Figure shows the cross section of a wall made of white pine of thickness `L_(a)` and brick of thickness `L_(d) = = 2.0 L_(a)` , sandwiching two layers of unknonw material with identical thicknesses and thermal conductivities. The thermal conductivity of the pine is `k_(a)` and that of the brick is `k_(d) ( = 5.0k_(a))`. The face area A of the wall is unknown. Thermal conduction through the wall has reached the steady state, the only known interface temperatures are `T_(1) = 25^(@)C, T_(2) = 20^(@)C`, and `T_(5 ) = - 10^(@)C`. What is interface temperature `T_(4)` ?

Explanation

(1) Temperature `T_(4)` helps determine the rate `P_(d)` at which energy is conducted through the brick, as given by Equation. However, we lack enough data to solve Equation for `T_(4)` . (2) Because the conduction is steady, the conduction rate `P_(d )` through the brick must equal the conduction rate `P_(n)` through the pine. That gets us going.
Calculations `:` Frm Equation and Figure, we can write
`p_(a) = k_(a) A(T_(1)-T_(2))/( L_(a))` and `P_(d) = k_(d) A( T_(4) -T_(5))/(L_(d))`
Setting `P_(a) = P_(d)` and solving for `T_(4)` yield
`T_(4) = ( k_(a) L_(d))/( k_(d) L_(a)) ( T_(1) -T_(2)) + T_(5)`
Letting `L_(d) = 2.0 L_(a)` and `k_(d) = 5.0 k_(a)` , and inserting th eknown temperature , we find
`T_(4) = (k_(a)( 2.0 L_(a)))/( ( 5.0 k_(a))L_(a)) ( 25^(@) C- 20^(@) C ) + ( - 10^(@)C)`
`= - 8.0^(@)C`