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

At steady state, the rate of flow of heat through first slab = rate of flow of heat through second slab.k1AT1−Tl1=k2AT−T2l2                ....(1)where 7 = temperature of junction From eq. (1)         k1l2T1−k1l2T=k2l1T−k2l1T2or     k1l2T1+k2l1T2=k2l1T+k1l2TSolving we getT=k1l2T1+k2l1T2k1l2+k2l1