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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Two plates each of area A, thicknesses L1 and L2 and thermal conductivities K1 and K2 are joined to form a single plate of thickness (L1 + L2). If the temperatures of the free surfaces are T1 and T2, Find the temperature of interface:

(A) [T1L2/K2 + T2L1/K1] / [L1/K1 - L2/K2]

(B) [T1L2/K2 + T2L1/K1] / [L1/K1 + L2/K2]

(C) [T1L1/K1 + T2L2/K2] / [L1/K1 + L2/K2]

(D)[T1L1/K1 - T2L2/K2] / [L1/K1 + L2/K2]

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1 Answer

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If T is the common temperature of the interface, then as in series rate of flow of heat remains same, i.e, H = H1(=H2)
(T1 – T2)/(R1+R2) = T1 – T / R1, i.e, T = (T1R2 + T2R1) / (R1 + R2)
Therefore, T = [T1L2/K2 + T2L1/K1] / [L1/K1 + L2/K2] (as R = L/KA)
answered Mar 10, 2014 by balaji.thirumalai
 

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