Thermal expansion refers to the increase in the dimensions of a material when it is heated. This occurs because particles in a substance move more energetically at higher temperatures, causing them to occupy more space. Thermal expansion can occur in length, area, or volume depending on the object's shape and constraints.
Linear Expansion
Linear expansion is the increase in length of a solid object due to an increase in temperature. It is calculated by:
\[
\Delta L = \alpha L_0 \Delta T
\]
where:
\( \Delta L \) = change in length
\( \alpha \) = coefficient of linear expansion
\( L_0 \) = original length
\( \Delta T \) = temperature change
Area Expansion
Area expansion is the increase in surface area of a material as it heats. It is approximately given by:
\[
\Delta A = 2\alpha A_0 \Delta T
\]
where:
\( \Delta A \) = change in area
\( \alpha \) = coefficient of linear expansion
\( A_0 \) = original area
\( \Delta T \) = temperature change
Note: The factor 2 comes from the fact that area is a two-dimensional quantity.
Volume Expansion
Volume expansion applies to solids, liquids, and gases and describes the increase in volume with temperature. The formula is:
\[
\Delta V = \beta V_0 \Delta T
\]
where:
\( \Delta V \) = change in volume
\( \beta \) = coefficient of volume expansion (approximately \( 3\alpha \) for solids)
\( V_0 \) = original volume
\( \Delta T \) = temperature change
Problem:
A steel rod of length 1.5 m is heated from 20°C to 80°C. If the coefficient of linear expansion of steel is \( 1.2 \times 10^{-5} \, \text{°C}^{-1} \), find the increase in length.
Solution:
Using: \( \Delta L = \alpha L_0 \Delta T \)
\( \alpha = 1.2 \times 10^{-5} \)
\( L_0 = 1.5 \, \text{m} \)
\( \Delta T = 80 - 20 = 60 \, \text{°C} \)
\[
\Delta L = (1.2 \times 10^{-5}) \times 1.5 \times 60 = 0.00108 \, \text{m} = 1.08 \, \text{mm}
\]
So, the rod expands by 1.08 mm.
Problem:
A metal cube has an initial volume of 0.02 m³ and is heated by 100°C. The coefficient of volume expansion is \( 3.6 \times 10^{-5} \, \text{°C}^{-1} \). Find the change in volume.
Solution:
Using: \( \Delta V = \beta V_0 \Delta T \)
\( \beta = 3.6 \times 10^{-5} \)
\( V_0 = 0.01 \, \text{m}^3 \)
\( \Delta T = 100 \, \text{°C} \)
\[
\Delta V = 3.6 \times 10^{-5} \times 0.02 \times 100 = 0.000072 \, \text{m}^3
\]
So, the volume increases by 7.2 × 10⁻⁵ m³.