Two walls of thicknes `l_(1)` and `l_(2)` and ther-mal condctivities `K_(1)` and `K_(2)` are in contact In the steady state, if the temperature at the outer faces are `T_(1)` and `T_(2)` find the temperature at the common wall .

Explanation

Let `theta` be the temperature of the junction or interface In steady state`(Q)/(t)` = constant
`:.(K_(1)A(theta_(1)-theta))/(l_(1)) = (K_(2)A(theta-theta))/(l_(2))`
`((K_(1)theta_(1))/(l_(1))+(K_(2)theta_(2))/(l_(2)))/(K_(1)/(l_(1))+K_(2)/(l_(2)))=(K_(1)theta_(1)l_(2)+K_(2)theta_(2)l_(1))/(K_(1)l_(2)+K_(2)l_(1))`
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