thermometer rises, when the thermometer is put in slightly warm water. If we take out the thermometer from the warm water the level of mercury falls again. Similarly, in the case of gases, a balloon partially inflated in a cool room may expand to full size when placed in warm water. On the other hand, a fully inflated balloon when immersed in cold water would start shrinking due to the contraction of the air inside.

What is Thermal Expansion?

It is our common experience that most substances expand on heating and contract on cooling. A change in the temperature of a body causes a change in its dimensions. The increase in the dimensions of a body due to the increase in its temperature is called thermal expansion.

There are three types of expansions(shown in Fig. 11.5) that can take place in solids:

• Linear Expansion: The expansion in length of a body due to an increase in its temperature is called linear expansion.
• Area Expansion: The expansion in area of a body due to an increase in its temperature is called area expansion.
• Volume Expansion: The expansion in volume of a body due to an increase in its temperature is called volume expansion.

Linear Expansion

Linear expansion refers to the increment in the length of a solid on heating. If the substance is in the form of a long rod, a little variation in temperature, \(\Delta T\) will cause it to deform, the fractional change in length, \(\frac{\Delta l}{l}\), is directly proportional to \(\Delta T\).

\(\frac{\Delta l}{l}=\alpha_1 \Delta T \dots(11.4)\)

where \(\alpha_1\) is known as the coefficient of linear expansion (or linear expansivity) and is characteristic of the material of the rod. The unit of \({\alpha_1}\) is per degree Celsius \(\left({ }^{\circ} \mathrm{C}^{-1}\right)\) in the CGS and per Kelvin \(\left(\mathrm{K}^{-1}\right)\) in the SI system.

In Table \(11.1\), typical average values of the coefficient of linear expansion for some material in the temperature range \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) are given. From this Table, compare the value of \(\alpha_1\) for glass and copper. We find that copper expands about five times more than glass for the same rise in temperature. Normally, metals expand more and have relatively high values of \(\alpha_1\).

Area Expansion

Area expansion refers to the increment in the surface area of the substance on heating. There is a little variation in temperature, \(\Delta T\) will cause it to deform, the fractional change in surface area, \({\Delta A / A}\), is directly proportional to \(\Delta T\). The expression for the Area expansion is:

\(
\frac{\Delta A}{A}=\alpha_A \cdot \Delta T
\)

Where \(\alpha_A\) is the coefficient of area expansion of the given solid.

Volume Expansion

Similarly, we consider the fractional change in volume, \(\frac{\Delta V}{V}\), of a substance for temperature change \(\Delta T\) and define the coefficient of volume expansion (or volume expansivity), \(\alpha_{V}\) as
\(
\alpha_{{v}}=\left(\frac{\Delta V}{V}\right) \frac{1}{\Delta T} \dots(11.5)
\)
Here \(\alpha_{{v}}\) is also a characteristic of the substance but is not strictly a constant. It depends in general on temperature (FIg 11.6). It is seen that \(\alpha_v\) becomes constant only at a high temperature.

Table \(11.2\) gives the values of coefficient of volume expansion of some common substances in the temperature range \(0-100^{\circ} \mathrm{C}\). You can see that thermal expansion of these substances (solids and liquids) is rather small, with materials, like pyrex glass and invar (a special iron-nickel alloy) having particularly low values of \(\alpha_v\). From this Table, we find that the value of \(\alpha_{\mathrm{v}}\) for alcohol (ethanol) is more than mercury and expands more than mercury for the same rise in temperature.

Anomalous Expansion of Water

Water exhibits an anomalous behaviour; it contracts on heating between \(0^{\circ} \mathrm{C}\) and \(4^{\circ} \mathrm{C}\). The volume of a given amount of water decreases as it is cooled from room temperature until its temperature reaches \(4^{\circ} \mathrm{C}\), [F1g. 11.7(a)]. Below \(4^{\circ} \mathrm{C}\), the volume increases, and therefore, the density decreases [Fig. 11.7(b)].

This means that water has the maximum density at \(4^{\circ} \mathrm{C}\). This property has an important environmental effect: bodies of water, such as lakes and ponds, freeze at the top first. As a lake cools toward \(4^{\circ} \mathrm{C}\), water near the surface loses energy to the atmosphere, becomes denser, and sinks; the warmer, less dense water near the bottom rises. However, once the colder water on top reaches temperature below \(4{ }^{\circ} \mathrm{C}\), it becomes less dense and remains at the surface, where it freezes. If water did not have this property, lakes, and ponds would freeze from the bottom up, which would destroy much of their animal and plant life.

Coefficient of volume expansion for an ideal gas

Gases, at ordinary temperature, expand more than solids and liquids. For liquids, the coefficient of volume expansion is relatively independent of the temperature. However, for gases it is dependent on temperature. For an ideal gas, the coefficient of volume expansion at constant pressure can be found from the ideal gas equation: \(P V=\mu R T\)
At constant pressure \(P \Delta V=\mu R \Delta T\)

\(\frac{\Delta V}{V}=\frac{\Delta T}{T}\)

1.e., \(\alpha_v=\frac{1}{T} \dots(11.6)\) for ideal gas

At \(0{ }^{\circ} \mathrm{C}, \alpha_v=3.7 \times 10^{-3} \mathrm{~K}^{-1}\), which is much larger than that for solids and liquids. Equation (11.6) shows the temperature dependence of \(\alpha_v\); it decreases with increasing temperature. For a gas at room temperature and constant pressure, \(\alpha\) is about \(3300 \times 10^{-6} \mathrm{~K}^{-1}\), as much as order(s) of magnitude larger than the coefficient of volume expansion of typical liquids.

