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Thermal Expansion - Linear Expansion

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Estimated time: 21 minutes
  • Introduction
  • Definition: Linear Expansion
  • Definition: Coefficient of Linear Expansion
  • Deriving the Formula
  • Coefficient of Linear Expansion (α) for Common Materials
  • Example 1
  • Example 2
  • Real-Life Examples
  • Key Points: Linear Expansion
Maharashtra State Board: Class 11

Introduction

On a scorching summer day, the temperature of metal railway rails can rise by 30–40 °C above the laying temperature. Since steel expands when heated, engineers leave small gaps (called expansion joints) between rail sections. Without these gaps, the rails would push against each other, buckle, and potentially derail trains.

Railway track expansion gap showing the joint between two rail sections

Fig: Expansion gap between two railway rail sections — a real-life application of linear expansion.

This everyday phenomenon is an example of linear expansion — the increase in length of a solid when its temperature rises. Let's understand the science behind it.

Maharashtra State Board: Class 11

Definition: Linear Expansion

Linear expansion is the increase in length of a solid due to an increase in temperature. It occurs primarily in one dimension — along the length of the object (e.g., a rod, wire, or rail).

Maharashtra State Board: Class 11

Definition: Coefficient of Linear Expansion

The coefficient of linear expansion of a solid is thus defined as the increase in the length per unit original length at 0 °C for a one-degree centigrade rise in temperature.

Maharashtra State Board: Class 11

Deriving the Formula

Let's build the mathematical relationship from experimental observations:

1. Observe the proportionality: Experiments show that for a rod of length l, the fractional change in length is directly proportional to the change in temperature:

 
\[\frac{\Delta l}{l}\propto\Delta T\]

2. Introduce the constant α: Replace the proportionality sign with a constant called the coefficient of linear expansion (α):

\[\frac{\Delta l}{l}=\alpha\Delta T\]

3. Rearrange to define α: If the rod has length l₀ at 0 °C and length lT at temperature T:

α = \[\frac{l_T-l_0}{l_0\times(T-T_0)}\]

where Δl = lT − l0 (change in length) and ΔT = T − T0 (rise in temperature).

4. Generalise for any initial temperature: Since α is nearly constant for a given material, we don't need the initial temperature to be 0 °C. For any two temperatures T₁ and T₂:

α = \[\frac{l_2-l_1}{l_1\times(T_2-T_1)}\]

This is the formula you will use in most numerical problems.

Maharashtra State Board: Class 11

Coefficient of Linear Expansion (α) for Common Materials

The value of α depends on the strength of interatomic bonds. Stronger bonds resist separation → lower α. Weaker bonds allow atoms to spread apart more easily → higher α.

Materials α (K⁻¹)
Carbon (diamond) 0.1 × 10⁻⁵
Glass 0.85 × 10⁻⁵
Iron 1.2 × 10⁻⁵
Steel 1.3 × 10⁻⁵
Gold 1.4 × 10⁻⁵
Copper 1.7 × 10⁻⁵
Silver 1.9 × 10⁻⁵
Aluminium 2.5 × 10⁻⁵
Sulphur 6.1 × 10⁻⁵
Mercury 6.1 × 10⁻⁵
Water 6.9 × 10⁻⁵
Carbon (graphite) 8.8 × 10⁻⁵
Maharashtra State Board: Class 11

Example 1

Problem: The length of a metal rod at 27 °C is 4 cm. The length increases to 4.02 cm when the metal rod is heated upto 387 °C. Determine the coefficient of linear expansion of the metal rod.

Given:
T₁ = 27 °C
T₂ = 387 °C
l₁ = 4 cm = 4 × 10⁻² m
l₂ = 4.02 cm = 4.02 × 10⁻² m

Using Eq.:

α = \[\frac{l_2-l_1}{l_1\times(T_2-T_1)}=\frac{(4.02-4.0)\times10^{-2}}{4\times10^{-2}\times(387-27)}\]

α = \[\frac{0.02\times10^{-2}}{4\times10^{-2}\times360}=\frac{0.02}{1440}\]

α = 1.39 × 10⁻⁵ °C⁻¹

Interpretation: This value is close to α for iron (1.2 × 10⁻⁵) and steel (1.3 × 10⁻⁵), suggesting the rod is likely made of an iron-based alloy.
Maharashtra State Board: Class 11

Example 2

Problem: The length of an iron rod at a temperature of 27 °C is 4.256m. Find the temperature at which the length of the same rod increases to 4.268 m.(α for iron = 1.2×10-5 K-1)

Given:
T₁ = 27 °C
l₁ = 4.256 m
l₂ = 4.268 m
α = 1.2 × 10⁻⁵ K⁻¹

Rearranging Eq. for T₂:

\[T_2=T_1+\frac{l_2-l_1}{l_1\times\alpha}\]

\[T_2=27+\frac{4.268-4.256}{4.256\times1.2\times10^{-5}}\]

\[T_2=27+\frac{0.012}{5.1072\times10^{-5}}=27+234.96\]
 
Interpretation: The rod needs to be heated by ~235 °C (from 27 °C to ~262 °C) to expand by just 1.2 cm. This shows how small linear expansion is for solids — even hundreds of degrees cause only millimetre-level changes!
Maharashtra State Board: Class 11

Real-Life Examples

  • Railway Track Gaps: Small gaps between rail sections allow metal to expand on hot days without buckling or bending.
  • Bridge Expansion Joints: Zigzag metal joints in bridges (like the Golden Gate Bridge) absorb seasonal expansion and contraction.

Bridge expansion joint close-up showing zigzag metal pattern

Fig: Close-up of a bridge expansion joint — the zigzag metal strip accommodates thermal expansion and contraction.

  • Shrink Fitting: A steel wheel is heated to expand, slipped over an axle, then cooled — it contracts and grips tightly.
  • Mercury Thermometers: Mercury's high α makes it expand visibly with small temperature changes — perfect for measurement.
  • Power Line Sag: Overhead electrical wires sag in summer (expansion) and become taut in winter (contraction).
  • Bimetallic Strips: Two metals with different α values bonded together — bends when heated. Used in thermostats and fire alarms.
Maharashtra State Board: Class 11

Key Points: Linear Expansion

  • Linear expansion = increase in length due to heating
  • Master formula: α = (l₂ − l₁) / [l₁ × (T₂ − T₁)]
  • α depends on the material — stronger bonds → smaller α
  • Unit of α: K⁻¹ or °C⁻¹ (numerically identical)
  • For solids, α is very small (~10⁻⁵), so expansion is typically in millimetres
  • Real-life: railway gaps, bridge joints, shrink fitting, bimetallic strips, thermometers
  • Relationship: α : β : γ = 1 : 2 : 3 (linear : areal : volumetric)

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