A wall has two layers A and B, each made of different material. Both the layers have the same thickness. The thermal conductivity of the material of A is twice that of B . Under thermal equilibrium, the temperature difference across the wall is `36^@C.` The temperature difference across the layer A is

Options

A. `12^(@)C`
B. `18^(@)C`
C. `6^(@)C`
D. `24^(@)C`

Answer

A

Explanation

Here, `K_(A) = 2K_(B), T_(A)-T_(B) = 36^(@)C`
Let T be the tempeature at the junction.
As `((Delta Q)/(Delta t))_(A) = ((Delta Q)/(Delta t))_(B)`
`:. (K_(A) A(T_(A)-T))/(x) = (K_(B) A (T- T_(B)))/(x)`
or `2K_(B) (T_(A)-T) = K_(B) (T-T_(B)) or 2(T_(A) - T)= T-T_(B)`
Add `(T_(A) -T)` on both sides, we get
`3(T_(A)-T) = T_(A) -T + T-T_(B) or 3(T_(A)-T) = T_(A) -T_(B)` or `T_(A)-T = (T_(A) -T_(B))/(3) = (36)/(3) = 12^(@)C`
`:.` Temperature difference across the layer `A = T_(A) -T = 12^(@)C`