Despite the difference between heat and temperature, adding heat in a body increases its temperature and removing heat decreases its temperature. You may have noticed that every material expand when heated (added heat to it) other than the anomalous behaviour of some materials such as water discussed below. Neglecting other materials which contract on heating, as the temperature of a material increases, the dimensions of the material increase. You may notice that a tight lid can be loosen by pouring hot water on it.
What is linear expansion?
Linear expansion of a material is its expansion along one dimension. We consider an example of a rod and find the change in length of the rod due to expansion. The initial length of the rod is \(L_i\) and the initial temperature is \(T_i\). As the temperature of the rod increases, its final temperature is \(T_f\) and final length is \(L_f\) (\(T_f>T_i\) and \(L_f>L_i\)). The rod undergoes linear expansion along its length. It has been experimentally found that the change in length \(ΔL=L_f−L_i\) is directly proportional to the initial length \(L_i\) and the change in temperature \(ΔT=T_f−T_i\) that is, \(\Delta L \propto {L_i}{\kern 1pt} \Delta T\) and
\[\Delta L = \alpha {L_i}{\kern 1pt} \Delta T \tag{1} \label{1}\]
where \(\alpha\) is the proportionality constant called coefficient of linear expansion. The above equation is valid only when the temperature change is not too high otherwise \(\alpha\) is not constant.
What is volume expansion?
Volume expansion is similar to linear expansion but in this case whole volume of the material expands not just in one dimension. The material we have undergoes a change in temperate \(\Delta T\) and its volume increases. So, we consider initial volume \(V_i\), initial temperature \(T_i\), final volume \(V_f\) and final temperate \(T_f\). Similar to the linear expansion above, the change in volume \(\Delta V = V_f - V_i\) due to the change in temperature \(\Delta T = T_f - T_i\) is directly proportional to the initial temperature \(V_i\) and the change in temperature \(\Delta T\) that is, \(\Delta V \propto {V_i}{\kern 1pt} \Delta T\) and
\[\Delta V = \beta {V_i}{\kern 1pt} \Delta T \tag{2} \label{2}\]
where \(\beta \) is the proportionality constant called coefficient of volume expansion and it is constant when the temperature change is not too high.
For solids the coefficient of volume expansion \(\beta\) is related to the coefficient of linear expansion by \(\beta = 3\alpha \). Can you prove this? Consider a cube of initial length \(l\) whose volume \(V\) is \(l^3\). Then, \(dV = 3{l^2}dl\). You know from Eq. \eqref{1} that \(dl = \alpha l{\kern 1pt} dT\), so you can write \(dV = 3{l^2}\alpha l{\kern 1pt} dT = 3{l^3}\alpha {\kern 1pt} dT = 3V\alpha {\kern 1pt} dT\):
\[\frac{dV}{{VdT}} = 3\alpha \]
From Eq. \eqref{2}, \(\beta = \frac{{dV}}{{VdT}}\) and hence
\[\beta = 3\alpha \tag{3} \label{3}\]
