One end of 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 at temperature `T_(2)`. The thermal conductivity of the material of the rod is `K` and emissivity of the rod is `e`. It is given that `T_(2)=T_(S)+DeltaT` where `DeltaT lt lt T_(S)`, `T_(S)` being the temperature of the surroundings. If `DeltaT prop (T_(1)-T_(S))`, find the proportionality constant. Consider that heat is lost only by radiation at the end where the temperature of the rod is `T_(2)`.

Answer

`(K) /(4eAsigmaLT_(S)^(3) + K)`

Explanation

Rate of heat conduction through rod rate of the heat lost from right end of the rod.
`therefore (KA(T_(1) - T_(2)))/(L) = eAsigma(T_(2)^(4) - T_(s)^(4))` ……. (i)
Given that `T_(2) = T_(s) + DeltaT`
`therefore T_(2)^(4) = (T_(s) + DeltaT)^(4) = T_(s)^(4) (1 + (DeltaT)/(T_(s)))^(4)`
Using binomial expansion, we have
`T_(2)^(4) = T_(s)^(4)(1 + 4(DeltaT)/(T_(s))) ("as" DeltaT lt lt T_(s))`
`therefore T_(2)^(4) - T_(s)^(4) = 4(DeltaT)(T_(s)^(3))`
Substituting in Eq.(i) , we have
`(K(T_(1) - T_(s) - DeltaT))/(L) = 4 esigma T_(s)^(3)DeltaT`
`implies K(T_(1) - T_(s))/(L) = (4esigmaT_(s)^(3) + (K)/(L))DeltaT`
`therefore DeltaT = K(T_(1) - T_(s))/((4esigmaLT_(s)^(3) + K))`
Comparing with the given relation, proportional constant = `(K)/(4esigmaLT_(s)^(3) + K)`