A wall is made of equally thick layers P and Q of different materials. Thermal conductivity of Q is half of that of the P . In the steady state , If the temperature difference across the wall is `24^(@) C` , then the temperature difference across the layer 'P' is .......

Options

A. `12^(@) C`
B. `16^(@) C`
C. `4^(@) C`
D. `8^(@)C`

Answer

D

Explanation

In the steady state, the rate of flow of heat through layer P and Q is same
`(k_(p)a(T_(1) - T_(0)))/(x) = (k_(Q) a (T_(0) - T_(2)))/(x)`
Here,
`K_(Q) =(K_(p))/(2)`
`K_(p) = 2 K_(Q)`
`(2K_(Q) a(T_(1)-T_(0)))/(x) = (K_(Q) a(T_(0) - T_(2)))/(x)`
`2 (T_(1) - T_(0)) = T_(0) - T_(2)`
`2T_(1) - 2T_(0) = T_(0) -T_(2)`
`2T_(1) - 3T_(0) + T_(2) = 0 " ". . . (i)`
According to the question,
`T_(1) - T_(2) = 24^(@) C " " . . . (ii)`
Now , from Eq. (i) .
` 2T_(1) - 3T_(0) + T_(1) - 24 = 0 `
`3T_(1) - 3T_(0) = 24`
`T_(1) - T_(0) = 8^(@) C`