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_(1))`. 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 wall is same.
Hence `(25-20)/(R_(1))= (20-T_(2))/R= (T_(3)-T_(4))/R=(T_(4)+20)/R_(1)`
`implies 25-20 = T_(4) +24 implies T_(4)=-15^(@)C`
also, `20-T_(3)=T_(3)-T_(4) implies T_(3)= (20+T_(4))/(2) =2.5^(@)C`
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