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A wall is made up of two layers, \( A \) and \( B \). The thickness of the two layers is the same, but the materials are different. The thermal conductivity of \( A \) is double that of \( B \). If in thermal equilibrium, the temperature difference between the two ends is \( 36^{\circ} C \), then the difference in temperature between the two surfaces of \( A \) will be:1. \( 6^{\circ} C \)2. \( 12^{\circ} C \)3. \( 18^{\circ} C \)4. \( 24^{\circ} C \) |
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Answer» Correct option is (2) 12∘C Consider two walls 'A' and 'B' of thickness 't' each, and thermal conductivity of '2K' and 'K' respectively. The temperatures at the left of A = Ta The temperatures at the right of B = Tb The temperature at the junction = T Heat flow is a constant in steady state: \(Q = KA \frac{dT}{dx}\) = constant Equating heat flow for both walls we get: \(2KA \frac{T-T_a}{t - 0} = KA \frac{T_b - T}{2t - t}\) Which simplifies to: 3T = Tb + 2Ta Also given that the temperature difference between the walls is 36∘C Tb − Ta = 36 Combining the two equation in T, Ta, Tb and eliminating Tb We get T − Ta = 12∘C |
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