A steel tape $1\ \text{m}$ long is correctly calibrated for a temperature of $27.0\ ^\circ\text{C}$. The length of a steel rod measured by this tape is found to be $63.0\ \text{cm}$ on a hot day when the temperature is $45.0\ ^\circ\text{C}$. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is $27.0\ ^\circ\text{C}$? Coefficient of linear expansion of steel $= 1.20 \times 10^{-5}\ \text{K}^{-1}$.

A steel tape 1m long is correctly calibrated for a temperature of 27.0circC. The length of a steel rod measure

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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 tape expands with temperature, so its markings are no longer true at $45^\circ\text{C}$ . The measured $63.0\ \text{cm}$ is shorter than the actual rod length. The actual length at $45^\circ\text{C}$ is $63.0136\ \text{cm}$ , and at $27^\circ\text{C}$ it is $63.0\ \text{cm}$ (the rod and tape are at the same temperature, so the reading is correct).

The key idea here is that the measuring instrument itself changes size with temperature. A steel tape calibrated at $27.0^\circ\text{C}$ will expand when the temperature rises. Its markings — originally exactly 1 cm apart at $27^\circ\text{C}$ — become slightly farther apart at $45^\circ\text{C}$ . So when you read “63.0 cm” on the hot tape, you are actually measuring a longer object than 63.0 true centimetres.

The rod is also made of steel, so it too expands. But the question asks for the rod’s actual length on the hot day — that is, its real physical length, not what the tape says. And then it asks for the rod’s length back at the calibration temperature.

Let’s go step by step.

Understand what the tape reading means. At $27^\circ\text{C}$ , the tape is exactly 1 m long, and each centimetre marking is exactly 1 cm apart. At $45^\circ\text{C}$ , the tape has expanded. Its total length becomes:

L_{\text{tape}}(45) = L_0 \left(1 + \alpha \Delta T\right)

where $L_0 = 100\ \text{cm}$ , $\alpha = 1.20 \times 10^{-5}\ \text{K}^{-1}$ , and $\Delta T = 45.0 - 27.0 = 18.0\ \text{K}$ .

L_{\text{tape}}(45) = 100 \left(1 + 1.20 \times 10^{-5} \times 18\right) = 100 \left(1 + 2.16 \times 10^{-4}\right) = 100.0216\ \text{cm}

So each “1 cm” marking on the hot tape is actually $1.000216\ \text{cm}$ apart.

Find the actual length of the rod at $45^\circ\text{C}$ . The tape reads $63.0\ \text{cm}$ . But each of those “cm” units is really $1.000216\ \text{cm}$ . So the true physical length of the rod at $45^\circ\text{C}$ is:

L_{\text{rod}}(45) = 63.0 \times 1.000216 = 63.0 + 63.0 \times 2.16 \times 10^{-4}

= 63.0 + 0.013608 \approx 63.0136\ \text{cm}

Tip

You can also think: the tape reading is less than the true length because the tape’s units have stretched. So the correction is additive and positive.

Now find the rod’s length at $27^\circ\text{C}$ . The rod is made of steel, so it also expands and contracts. We know its length at $45^\circ\text{C}$ is $63.0136\ \text{cm}$ . To find its length at $27^\circ\text{C}$ , we reverse the expansion:

L_{\text{rod}}(27) = \frac{L_{\text{rod}}(45)}{1 + \alpha \Delta T}

where $\Delta T = 18\ \text{K}$ again (cooling down).

$L_{\text{rod}}(27) = \frac{63.0136}{1 + 2.16 \times 10^{-4}} \approx 63.0136 \times (1 - 2.16 \times 10^{-4})$ …

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