The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x respectively. Temperatures on the opposite faces of composite slab are `T_(1) and T_(2)` where `T_(2)gtT_(1)`, as shown in fig. what is the rate of flow of heat through the slab in a steady state?

Options

A. 1
B. `1/2`
C. `2/3`
D. `1/3`

Answer

D

Explanation

Let the temperature of common interface be T K
Rate of heat flow `H = (Q)/(t) = (KA DeltaT)/(l)`
`:. H = (Q/t)_(1) = (2KA(T-T_(1)))/(4x)`
and `H_(2) = (Q/t)_(2) = (KA(T_(2) - T))/(x)`
In steady state the rate of heat flow should be same in whole system i.e.,
`H_(1) = H_(2)`
or `(2Ka(T-T_(1)))/(4x) = (KA(T_(2)-T))/(x)`
`(T-T_(1))/(2) = T_(2) - T` or `T - T_(1) = 2T_(2) - 2T`
`T = (2T_(2) + T_(1))/(3) "....."(i)`
Hence, heat flow composite slab is
`H = [(Ka(T_(2)-T))/(x)] = (KA)/(x) (T_(2) - (2T_(2) - T_(1))/(3))` [From eqn. (i)]
`= (KA)/(3x) (T_(2) - T_(1))`
According to question, `H = [(A(T_(2) - T_(1))K)/(x)]f"....."(iii)`
By comparing eqs. (ii) and (iii), we get `f = 1/3`