One end of a thermally insulated rod is kept at a temperature T1 and the other at T2. The rod is composed of two sections of lengths ℓ1 and ℓ2 and thermal conductivities k1 and k2 respectively. The temperature at the interface of the two sections is
(A)(k2ℓ2T1+k1ℓ1T2)/(k1ℓ1+k2ℓ2)
(B)(k2ℓ1T1+k1ℓ1T2)/(k2ℓ1+k1ℓ2)
(C)(k1ℓ2T1+k2ℓ1T2)/(k1ℓ2+k2ℓ1)✓ Correct
(D)(k1ℓ1T1+k2ℓ2T2)/(k1ℓ1+k2ℓ2)
Answer: (C)
Solution
In the steady state the same heat current flows through both sections, so at the junction temperature T we have ℓ1k1(T1−T)=ℓ2k2(T−T2). Cross-multiplying and collecting the terms in T gives T=k1ℓ2+k2ℓ1k1ℓ2T1+k2ℓ1T2, which is a conductance-weighted average of the two end temperatures.
JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option