Thermal expansion — linear, area, volume with coefficient relationships

Question

A metal rod of length 1 m is heated from 20 degrees C to 120 degrees C. If the coefficient of linear expansion $\alpha = 1.2 \times 10^{-5} \text{ /°C}$ , find the change in length, area, and volume. Also derive the relationship between $\alpha$ , $\beta$ (area), and $\gamma$ (volume) coefficients.

Thermal Expansion Type Selection

flowchart TD
    A["Object heated — temperature increases by ΔT"] --> B["What dimension changes?"]
    B --> C["Length only — Linear Expansion"]
    B --> D["Area — Superficial Expansion"]
    B --> E["Volume — Cubical Expansion"]
    C --> F["ΔL = αLΔT"]
    D --> G["ΔA = βAΔT"]
    E --> H["ΔV = γVΔT"]
    F --> I["α = coefficient of linear expansion"]
    G --> J["β = 2α"]
    H --> K["γ = 3α"]
    I --> L["Relationship: α : β : γ = 1 : 2 : 3"]

Solution — Step by Step

When a rod is heated, its length increases proportionally:

\Delta L = \alpha L_0 \Delta T

where $L_0$ is the original length, $\alpha$ is the coefficient of linear expansion, and $\Delta T$ is the temperature change.

For our problem:

\Delta L = 1.2 \times 10^{-5} \times 1 \times (120 - 20) = 1.2 \times 10^{-5} \times 100

\boxed{\Delta L = 1.2 \times 10^{-3} \text{ m} = 1.2 \text{ mm}}

The new length: $L = L_0(1 + \alpha \Delta T) = 1.0012 \text{ m}$ .

For a flat surface (like a metal sheet), the area expansion coefficient $\beta$ is:

\Delta A = \beta A_0 \Delta T

Deriving $\beta = 2\alpha$ :

Consider a square plate of side $L$ . After heating:

L' = L(1 + \alpha \Delta T)

New area:

A' = (L')^2 = L^2(1 + \alpha \Delta T)^2 = A_0(1 + 2\alpha \Delta T + \alpha^2 \Delta T^2)

Since $\alpha \Delta T$ is very small (order of $10^{-3}$ ), the $\alpha^2 \Delta T^2$ term is negligible:

A' \approx A_0(1 + 2\alpha \Delta T)

So $\beta = 2\alpha$ .

For our rod (assuming initial area $A_0$ ):

\beta = 2 \times 1.2 \times 10^{-5} = 2.4 \times 10^{-5} \text{ /°C}

Similarly, for volume expansion:

\Delta V = \gamma V_0 \Delta T

Deriving $\gamma = 3\alpha$ :

For a cube of side $L$ :

V' = (L')^3 = L^3(1 + \alpha \Delta T)^3 \approx V_0(1 + 3\alpha \Delta T)

(ignoring higher-order terms)

So $\gamma = 3\alpha$ .

\gamma = 3 \times 1.2 \times 10^{-5} = 3.6 \times 10^{-5} \text{ /°C}

\alpha : \beta : \gamma = 1 : 2 : 3

This means if you know any one coefficient, you can find the other two. This relationship holds for isotropic materials (same properties in all directions).

\Delta L = \alpha L_0 \Delta T \qquad (\text{linear})

\Delta A = \beta A_0 \Delta T \qquad \beta = 2\alpha \quad (\text{superficial})

\Delta V = \gamma V_0 \Delta T \qquad \gamma = 3\alpha \quad (\text{cubical})

Why This Works

Thermal expansion occurs because heating increases the average vibrational amplitude of atoms in the lattice. The interatomic potential is asymmetric — it is steeper on the compression side and shallower on the stretching side. So as vibrations increase, the average position shifts outward, increasing the equilibrium distance between atoms. This is why most materials expand when heated.

For JEE/NEET, the most frequently tested application is the expansion of a hole. When a metal plate with a hole is heated, the hole gets larger, not smaller. Think of it as the entire plate scaling up uniformly — every dimension increases, including the hole diameter. Students intuitively expect the hole to shrink, but that is wrong.

Alternative Method — Apparent Expansion of Liquids

Liquids do not have a definite shape, so we can only measure cubical expansion ( $\gamma$ ). When a liquid is heated in a container, the container also expands. The apparent expansion is what we measure:

\gamma_{\text{apparent}} = \gamma_{\text{liquid}} - \gamma_{\text{container}}

The real expansion of the liquid is:

\gamma_{\text{real}} = \gamma_{\text{apparent}} + \gamma_{\text{container}}

Common Mistake

Students forget that the $1:2:3$ relationship between $\alpha$ , $\beta$ , and $\gamma$ is an approximation valid only when $\alpha \Delta T \ll 1$ (which is true for most practical cases). In problems with extremely large temperature changes, the higher-order terms ( $\alpha^2$ , $\alpha^3$ ) may not be negligible. However, for JEE and NEET, the approximation always holds — just use $\beta = 2\alpha$ and $\gamma = 3\alpha$ directly.