Thermal Expansion
When heat is absorbed by different substances, the internal energy of the substance will increase. This makes the molecules move faster, meaning the kinetic energy of the substance increases, and tend to move farther apart from each other. This phenomenon is known as the thermal expansion.
Change in size and phase change of a substance are the commonly observed effects of heat or temperature increase. The following equation represents the change in a solid substance when exposed to increase in temperature.
Linear Expansion
\(\mathbf{\Delta L = \alpha L_0 \Delta T}\) 
\(\Delta L\) = increase in length \(\alpha\) = coefficient of linear expansion \(L_0\) = initial length \(\Delta T\) = increase in temperature 
Area Expansion
\(\mathbf{\Delta A = \gamma A_0 \Delta T}\) 
\(\Delta A\) = increase in area \(\gamma\) = coefficient of area expansion \(A_0\) = initial area \(\Delta T\) = increase in temperature 
Volume Expansion
\(\mathbf{\Delta V = \beta V_0 \Delta T}\)

\(\Delta V\) = increase in volume \(\beta\) = coefficient f volume expansion \(V_0\) = initial volume \(\Delta T\) = increase in temperature 
The coefficient of expansion is a constant with a unit of per ºC(/ºC) or per K (/K). The coefficient of volume expansion and the coefficient of linear expansion is related as \(\beta=3 \alpha\).
Example 1.
An aluminum bar, 65cm long at 20 ºC is heated at 50 ºC. What is the change in length of the bar? The coefficient of expansion of the aluminum is 25 x 10^{6}/ºC.
Solution:
To solve for the change in length of the bar, we use the formula for linear expansion which is \(\mathbf{\Delta L = \alpha L_0 \Delta T}\). Substitute the given values to the equation, we have
\(\Delta L = (25 \times 10^{6}/^\circ C)(65\;cm)(50^\circ C  20^\circ C)=0.049\;cm\)
Thus, the change in length of the aluminum bar is 0.049 cm.
Example 2.
A rod made of lead has a length of 4.5 m at 10 ºC. After an increase in the temperature of the rod, the rod lengthens by 0.00075 m. What is the rod's final temperature? The coeffiecient of expansion of lead is 29 x 10^{6}/ºC.
Given:
\(L_0\) = 4.5 m
\(T_i\) = 10 ºC
\(\Delta L\) = 0.00075 m
\(\alpha\) = 29 x 10^{6}/ºC
Solution:
Using the formula for linear expansion \(\mathbf{\Delta L = \alpha L_0 \Delta T}\) and dividing both sides by \(\alpha L_0\) to eleminate the expression on the left side of the equations and be left with \(\Delta T\), we have
\(\mathbf{\Delta T={\Delta L \over \alpha L_0}}\)
and rewriting the temperature in terms of the intitial and final temperature, the equation above becomes
\(\mathbf{T_f  T_i={\Delta L \over \alpha L_0}}\).
Add \(T_i\) to both sides of the equation to cancel it on the right side, so what is left is the variable for final temperature, \(T_f\).
\(\mathbf{T_f ={\Delta L \over \alpha L_0}+T_i}\)
Substitute all the given values to solve for the final temperature of the rod,
\(T_f ={0.00075\;m \over (29 \times 10^{6}/^\circ C)(4.5\;m)}+10^\circ C\\ \quad={0.00075\;m \over 0.0001305\;m/^\circ C}+10^\circ C\\ \quad=5.75^\circ C + 10^\circ C\\ \quad=15.75^\circ C \)
The final temperature of the lead rod is 15.75 ºC.