Three rods of identical cross-section and lengths are made of three different materials of thermal c

Q: Why is the rate of heat transfer exactly the same in all three rods?

Because the system is in a 'steady state'. If the heat entering a rod was greater than the heat leaving it, energy would accumulate, and the temperature of the rod would rise. Steady state means temperatures are constant, so heat in must equal heat out.

Q: What if the rods had different cross-sectional areas or lengths?

You would simply keep $A_1, A_2, A_3$ and $L_1, L_2, L_3$ in the Fourier's law equation. Instead of $K\Delta T$ being constant, the entire term $\frac{KA\Delta T}{L}$ would be equated for all three rods.

Q: Why did we ignore the negative sign that is sometimes written in Fourier's law?

The negative sign in $H = -KA \frac{dT}{dx}$ just indicates that heat flows in the direction of decreasing temperature. By explicitly taking $\Delta T$ as $(T_{\text{hot}} - T_{\text{cold}})$, we are calculating the magnitude of the heat flow, which is always positive.

Q: Can we find the exact numerical values of $K_1, K_2,$ and $K_3$ from this data?

No, we cannot. Because we are not given the actual numerical values for the heat current $H$, the area $A$, or the length $L$, we can only determine the relative ratios of the thermal conductivities, not their absolute values.

Q: What does 'no loss of energy from the surface' actually mean physically?

It means the rods are perfectly insulated along their curved sides (like wrapping them in thick wool). Therefore, 100% of the thermal energy entering the left face of a rod must exit through its right face, ensuring the heat current $H$ remains perfectly constant along the entire length.

Question

Three rods of identical cross-section and lengths are made of three different materials of thermal conductivity $K_1$, $K_2$ and $K_3$, respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at $100^{\circ} ext{C}$ and the other at $0^{\circ} ext{C}$ (see figure). If the joints of the rod are at $70^{\circ} ext{C}$ and $20^{\circ} ext{C}$ in steady state and there is no loss of energy from the surface of the rod, the correct relationship between $K_1$, $K_2$ and $K_3$ is Options: (A) $K_1 : K_3 = 2 : 3; K_2 : K_3 = 2 : 5$ (B) $K_1 < K_2 < K_3$ (C) $K_1 : K_2 = 5 : 2; K_1 : K_3 = 3 : 5$ (D) $K_1 > K_2 > K_3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are looking at a pipeline, but instead of water, heat is flowing through it. We have three distinct metallic rods joined end-to-end, forming a single long rod. The left end is exposed to a boiling hot reservoir at $100^{\circ}	ext{C}$, while the right end is submerged in freezing ice at $0^{\circ}	ext{C}$.

The problem states two critical conditions: the system is in a **steady state**, and there is **no loss of energy** from the curved surfaces of the rods. This means that every single joule of heat that enters the first rod must travel all the way through the second and third rods to exit at the cold end. There are no leaks.

### The Master Equation: Fourier's Law
To quantify this heat flow, we use Fourier's Law of Heat Conduction. The rate of heat transfer, or heat current $H$, is given by:

$$H = \frac{KA\Delta T}{L}$$

where $K$ is the thermal conductivity of the material, $A$ is the cross-sectional area, $\Delta T$ is the temperature difference across the rod, and $L$ is its length.

Here is where the problem hands us a massive shortcut. We are told that all three rods have **identical cross-sections and lengths**. Therefore, $A$ and $L$ are constants for all three sections. This simplifies our heat current equation beautifully. We can now say that the heat current is directly proportional to the product of thermal conductivity and the temperature difference:

$$H \propto K\Delta T$$

### Equating the Heat Currents
Because the system is in a steady state, the rate of heat flow through each rod must be exactly the same. If it weren't, heat would pile up at the junctions, causing their temperatures to change—which violates the definition of a steady state.

So, we can confidently write:

$$H_1 = H_2 = H_3$$

Now, let's substitute our simplified proportionalities into this equality. We know the temperatures at every junction: $100^{\circ}	ext{C}$, $70^{\circ}	ext{C}$, $20^{\circ}	ext{C}$, and $0^{\circ}	ext{C}$.

For the first rod, the temperature drops from $100^{\circ}	ext{C}$ to $70^{\circ}	ext{C}$. For the second, it drops from $70^{\circ}	ext{C}$ to $20^{\circ}	ext{C}$. And for the third, it drops from $20^{\circ}	ext{C}$ to $0^{\circ}	ext{C}$. Plugging these in, we get:

$$K_1(100 - 70) = K_2(70 - 20) = K_3(20 - 0)$$

### Final Calculation
Let's perform the simple subtractions inside the brackets:

$$30 K_1 = 50 K_2 = 20 K_3$$

To make this relationship cleaner, we can divide the entire equation by $10$:

$$3 K_1 = 5 K_2 = 2 K_3$$

This is our golden relationship. The options provided in the question are formatted as ratios, so let's extract the ratios from our equation.

First, let's compare the first and third rods:

$$3 K_1 = 2 K_3 \implies \frac{K_1}{K_3} = \frac{2}{3}$$

This gives us the ratio $K_1 : K_3 = 2 : 3$.

Next, let's compare the second and third rods:

$$5 K_2 = 2 K_3 \implies \frac{K_2}{K_3} = \frac{2}{5}$$

This gives us the ratio $K_2 : K_3 = 2 : 5$.

Looking at the given options, these ratios perfectly match option (a). The elegance of this problem lies in recognizing that steady state implies a constant heat current, allowing us to bypass complex calculations and rely purely on proportionalities.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Concept of Steady State$

$Fourier’s Law of Heat Conduction$

$Setting up the Equations$

$Simplifying the Relation$

$Finding the Ratios$

$\text{Conclusion & Extension}$

The Sigma Insight: Heat Transfer

Analyzing the Setup Imagine you are looking at a pipeline, but instead of water, heat is flowing through it

The Master Equation

Equating the Heat Currents Because the system is in a steady state, the rate of heat flow through each rod must be exactly the same

Final Calculation

Similar Questions

Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at $0^{\circ} C$ and $9 0^{\circ} C$ respectively. The temperature of junction of the three rods will be

One end of a thermally insulated rod is kept at a temperature $T_{1}$ and the other at $T_{2}$ . The rod is composed of two sections of lengths $l_{1}$ and $l_{2}$ and thermal conductivities $K_{1}$ and $K_{2}$ respectively. The temperature at the interface of the two sections is

A rod $C D$ of thermal resistance $10.0 kW^{- 1}$ is joined at the middle of an identical rod $A B$ as shown in figure. The end $A, B$ and $D$ are maintained at $20 0^{\circ} C, 10 0^{\circ} C$ and $12 5^{\circ} C$ respectively. The heat current in $C D$ is $P$ watt. The value of $P$ is ....... .

Select Answer:

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; } Concept of Steady State

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fourier’s Law of Heat Conduction

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Conclusion & Extension}