A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The v

Q: Why is the rate of heat flow constant in steady state?

In steady state, no part of the bar absorbs heat to raise its own temperature. All heat entering a cross-section must exit it, making the heat current $H$ constant throughout the length.

Q: What if the bar was not insulated from the sides?

If heat escapes from the sides, $H$ decreases along the length. This makes the temperature gradient less steep over distance, resulting in an exponential decay curve rather than a straight line.

Q: Why is there a negative sign in Fourier's law?

Heat flows from higher to lower temperature. Since temperature decreases as $x$ increases, the gradient $d\theta/dx$ is inherently negative. The minus sign in the formula ensures the heat flow $H$ is calculated as a positive quantity.

Q: Does the material of the bar affect the shape of the graph?

No, as long as the thermal conductivity $k$ is constant, the graph remains a straight line. However, a different material with a different $k$ would change the *slope* of that straight line.

Q: What happens before the steady state is reached?

Before steady state (known as the transient state), the temperature at various points changes with time. The graph would be a curve that gradually straightens out as the system approaches thermal equilibrium.

Question

A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature $	heta$ along the length $x$ of the bar from its hot end is best described by which of the following figure.

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Ends

Imagine you are observing a long metallic bar. One end is thrust into a blazing furnace, while the other rests in a block of ice. Naturally, heat begins to march down the length of the bar, flowing from the hot end to the cold end.

The problem gives us a critical piece of information: the bar is in a **steady-state**. But what does that actually mean? In the beginning (the transient state), the bar is actively absorbing heat to raise its own temperature. However, once steady-state is achieved, the bar has reached its thermal capacity. It no longer absorbs heat. Every joule of heat energy that enters a cross-section of the bar immediately exits the other side. The temperature at any specific point $x$ becomes locked in time.

### Fourier's Law: The Master Equation

To understand how the temperature $	heta$ changes as we walk down the distance $x$ of the bar, we must consult the governing rule of thermal conduction: **Fourier's Law**.

Fourier's Law states that the rate of heat flow $H$ (which is $\frac{dQ}{dt}$) is directly proportional to the cross-sectional area $A$ and the temperature gradient $\frac{d	heta}{dx}$. Mathematically, it is written as:

$$H = -kA \frac{d	heta}{dx}$$

Here, $k$ is the thermal conductivity of the metal. The negative sign is crucial—it simply indicates that heat flows in the direction of decreasing temperature (down the thermal hill).

### The Magic of Steady State

Because we are in a steady-state, the rate of heat flow $H$ is perfectly constant throughout the entire length of the bar. It does not matter if you measure $H$ near the hot end or the cold end; it is the same value.

Let's rearrange Fourier's Law to isolate the temperature gradient:

$$\frac{d	heta}{dx} = -\frac{H}{kA}$$

Look at the right side of this equation. $H$ is constant. The thermal conductivity $k$ is a property of the metal and is constant. The cross-sectional area $A$ is uniform and constant. Therefore, the entire term $-\frac{H}{kA}$ is just one big constant!

### The Mathematical Verdict

If the derivative of a function is a constant, what is the original function? Let's integrate both sides with respect to $x$:

$$\int d	heta = -\int \left(\frac{H}{kA}ight) dx$$

$$	heta(x) = -\left(\frac{H}{kA}ight) x + C$$

At the very beginning of the bar ($x = 0$), the temperature is at its maximum, let's call it $	heta_H$. Plugging this in gives us our constant of integration $C = 	heta_H$. Our final equation becomes:

$$	heta(x) = -\left(\frac{H}{kA}ight) x + 	heta_H$$

Does this equation look familiar? It is the exact blueprint of a linear equation, $y = -mx + c$. The temperature $	heta$ drops linearly as the distance $x$ increases. If you plot this on a graph, it yields a perfect straight line with a negative slope.

This beautiful linear relationship is the hallmark of steady-state 1D heat conduction without any lateral heat loss!

Animated Solution for Physics - Properties of Solids and Liquids: A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature $θ$ along the length $x$ of the bar from its hot end is best described by which of the following figure.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Fourier’s Law of Heat Conduction$

$Applying Steady-State Condition$

$Integrating the Gradient$

$Analyzing the Graph$

$What if it wasn’t insulated?$

The Sigma Insight: Heat Transfer

The Setup

Fourier's Law

The Magic of Steady State

The Mathematical Verdict

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