The figure shows the cross-section of the outer wall of a house buit in a hill-resort to keep the house insulated from the freezing temperature of outside. The wall consists of teak wood of thickness `L_(1)` and brick of thickness `(L_(2) = 5L_(1))`, sandwitching two layers of an unknown material with identical thermal conductivites and thickness. The thermal conductivity of teak wood is `K_(1)` and that of brick is `(K_(2) = 5K)`. Heat conducion through the wall has reached a steady state with the temperature of three surfaces being known. `(T_(1) = 25^(@)C, T_(2) = 20^(@)C` and `T_(5) = - 20^(@)C)`. Find the interface temperature `T_(4)` and `T_(3)`.

Explanation

Let interface area be `A`. then thermal resistance of wood.
`R_(1) = (L_(1))/(K_(1)A)` and that of brick wall
`R_(2) = (L_(2))/(K_(2)A) = (5L_(1))/(5K_(1)A) = R_(1)`
Let thermal resistance of the each sand witch layer `= R`. Then the above wall can be visualised as a circuit

thermal current through each walls is same.
Hence `(25 -20)/(R_(1)) = (20 -T_(3))/(R ) = (T_(3)-T_(4))/(R ) = (T_(4)+20)/(R_(1))`
`rArr 25 -20 = T_(4) +20 rArr T_(4) =- 15^(@)C`
also, `20 - T_(3) = T_(3) - T_(4) rArr T_(3) = (20 +T_(4))/(2) = 2.5^(@)C`