Coefficient of Linear Expansion : Definition, Unit and Solved Examples

With new discoveries and innovations constantly being made, the study of Physics Topics remains a vibrant and exciting field of research.

What are Three Expansion of Solids?

Introduction : Expansion of Solids

Usually all solid substances expand on heating and contract on cooling. For a certain rise in temperature, this change is so small for solids, compared to that of liquids and gases, that the change is not always noticeable. But with the help of proper experiment, it can be shown that solids expand on heating. This phenomenon of expansion with change in temperature is called thermal expansion. Expansions in solids by the application of external forces has been discussed in the chapter Elasticity. In this chapter, we shall discuss thermal expansion of solids and liquids.

Thermal expansion in solids is of three types—

1. linear expansion (a change in length),
2. surface or superficial expansion (the change in surface area) and
3. volume or cubical expansion (change in volume).

The thermal expansion of different solids, for the same rise in temperature are different. For example, copper undergoes a greater thermal expansion than iron, for the same rise in tem-perature with respect to their initial length, surface or volume.

Generally, with the rise in temperature, a solid expands equally in all directions. But there are exceptions as well. A few crystals expand differently in different directions when heated. Again an alloy of iron and nickel, called invar, practically does not show any temperature.

Reason for thermal expansion of solids: From the simple considerations of the structure of a crystalline solid, it can be said that the atoms of the crystal are arranged in a regular array under the elastic force [Fig.].

Between two adjacent atoms this elastic force behaves like an almost inextensible spring. At a fixed temperature, two atoms thus maintain an average distance between them and vibrate. This average distance increases with rise in temperature. Hence, solids expand with the rise in temperature.

Coefficient of Linear Expansion

Experimentally it is observed that the linear expansion of a metal rod on heating is directly proportional to

1. the initial length of the rod and
2. the rise in temperature of the rod.

Let l₁ be the length of a rod at temperature t₁ and the length becomes l₂ at temperature t₂, (where t₂ > t₁).
∴ The increase in length = (l₂ – l₁) and the increase in temperature = (t₂ – t₁)

As expansion of length is directly proportional to the initial length and the rise in temperature, (l₂ – l₁) ∝ l₁, for the same temperature change; and (l₂ – l₁) ∝ (t₂ – t₁), for the same initial length.
∴ (l₂ – l₁) ∝ l₁(t₂ – t₁) when both l₁ and (t₂ – t₁) vary or, (l₂ – l₁) = αl₁(t₂ – t₁)
Here α is the constant of proportionality, whose value is dif-ferent for different materials and is called the coefficient of linear expansion of that material.

If the initial temperature t₁ = 0, and the corresponding length is l₀, then the length at temperature t₂ = t is, from equation (3),
l_t = l₀(l + αt) ……….. (4)

Also from equation (1), if l₁ = 1 and (t₂ – t₁) = 1 then α = (l₂ – l₁). From this, the coefficient of linear expansion can be defined.

Definition: The increase in length for unit rise in temperature for a unit length of a solid is called the coefficient of linear expansion of the material of the solid.

The coefficient of linear expansion, α, is not a constant. In the above discussion we assume that the value of a does not depend on the temperature of the body. But it is not exactly true. The value of α depends on the temperature of the body. A solid of fixed length, for unit degree rise in temperature, expands a little differently in different temperature ranges. In fact, the value of α, calculated using equation (2) is the average value of α between the temperatures t₁ and t₂.

For more precise calculations, initial length of the material is taken as its length at 0°C . Thus, the expression for α modifies as
α \(=\frac{\text { increase in length for } 1^{\circ} \mathrm{C} \text { rise in temperature }}{\text { length at } 0^{\circ} \mathrm{C}}\)

Unit of coefficient of linear expansion :

From relation (2), unit of α \(=\frac{\text { unit of length }}{\text { unit of length } \times \text { unit of temperature }}\)
\(=\frac{1}{\text { unit of temperature }}\)

This leads to two conclusions:

i) Value of α is independent of the unit of length.
In equation (2), since \(\frac{l_2-l_1}{l_1}\) is a ratio of two lengths the coefficient of linear expansion does not depend on the unit of length.

For example, α for iron is 12 × 10^-6°C^-1 means that an iron rod of length 1 cm or 1 ft or 1 m, when heated through 1°C, will expand by 12 × 10^-6 cm or 12 × 10^-6 ft or 12 × 10^-6 m respectively.

ii) Value of α depends on the unit of temperature.
Unit of α is per °C or per °F.
As the change in temperature by 1°F = \(\frac{5}{9}\)°C change in temperature,
α_F = \(\frac{5}{9}\)α_C, where α_F = value of α in Fahrenheit and
α_C = value of α in Celsius scale.

For example, coefficient of linear expansion of iron
= 12 × 10^-6 °C^-1 = \(\frac{5}{9}\) × 12 × 10^-6 °F^-1
= 6.67 × 10^-6 °F^-1.

Coefficients of linear expansion of some solids at 20°C

Numerical Examples

Example 1.
A steel rod is 1.5 m long at 20°C. What will be the increase in its length if it is heated up to 100°C ? a for steel = 11 × 10^-6 °C^-1.
Solution:
Here, l₁ = 1.5 m = 150 cm, α = 11 × 10^-6°C^-1,
t₁ = 20°C and t₂ = 100°C
Increase in length = l₂ – l₁ = l₁α(t₂ – t₁)
= 150 × 11 × 10^-6 × (100 – 20)
= 150 × 11 × 10^-6 × 80
= 0.132 cm.

Example 2.
The length of a zinc rod, when heated from 20°C to 80°C, increases by 0.6 cm. If the coefficient of linear expansion of zinc is 27 × 10^-6°C^-1, what is the initial length of the rod?
Solution:

Example 3.
The lengths of a copper rod are 200.166 cm and 200.664 cm at 50°C and 200°C respectively. What is the coefficient of linear expansion of copper?
Solution:
α = \(\frac{l_2-l_1}{l_1\left(t_2-t_1\right)}\)
Given, l₁ = 200.166 cm, l₂ = 200.664 cm, t₁ = 50°C and t₂ = 200°C
∴ α = \(\frac{200.664-200.166}{200.166(200-50)}\) = \(\frac{0.498}{200.166 \times 150}\)
= 16.6 × 10^-6°C^-1.

Example 4.
A brass rod has a length of 150cm at 40°C. What will be its length at 100°C? The coefficient of linear expansion of brass is 18 × 10^-6°C^-1.
Solution:
l₂ = l₁{l + α(t₂ – t₁)}
Given, l₁ = 150 cm, t₁ = 40°C, t₂ = 100°C, α = 18 × 10^-6°C^-1
∴ l₂ = 150{1 + 18 × 10^-6(100 – 40))
= 150{1 + 108 × 10^-5} = 150 × 1.00108
= 150.162 cm.

Example 5.
Coefficient of linear expansion of aluminium is 19 × 10^-6 per degree celsius. What will be the value of this coefficient in Fahrenheit scale?
Solution:
α_F = \(\frac{5}{9}\)α_C = 19 × 10^-6 × \(\frac{5}{9}\) = 10.56 × 10^-6°F^-1.