The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially, each of th

Q: Why can't we just use the average temperature of the wire for thermal expansion?

Actually, for a linear temperature gradient, using the average temperature $\frac{T_{initial} + T_{final}}{2}$ does give the correct answer! However, integrating is the fundamental and foolproof method, especially if the cross-section or conductivity were to vary.

Q: Why is the rate of heat flow the same in both wires?

Because the system is in a steady state and thermally insulated from the surroundings. This means no heat is accumulating anywhere, and no heat is escaping from the sides. What enters one wire must flow through the next.

Q: What does 'thermally insulated from its surroundings' mean here?

It means heat only flows along the length of the wires (1D conduction). There is no heat loss from the curved surface of the wires to the surrounding air.

Q: Why did we take $\Delta T_x = T_x - 10$ and not just $T_x$?

Thermal expansion depends on the *change* in temperature from the initial state. The wire was initially at $10^\circ\text{C}$ uniformly. So the change for any element is its final temperature $T_x$ minus its initial temperature $10^\circ\text{C}$.

Q: Is the temperature gradient always linear?

Yes, for a uniform rod with constant thermal conductivity and no internal heat generation, the steady-state temperature profile is always a straight line.

Question

The ends $Q$ and $R$ of two thin wires, $PQ$ and $RS$, are soldered (joined) together. Initially, each of the wire has a length of $1	ext{ m}$ at $10^\circ	ext{C}$. Now, the end $P$ is maintained at $10^\circ	ext{C}$, while the end $S$ is heated and maintained at $400^\circ	ext{C}$. The system is thermally insulated from its surroundings. If the thermal conductivity of wire $PQ$ is twice that of the wire $RS$ and the coefficient of linear thermal expansion of $PQ$ is $1.2 	imes 10^{-5}	ext{ K}^{-1}$, the change in length of the wire $PQ$ is

Options:

(A) 0.78 mm

(B) 0.90 mm

(C) 1.56 mm

(D) 2.34 mm

Sigma Percentile · Accepted Answer

This problem is a beautiful fusion of two fundamental concepts in thermal physics: **steady-state heat conduction** and **thermal expansion**. It challenges us to think beyond uniform heating and deal with a situation where the temperature varies continuously along the length of a material.

Imagine you are standing at the junction of these two wires. On one side, you have a cooler region, and on the other, a blazing hot furnace. Heat is flowing right through you! Let's break down how we can find exactly how much the cooler wire expands under this temperature gradient.

### Analyzing the Setup

We are given two wires, $PQ$ and $RS$, connected in series. The left end $P$ is maintained at a cool $10^\circ	ext{C}$, while the right end $S$ is kept at a scorching $400^\circ	ext{C}$.

Because the system is thermally insulated from its surroundings, heat can only flow along the length of the wires. This is a classic case of 1D steady-state heat conduction. In a steady state, the rate at which heat flows through wire $PQ$ must be exactly equal to the rate at which it flows through wire $RS$.

### The Master Equation: Steady State Conduction

Let's denote the temperature at the junction (where $Q$ and $R$ meet) as $T$. We can use Fourier's law of heat conduction to set up our master equation. The rate of heat flow is given by $\frac{dQ}{dt} = \frac{KA\Delta T}{L}$.

Equating the heat flow rates for both wires, we get:
$$ \frac{K_{PQ} A (T - T_P)}{L_{PQ}} = \frac{K_{RS} A (T_S - T)}{L_{RS}} $$

We are given that the thermal conductivity of $PQ$ is twice that of $RS$, so $K_{PQ} = 2K$ and $K_{RS} = K$. Both wires have the same length ($1	ext{ m}$) and cross-sectional area $A$. Substituting these values, the equation simplifies beautifully:
$$ \frac{2KA(T - 10)}{1} = \frac{KA(400 - T)}{1} $$

Canceling out the common terms $K$ and $A$, we are left with a simple linear equation:
$$ 2(T - 10) = 400 - T $$
$$ 3T = 420 \implies T = 140^\circ	ext{C} $$

So, the junction settles at a steady temperature of $140^\circ	ext{C}$.

### The Temperature Gradient

Now, we need to find the expansion of wire $PQ$. The tricky part here is that the temperature is not uniform across the wire. It's $10^\circ	ext{C}$ at end $P$ and $140^\circ	ext{C}$ at the junction $Q$.

Because the wire is uniform, the temperature increases linearly along its length. We can express the temperature $T_x$ at any distance $x$ from end $P$ as:
$$ T_x = 10 + \left(\frac{140 - 10}{1}ight)x = 10 + 130x $$

This means the change in temperature for any point $x$ from its initial state of $10^\circ	ext{C}$ is $\Delta T_x = T_x - 10 = 130x$.

### Integrating for Total Expansion

Since the temperature change varies with $x$, different parts of the wire expand by different amounts. To find the total expansion, we must consider an infinitesimally small element of length $dx$ at a distance $x$.

The expansion of this tiny element is given by $dy = \alpha \Delta T_x dx$. Substituting our expression for $\Delta T_x$, we get $dy = \alpha (130x) dx$.

To find the total expansion $\Delta y$ of the entire wire $PQ$, we integrate this expression from $x = 0$ to $x = 1$:
$$ \Delta y = \int_0^1 \alpha (130x) dx $$
$$ \Delta y = 130\alpha \left[ \frac{x^2}{2} ight]_0^1 = 65\alpha $$

### Final Calculation

We are almost there! We just need to plug in the given value for the coefficient of linear expansion, $\alpha = 1.2 	imes 10^{-5}	ext{ K}^{-1}$.

$$ \Delta y = 65 	imes 1.2 	imes 10^{-5}	ext{ m} $$
$$ \Delta y = 78 	imes 10^{-5}	ext{ m} $$

To match the options, we convert this to millimeters by multiplying by $10^3$:
$$ \Delta y = 0.78	ext{ mm} $$

**Pro-Tip:** Because the temperature varies *linearly*, you could actually just use the average temperature of the wire! The average temperature is $\frac{10 + 140}{2} = 75^\circ	ext{C}$. The change in average temperature is $75 - 10 = 65^\circ	ext{C}$. Plugging this into $\Delta L = L \alpha \Delta T_{avg}$ gives $1 	imes 1.2 	imes 10^{-5} 	imes 65 = 0.78	ext{ mm}$. Boom! Same answer, half the time. But knowing the integration method is crucial for when things aren't linear.

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Visualized Solution

$Visualizing the Setup$

$Steady State Heat Conduction$

$Equating Heat Flow Rates$

$Solving for Junction Temperature$

$Calculating T$

$Temperature Gradient in P Q$

$Thermal Expansion of an Element$

$Integrating for Total Expansion$

$Final Calculation$

$The Final Answer$

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Master Equation

The Temperature Gradient

Integrating for Total Expansion

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Steady State Heat Conduction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Heat Flow Rates

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Junction Temperature

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Temperature Gradient in PQ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Thermal Expansion of an Element

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integrating for Total Expansion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Final Answer