A student records the initial length $l$, change in temperature $\Delta T$ and change in length $\Delta l$ of a rod as follows: | S.No. | $l$ (m) | $\Delta T$ ($^\circ$C) | $\Delta l$ (m) | | --- | --- | --- | --- | | 1. | 2 | 10 | $4 \times 10^{-4}$ | | 2. | 1 | 10 | $4 \times 10^{-4}$ | | 3. | 2 | 20 | $2 \times 10^{-4}$ | | 4. | 3 | 10 | $6 \times 10^{-4}$ | If the first observation is correct, what can you say about observations 2, 3 and 4.

A student records the initial length l, change in temperature Δ T and change in length Δ l of a rod as follows

NCERT Exemplar · Q14

A student records the initial length $l$ , change in temperature $\Delta T$ and change in length $\Delta l$ of a rod as follows:

S.No.	$l$ (m)	$\Delta T$ ( $^\circ$ C)	$\Delta l$ (m)
1.	2	10	$4 \times 10^{-4}$
2.	1	10	$4 \times 10^{-4}$
3.	2	20	$2 \times 10^{-4}$
4.	3	10	$6 \times 10^{-4}$

If the first observation is correct, what can you say about observations 2, 3 and 4.

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Concept understanding — Thermal Expansion Coefficient

Thermal Expansion Coefficient: From Intuition to Precision

The Intuition: What Happens When Things Get Hot?

Think about a metal railway track on a summer day. The track is laid in sections with small gaps between them. On a hot afternoon, those gaps get smaller — sometimes the track even buckles. Why? Because the metal expands when heated.

Or consider a mercury thermometer. The liquid mercury sits in a bulb at the bottom. When your body warms the bulb, the mercury expands and rises up the narrow tube. The hotter you are, the higher it climbs.

This is thermal expansion: most materials get bigger when heated and smaller when cooled. The atoms inside vibrate more vigorously as temperature rises, pushing their neighbours slightly farther apart. The entire object grows in every direction.

But different materials expand by different amounts. A steel rod and an aluminium rod of the same length, heated by the same amount, will not end up the same length. Aluminium expands more. So we need a number that tells us how much a given material expands per degree of temperature change. That number is the thermal expansion coefficient.

The Precise Statement: Defining the Coefficient

There are actually three coefficients, depending on whether we care about length, area, or volume. For a first meeting, we focus on the most common one: the linear thermal expansion coefficient, denoted by the Greek letter $\alpha$ (alpha).

\alpha = \frac{1}{L_0} \cdot \frac{\Delta L}{\Delta T}

Where:

$L_0$ is the original length of the object (at some starting temperature)
$\Delta L$ is the change in length (final length minus original length)
$\Delta T$ is the change in temperature (final temperature minus initial temperature)

What this formula says in plain English: The coefficient $\alpha$ is the fractional change in length per degree of temperature change. If $\alpha = 2.5 \times 10^{-5} \, \text{per} \, ^\circ\text{C}$ , it means that for every $1^\circ\text{C}$ rise in temperature, the material expands by $0.0025\%$ of its original length.

How to Use It: The Working Formula

From the definition, we can rearrange to get the practical formula:

\Delta L = \alpha \, L_0 \, \Delta T

So the final length $L$ after a temperature change is:

L = L_0 + \Delta L = L_0 (1 + \alpha \, \Delta T)

Tip

For small temperature changes (say, less than $100^\circ\text{C}$ ), this linear approximation is excellent. For very large changes, the coefficient itself may change slightly with temperature, but at the introductory level we treat $\alpha$ as constant.

A Concrete Example

A steel bridge girder is $50.00 \, \text{m}$ long at $20^\circ\text{C}$ . The linear expansion coefficient of steel is $\alpha = 1.2 \times 10^{-5} \, /^\circ\text{C}$ . How much longer is it on a $40^\circ\text{C}$ day?

Step 1: Identify the quantities.

$L_0 = 50.00 \, \text{m}$
$\Delta T = 40 - 20 = 20^\circ\text{C}$
$\alpha = 1.2 \times 10^{-5} \, /^\circ\text{C}$

Step 2: Apply the formula.

\Delta L = \alpha \, L_0 \, \Delta T = (1.2 \times 10^{-5})(50.00)(20)

Step 3: Calculate.

