Three rods of copper, brass and steel are welded together to form a Y-shaped structure. Area of cros

Q: Why do we assume heat flows from copper to brass and steel?

Copper is at $100^\circ\text{C}$ while brass and steel are at $0^\circ\text{C}$. Heat naturally flows from higher temperature to lower temperature.

Q: What does 'steady state' mean here?

Steady state means the temperature at any point in the rods does not change with time. Therefore, no heat is absorbed by the junction, and incoming heat equals outgoing heat.

Q: Why did we cancel the area $A$ during the calculation of $T$?

Since all three rods have the exact same cross-sectional area of $4 \text{ cm}^2$, it appears as a common factor in every term of the heat balance equation and can be factored out and cancelled.

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

If the areas were different, you would simply substitute their respective values into the equation $H = \frac{KA\Delta T}{L}$ and they wouldn't cancel out.

Q: Are the CGS units for thermal conductivity a problem here?

No, because all other given quantities (lengths in cm, area in $\text{cm}^2$, temperature in $^\circ\text{C}$) are also in CGS units. The final heat current will naturally be in cal/s.

Question

Three rods of copper, brass and steel are welded together to form a Y-shaped structure. Area of cross-section of each rod = $4 	ext{ cm}^2$. End of copper rod is maintained at $100^\circ	ext{C}$ whereas ends of brass and steel are kept at $0^\circ	ext{C}$. Lengths of the copper, brass and steel rods are 46, 13 and 12 cm, respectively. The rods are thermally insulated from surrounding except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units, respectively. Rate of heat flow through copper rod is

Options:

(A) 1.2 cal/s

(B) 2.4 cal/s

(C) 4.8 cal/s

(D) 6.0 cal/s

Sigma Percentile · Accepted Answer

Welcome to a classic problem of thermal conduction! Imagine you are standing in front of a Y-shaped structure...

### The Y-Shaped Thermal Junction

We are given three rods made of copper, brass, and steel, all welded together at a single central junction. The copper rod is connected to a heat source at $100^\circ	ext{C}$, while the brass and steel rods are connected to heat sinks at $0^\circ	ext{C}$.

Think of heat flow exactly like water flowing through pipes. The temperature difference acts as the "pressure" driving the heat, and the thermal conductivity $K$ determines how easily the material allows this flow. Since copper is at the highest temperature, heat will naturally flow from the copper rod into the junction, and from there, it will split and flow outwards through the brass and steel rods.

### The Principle of Steady State

The problem states that the rods are in a **steady state**. This is a crucial physical constraint. In a steady state, the temperature at any point in the system no longer changes with time.

This implies that the central junction is not storing or losing any net thermal energy. Therefore, according to the principle of conservation of energy (similar to Kirchhoff's Current Law in electricity), the total rate of heat entering the junction must exactly equal the total rate of heat leaving it.

Let the temperature of the central junction be $T$. We can write the heat balance equation as:

$$H_{Cu} = H_{Br} + H_{St}$$

### Setting Up the Master Equation

We know that the rate of heat flow $H$ through a rod of length $L$, cross-sectional area $A$, and thermal conductivity $K$ is given by Fourier's Law of Heat Conduction:

$$H = \frac{KA(T_{hot} - T_{cold})}{L}$$

Let's apply this master equation to our three rods. For the copper rod, heat flows from $100^\circ	ext{C}$ to $T$. For the brass and steel rods, heat flows from $T$ to $0^\circ	ext{C}$. Substituting these into our heat balance equation, we get:

$$\frac{K_{Cu}A(100 - T)}{L_{Cu}} = \frac{K_{Br}A(T - 0)}{L_{Br}} + \frac{K_{St}A(T - 0)}{L_{St}}$$

### The Algebra of Heat Flow

Notice a beautiful simplification here: the cross-sectional area $A = 4 	ext{ cm}^2$ is identical for all three rods. Because it appears in every term, we can factor it out and cancel it completely from the equation!

Now, let's substitute the given values for thermal conductivities and lengths:

$$\frac{0.92(100 - T)}{46} = \frac{0.26(T - 0)}{13} + \frac{0.12(T - 0)}{12}$$

Don't let the decimals intimidate you. The numbers are specifically chosen to simplify elegantly. Let's compute the fractions:

- For copper: $\frac{0.92}{46} = 0.02$
- For brass: $\frac{0.26}{13} = 0.02$
- For steel: $\frac{0.12}{12} = 0.01$

Substituting these simplified coefficients back into our equation:

$$0.02(100 - T) = 0.02T + 0.01T$$

$$0.02(100 - T) = 0.03T$$

To make the algebra even cleaner, let's multiply the entire equation by 100 to remove the decimals:

$$2(100 - T) = 3T$$

$$200 - 2T = 3T$$

$$5T = 200$$

$$T = 40^\circ	ext{C}$$

We have successfully found the junction temperature! It stabilizes exactly at $40^\circ	ext{C}$.

### The Final Calculation

The question asks for the rate of heat flow specifically through the **copper rod**. Now that we know the junction temperature $T = 40^\circ	ext{C}$, we can plug it back into the individual heat flow expression for copper.

$$H_{Cu} = \frac{K_{Cu}A(100 - T)}{L_{Cu}}$$

$$H_{Cu} = \frac{0.92 	imes 4 	imes (100 - 40)}{46}$$

We already know that $\frac{0.92}{46} = 0.02$, so the expression simplifies to:

$$H_{Cu} = 0.02 	imes 4 	imes 60$$

$$H_{Cu} = 0.08 	imes 60$$

$$H_{Cu} = 4.8 	ext{ cal/s}$$

The rate of heat flow through the copper rod is **$4.8 	ext{ cal/s}$**. This perfectly matches option (c).

This problem is a fantastic demonstration of how thermal circuits behave exactly like electrical circuits. By mastering the steady-state heat balance, you can solve any complex network of thermal conductors!

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Visualized Solution

$Y -Shaped Structure$

$Steady State Heat Flow$

$Fourier’s Law of Heat Conduction$

$Substituting the Values$

$Simplifying the Equation$

$Solving for Junction Temperature T$

$Calculating Rate of Heat Flow$

$Final Answer$

$The Way Forward$

The Sigma Insight: Heat Transfer

The Y-Shaped Thermal Junction

The Principle of Steady State

Setting Up the Master Equation

The Algebra of Heat Flow

The 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; } Y-Shaped Structure

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

.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; } Substituting the Values

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

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

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

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

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