Question:
Three identical conductors A, B, and C are in contact. Thermal conductivities are k, 2k, and k/2 respectively. A is at $100^\circ C$ and C is at $0^\circ C$. During steady state, the junction temperature between A and B is nearly:
Three identical conductors A, B, and C are in contact. Thermal conductivities are k, 2k, and k/2 respectively. A is at $100^\circ C$ and C is at $0^\circ C$. During steady state, the junction temperature between A and B is nearly:
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Steady state means the heat flowing through each section is equal.
Updated On: Jun 10, 2026
- $37^\circ C$
- $71^\circ C$
- $29^\circ C$
- $63^\circ C$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
In steady state, heat flow rate $H = \frac{kA\Delta T}{L}$ is constant.
Step 2: Analysis
Since A, B, and C are identical ($A, L$ same), let $T_1$ be junction A-B and $T_2$ be junction B-C. $H = k(100-T_1) = 2k(T_1-T_2) = (k/2)(T_2-0)$. From $H$, $100-T_1 = \frac{T_2}{2} \implies T_2 = 200-2T_1$. Substitute into $2(T_1-T_2) = T_2/2$: $4(T_1-T_2) = T_2 \implies 4T_1 = 5T_2$. $4T_1 = 5(200-2T_1) = 1000 - 10T_1 \implies 14T_1 = 1000 \implies T_1 \approx 71.4^\circ C$.
Step 3: Conclusion
The temperature is nearly $71^\circ C$.
Final Answer: (B)
In steady state, heat flow rate $H = \frac{kA\Delta T}{L}$ is constant.
Step 2: Analysis
Since A, B, and C are identical ($A, L$ same), let $T_1$ be junction A-B and $T_2$ be junction B-C. $H = k(100-T_1) = 2k(T_1-T_2) = (k/2)(T_2-0)$. From $H$, $100-T_1 = \frac{T_2}{2} \implies T_2 = 200-2T_1$. Substitute into $2(T_1-T_2) = T_2/2$: $4(T_1-T_2) = T_2 \implies 4T_1 = 5T_2$. $4T_1 = 5(200-2T_1) = 1000 - 10T_1 \implies 14T_1 = 1000 \implies T_1 \approx 71.4^\circ C$.
Step 3: Conclusion
The temperature is nearly $71^\circ C$.
Final Answer: (B)
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