One end of a rod of length `L` and cross-sectional area `A` is kept in a furance of temperature `T_(1)`. The other end of the rod is kept at a temperature `T_(2)`. The thermal conductivity of the material of the rod is `K` and emissity of the rod is e, it is given that `T_(2) = T_(S) + DeltaT lt lt T_(S)` being the temperature of the surroundings. if `DeltaT alpha (T_(1) -T_(S))` , find the proportionally constant that heat is lost only by radiation at the end where the temp, of the rod is `T_(2)`.

Explanation

Rate of Heat gained by rod through conduction , `Q = (KA(T_(1) - T_(2)))/(L)`

Rate of Heat lost by rod through radiation , `Q_(2) = e A sigma (T_(2)^(4) - T_(s)^(4))`
In steady state `Q_(1) = Q_(2)`
`therefore (KA(T_(1)-T_(2)))/(L) = e A sigma (T_(2)^(4) - T_(s)^(4))..........(1)`
Given `T_(2) = T_(s) + DeltaT " " therefore T_(2)^(4) = (T_(s) + DeltaT)^(4) = T_(S)^(4) + (1 + (DeltaT)/(T_(s)))^(4)`
As `Delta T lt lt T_(s)` , so using bionmial theorem .
` T_(2)^(4) = T_(s)^(4) ( 1+(4Delta T)/(T_(s))) rArr T_(2)^(4) - T_(s)^(4) = 4T_(s)^(3) DeltaT `
substituting this value in (1) , we get `(KA(T_(1) - T_(2)))/(L) = e A sigma. (4T_(s)^(3) DeltaT) ` Also `T_(2) = T_(s) + Delta T`
`therefore (KA(T_(1) - T_(s) - DeltaT))/(L) = e A sigma (4T_(s)^(3) DeltaT) rArr (K (T_(1) - T_(s)))/(L) - (K)/(L) DeltaT = 4esigma T_(s)^(3) Delta T`
(or) `(K(T_(1) - T_(s)))/(L) = (4esigma T_(s)^(3) + (K)/(L)) DeltaT , Delta T = (K(T_(1) - T_(S)))/((4 e sigmaL T_(s)^(3) + K))` But given `Delta T = C(T_(1) - T_(s))`
`therefore ` Constant of proportioanlity C is `(K)/(K + 4 e sigma L T_(s)^(3))`