Relation between \(\left(\alpha_v\right)\) and \(\left(\alpha_1\right)\)

There is a simple relation between the coefficient of volume expansion \(\left(\alpha_v\right)\) and the coefficient of linear expansion \(\left(\alpha_1\right)\). Imagine a cube of length, \(l\), that expands equally in all directions, when its temperature increases by \(\Delta T\). We have
\(\Delta l=\alpha_1 l \Delta T\)
so, \(\Delta V=(l+\Delta l)^3-l^3 \simeq 3 l^2 \Delta l \dots(11.7)\)
In Equation (11.7), terms in \((\Delta l)^2\) and \((\Delta l)^3\) have been neglected since \(\Delta l\) is small compared to \(l\).
So
\(
\Delta V=\frac{3 V \Delta l}{l}=3 V \alpha_1 \Delta T \dots(11.8)
\)
which gives
\(
\alpha_v=3 \alpha_1
\)

Example 1: Derive the relationship between the coefficient of area expansion \(\left(\alpha_A\right)\) and the coefficient of linear expansion \(\left(\alpha_l\right)\).

Solution:

Consider a cube with a length of I that expands evenly in all directions as its temperature rises by \(T\). The original area will be length square and the new area, after a temperature increase is,
\(
\begin{aligned}
& A+\Delta A=(l+\Delta l)^2 \\
& A+\Delta A=l^2+(\Delta l)^2+2 l \Delta l
\end{aligned}
\)
The terms in \((\Delta l)^2\) have been neglected since \(\Delta l\) is small compared to \(l\).
\(
\begin{aligned}
& A+\triangle A \approx l^2+2 l \Delta l \\
& \text { or } A+\triangle A=A+(2 A \Delta l) / l \\
& \text { or } \triangle A \approx 2 A \Delta l / l
\end{aligned}
\)
Rearrange the above equation,
or \(\Delta A / A \approx 2(\Delta l / l)\)
\(\operatorname{or}_A=2 \alpha_l\)
Therefore, the coefficient of area expansion is twice the coefficient of linear expansion.

Example 2: What will be the pressure that has to be applied to the ends of a steel wire of length \(10 \mathrm{~cm}\) to keep its length constant when its temperature is raised by \(100^{\circ} \mathrm{C}\). (For steel Young’s modulus is \(2 \times 10^{11} \mathrm{Nm}^{-2}\) and the coefficient of thermal expansion is \(1.1 \times 10^{-5} \mathrm{~K}^{-1}\) ).

Solution:

Given,
\(
\begin{aligned}
& \Delta T=100^{\circ} \mathrm{C}, \\
& Y=2 \times 1011 \mathrm{Nm}^{-2} \\
& \alpha=1.1 \times 10^{-5} \mathrm{~K}^{-1}
\end{aligned}
\)
The expression for thermal strain is
\(
\begin{aligned}
\Delta F / A & =Y(\Delta l / l) \\
& =Y \alpha \Delta T
\end{aligned}
\)
Thermal stress in a rod is the pressure due to the thermal strain.
Substitute the value in the above expression.
\(
\begin{gathered}
\text { Thermal strain }=\left(2 \times 10^{11} \mathrm{Nm}^{-2}\right) \times\left(1.1 \times 10^{-5} \mathrm{~K}^{-1}\right) \times\left(100^{\circ} \mathrm{C}\right) \\
=2.2 \times 10^8 \mathrm{~Pa}
\end{gathered}
\)

Example 3: Two wires are made of the same material and have the same volume. However, wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Dx on applying force \(F\), how much force is needed to stretch wire 2 by the same amount?

Solution:

Young’s modulus is the same for the same substance.
The area of the cross-section for wire 1 is \(A\) and wire 2 is \(3 A\).
The volume of wires 1 and 2 is
\(
\begin{aligned}
& V_1=V_2 \\
& \left(A \times l_1\right)=\left(3 A \times l_2\right) \\
& l_2=l_1 / 3 \\
& Y=(F / A) /(\Delta l / l) \\
& \text { or } F_1=Y A\left(\Delta l_1 / l_1\right) \\
& \text { Similarly, } \\
& F_2=Y 3 A\left(\Delta l_2 / l_2\right)
\end{aligned}
\)
Wire 2 is stretched by the same amount, therefore,
\(
\begin{aligned}
& \Delta l_1=\Delta l_2=x \\
& F_2=Y 3 A x /\left(l_1 / 3\right) \\
& F_2=9\left(Y A x / l_1\right) \\
& F_2=9 F_1
\end{aligned}
\)

Example 4: A blacksmith fixes an iron ring on the rim of the wooden wheel of a horse cart. The diameter of the rim and the iron ring are \(5.243 \mathrm{~m}\) and \(5.231 \mathrm{~m}\), respectively at \(27^{\circ} \mathrm{C}\). To what temperature should the ring be heated so as to fit the rim of the wheel?

Solution:

Given,
\(
\begin{gathered}
 \begin{array}{l}
T_1=27{ }^{\circ} \mathrm{C} \\
L_{\mathrm{T} 1}=5.231 \mathrm{~m} \\
L_{\mathrm{T} 2}=5.243 \mathrm{~m} \\
\text { So, } L_{\mathrm{T} 2}=L_{\mathrm{T} 1}\left[1+\alpha_1\left(T_2-T_1\right)\right]
\end{array} \\
5.243 \mathrm{~m}=5.231 \mathrm{~m}\left[1+1.20 \times 10^{-5} \mathrm{~K}^{-1}\left(T_2-27^{\circ} \mathrm{C}\right)\right] \\
\text { or } T_2=218^{\circ} \mathrm{C} .
\end{gathered}
\)