A brass rod of length $50\ \text{cm}$ and diameter $3.0\ \text{mm}$ is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $250\ ^\circ\text{C}$, if the original lengths are at $40.0\ ^\circ\text{C}$? Is there a 'thermal stress' developed at the junction? The ends of the rod are free to expand (Coefficient of linear expansion of brass $= 2.0 \times 10^{-5}\ \text{K}^{-1}$, steel $= 1.2 \times 10^{-5}\ \text{K}^{-1}$).

> [!TLDR] > Each rod expands independently according to its own coefficient of linear expansion. The total change in length is the sum of the individual expansions. Since the ends are free, no thermal stress develops at the junction. The combined expansion is $0.44\ \text{cm}$. ## Why this approach works When two rods of different materials are joined end-to-end and heated, each rod expands according to its own coefficient of linear expansion. Because the rods are free at both ends, there is no constraint that would force one rod to stretch or compress the other — they simply grow side by side. The total change in length is therefore just the sum of the two independent expansions. The key idea: **thermal expansion is additive for rods in series when ends are free**. No stress arises because nothing prevents the rods from expanding freely. --- ### Step-by-step solution **1. Identify the given data** - Length of brass rod: $L_b = 50\ \text{cm}$ - Length of steel rod: $L_s = 50\ \text{cm}$ - Diameter of both rods: $3.0\ \text{mm}$ (not needed for length change, but relevant for stress — we'll see why) - Initial temperature: $T_i = 40.0\ ^\circ\text{C}$ - Final temperature: $T_f = 250\ ^\circ\text{C}$ - Temperature change: $\Delta T = T_f - T_i = 210\ ^\circ\text{C} = 210\ \text{K}$ (since a change of $1\ ^\circ\text{C}$ equals $1\ \text{K}$) - Coefficient of linear expansion for brass: $\alpha_b = 2.0 \times 10^{-5}\ \text{K}^{-1}$ - Coefficient of linear expansion for steel: $\alpha_s = 1.2 \times 10^{-5}\ \text{K}^{-1}$ **2. Recall the formula for linear expansion** For a rod of original length $L$, the change in length when temperature changes by $\Delta T$ is: $$ \Delta L = \alpha \, L \, \Delta T $$ This formula assumes the rod is free to expand — no external forces opposing it. **3. Calculate expansion of the brass rod** $$ \Delta L_b = \alpha_b \, L_b \, \Delta T = (2.0 \times 10^{-5})(50)(210) $$ First multiply: $2.0 \times 10^{-5} \times 50 = 1.0

A brass rod of length 50cm and diameter 3.0mm is joined to a steel rod of the same length and diameter. What i

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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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Each rod expands independently according to its own coefficient of linear expansion. The total change in length is the sum of the individual expansions. Since the ends are free, no thermal stress develops at the junction. The combined expansion is $0.44\ \text{cm}$ .

Why this approach works

When two rods of different materials are joined end-to-end and heated, each rod expands according to its own coefficient of linear expansion. Because the rods are free at both ends, there is no constraint that would force one rod to stretch or compress the other — they simply grow side by side. The total change in length is therefore just the sum of the two independent expansions.

The key idea: thermal expansion is additive for rods in series when ends are free. No stress arises because nothing prevents the rods from expanding freely.

Step-by-step solution

1. Identify the given data

Length of brass rod: $L_b = 50\ \text{cm}$
Length of steel rod: $L_s = 50\ \text{cm}$
Diameter of both rods: $3.0\ \text{mm}$ (not needed for length change, but relevant for stress — we'll see why)
Initial temperature: $T_i = 40.0\ ^\circ\text{C}$
Final temperature: $T_f = 250\ ^\circ\text{C}$
Temperature change: $\Delta T = T_f - T_i = 210\ ^\circ\text{C} = 210\ \text{K}$ (since a change of $1\ ^\circ\text{C}$ equals $1\ \text{K}$ )
Coefficient of linear expansion for brass: $\alpha_b = 2.0 \times 10^{-5}\ \text{K}^{-1}$
Coefficient of linear expansion for steel: $\alpha_s = 1.2 \times 10^{-5}\ \text{K}^{-1}$

2. Recall the formula for linear expansion

For a rod of original length $L$ , the change in length when temperature changes by $\Delta T$ is:

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

This formula assumes the rod is free to expand — no external forces opposing it.

3. Calculate expansion of the brass rod

\Delta L_b = \alpha_b \, L_b \, \Delta T = (2.0 \times 10^{-5})(50)(210)

First multiply: $2.0 \times 10^{-5} \times 50 = 1.0 \times 10^{-3}$

Then: $1.0 \times 10^{-3} \times 210 = 0.21\ \text{cm}$

So $\Delta L_b = 0.21\ \text{cm}$ .

4. Calculate expansion of the steel rod

\Delta L_s = \alpha_s \, L_s \, \Delta T = (1.2 \times 10^{-5})(50)(210)

First: $1.2 \times 10^{-5} \times 50 = 6.0 \times 10^{-4}$

Then: $6.0 \times 10^{-4} \times 210 = 0.126\ \text{cm}$

So $\Delta L_s = 0.126\ \text{cm}$ .

5. Total change in length of the combined rod

Since the rods are in series and both ends are free, the total expansion is simply:

\Delta L_{\text{total}} = \Delta L_b + \Delta L_s = 0.21 + 0.126 = 0.336\ \text{cm}

Rounding to two significant figures (matching the given data): $0.34\ \text{cm}$ . …

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Why this approach works

Step-by-step solution

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