A large cylindrical rod of length $$L$$ is made by joining two identical rods of copper and steel of length $$\left(\frac{L}{2}\right)$$ each. The rods are completely insulated from the surroundings. If the free end of copper rod is maintained at $$100^\circ C$$ and that of steel at $$0^\circ C$$ then the temperature of junction is (Thermal conductivity of copper is $$9$$ times that of steel)

Solution & Explanation

1. Understand the Setup and Heat Flow Rate

Since the two identical rods (copper and steel) are joined end-to-end and completely insulated from the surroundings, the heat transfer takes place under a steady-state condition. This implies that the rate of heat flow ($$H$$) through the copper rod must be exactly equal to the rate of heat flow through the steel rod:

$$H_{\text{copper}} = H_{\text{steel}}$$

2. Apply the Formula for Thermal Conduction

The rate of heat conduction is given by Fourier's law:

$$H = \frac{KA(\Delta T)}{l}$$

Where:

• $$K$$ = Thermal conductivity of the material
• $$A$$ = Cross-sectional area (identical for both rods)
• $$l$$ = Length of each rod ($$\frac{L}{2}$$, which is identical for both rods)
• $$\Delta T$$ = Temperature difference across the individual rod

Let $$\theta$$ be the temperature at the junction. We are given:

• Free end of copper temperature = $$100^\circ\text{C}$$
• Free end of steel temperature = $$0^\circ\text{C}$$
• Thermal conductivity relationship: $$K_{\text{copper}} = 9K_{\text{steel}}$$

3. Equate Heat Flows and Solve for $$\theta$$

Equating the heat flow rates for both segments:

$$\frac{K_{\text{copper}} \cdot A \cdot (100 - \theta)}{l} = \frac{K_{\text{steel}} \cdot A \cdot (\theta - 0)}{l}$$

Since the physical dimensions $$A$$ and $$l$$ are identical, they cancel out from both sides:

$$K_{\text{copper}}(100 - \theta) = K_{\text{steel}}(\theta)$$

Substitute $$K_{\text{copper}} = 9K_{\text{steel}}$$ into the simplified equation:

$$9K_{\text{steel}}(100 - \theta) = K_{\text{steel}}(\theta)$$

Divide both sides by $$K_{\text{steel}}$$:

$$9(100 - \theta) = \theta$$

$$900 - 9\theta = \theta$$

$$10\theta = 900 \implies \theta = 90^\circ\text{C}$$

Concept Check: Because copper has a much higher thermal conductivity than steel ($$9$$ times higher), it requires a significantly smaller temperature drop ($$10^\circ\text{C}$$) to transfer the exact same amount of heat that steel transfers over a much larger temperature drop ($$90^\circ\text{C}$$).

Correct Option Key: Option A ($$90^\circ\text{C}$$)