A wall has two layers A and B each made of different materials. The thickness of both the layers is the same. The thermal conductivity of A, `K_(A) = 3K_(B)`. The temperature different across the wall is `20^(@)C` in thermal equilibrium

Options

A. The temperature difference across A is `15^(@)C`.
B. Rate of heat transfer across A is more than across B
C. Rate of heat transfer across both is same.
D. Temperature difference aross A is `5^(@)C`

Answer

D

Explanation

`(Q)/(At) =(K(theta_(1)-theta_(2)))/(d)`= constant
`:. K_(A)((theta_(1)=theta)/(d)) = K_(B)((theta-theta_(2))/(d))`
`(K_A)/(K_B) = (theta-theta_(2))/(theta_(1)-theta) or 3 = (theta-theta_(2))/(theta_(1)=theta)`
or, `3theta_(1)+theta_(2) = 4 theta` ...(1)
Given `theta_(1)-theta_(2) = 4 theta` ..(2)
Solving (1) and (2) we have, `theta-theta_(2)=15^(@)C`
`:. theta_(1) - theta = theta_(1)-theta_(2)+theta_(2) - theta`
`=(theta_(1)-theta_(2)) - (theta-theta_(2))`
`=20^(@)C - 15^(@)C = 5^(@)C`.