\Delta L = 1.2 \times 10^{-5} \times 1000 = 0.012 \, \text{m} = 1.2 \, \text{cm}

So the girder expands by $1.2 \, \text{cm}$ . That is why bridges have expansion joints — without them, the structure would buckle.

Two Important Cousins: Area and Volume Expansion

For a thin sheet (like a metal plate), we care about area expansion. The area expansion coefficient is approximately $2\alpha$ . For a solid object, the volume expansion coefficient is approximately $3\alpha$ . These come from the same idea: if every linear dimension grows by a factor $(1 + \alpha \Delta T)$ , then area grows by $(1 + \alpha \Delta T)^2 \approx 1 + 2\alpha \Delta T$ , and volume by $(1 + \alpha \Delta T)^3 \approx 1 + 3\alpha \Delta T$ .

Watch out

These approximations ( $2\alpha$ and $3\alpha$ ) are valid only when $\alpha \Delta T$ is small compared to 1. For most solids and modest temperature changes, this is true. For gases, the expansion is much larger and a different treatment is needed.

What the Coefficient Tells Us About Materials

| Material | $\alpha$ (per $^\circ\text{C}$ ) | Behaviour |

|----------|--------------------------------|-----------| …

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The coefficient of linear thermal expansion is constant for a given material. Using the first observation to find this constant, we can then check the consistency of the other observations. Observations 2 and 3 are incorrect, while observation 4 is correct.

When a material is heated, its dimensions generally increase. This phenomenon is called thermal expansion. For a rod, the increase in length, or linear expansion, depends on three factors:

The original length of the rod.
The change in temperature.
The material of the rod, characterized by its coefficient of linear thermal expansion.

The relationship is given by the formula:

\Delta l = l \alpha \Delta T

Here, $\Delta l$ is the change in length, $l$ is the original length, $\Delta T$ is the change in temperature, and $\alpha$ (alpha) is the coefficient of linear thermal expansion. For a specific material, $\alpha$ is a constant. This means that if we know $\alpha$ for the rod from one observation, we can use it to verify other observations for the same rod.

Let's use the first observation to determine the value of $\alpha$ .

Calculate $\alpha$ from the first observation:

From the first row of the table:

$l = 2 \text{ m}$

$\Delta T = 10 \text{ }^\circ\text{C}$

$\Delta l = 4 \times 10^{-4} \text{ m}$

Using the formula $\Delta l = l \alpha \Delta T$ , we can solve for $\alpha$ :

$\alpha = \frac{\Delta l}{l \Delta T}$

$\alpha = \frac{4 \times 10^{-4} \text{ m}}{(2 \text{ m})(10 \text{ }^\circ\text{C})}$

$\alpha = \frac{4 \times 10^{-4}}{20} \text{ }^\circ\text{C}^{-1}$

$\alpha = 0.2 \times 10^{-4} \text{ }^\circ\text{C}^{-1}$

$\alpha = 2 \times 10^{-5} \text{ }^\circ\text{C}^{-1}$

This value of $\alpha$ should be constant for all observations of this rod. Now we will check observations 2, 3, and 4 using this value.
Check observation 2:

From the second row:

$l = 1 \text{ m}$

$\Delta T = 10 \text{ }^\circ\text{C}$

$\Delta l = 4 \times 10^{-4} \text{ m}$

Let's calculate the expected $\Delta l$ using our derived $\alpha$ :

$\Delta l_{\text{expected}} = l \alpha \Delta T$

$\Delta l_{\text{expected}} = (1 \text{ m})(2 \times 10^{-5} \text{ }^\circ\text{C}^{-1})(10 \text{ }^\circ\text{C})$

$\Delta l_{\text{expected}} = 2 \times 10^{-4} \text{ m}$

The recorded $\Delta l$ is $4 \times 10^{-4} \text{ m}$ , which is different from the expected value.

Therefore, observation 2 is incorrect.
Check observation 3:

From the third row:

$l = 2 \text{ m}$

$\Delta T = 20 \text{ }^\circ\text{C}$

$\Delta l = 2 \times 10^{-4} \text{ m}$

Let's calculate the expected $\Delta l$ using our derived $\alpha$ :

$\Delta l_{\text{expected}} = l \alpha \Delta T$ …

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