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)/(4esigmaLT_(s)^(3)+K))`

Explanation

`((K)/(4esigmaLT_(s)^(3)+K))`
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`
`thereforeT_(2)^(4)=(T_(s)+DeltaT)^(4)=T_(s)^(4)(1+(DeltaT)/(T))^(4)`
Using binomial expansion, we have
`T_(2)^(4)=T_(s)^(4)(1+4(DeltaT)/(T_(s)))(asDeltaTltltT_(S))`
`thereforeT_(2)^(4)-T_(s)^(4)=4(DeltaT)(T_(s)^(3))`
Substituting in Eq. (i), we have `(K(T_(1)-T_(s)-DeltaT))/(L)=4esigmaT_(s)^(3)DeltaT`
or `(K(T_(1)-T_(s)))/(L)=(4esigmaT_(s)^(3)+(K)/(L))DeltaTthereforeDeltaT=(K(T_(1)-T_(s)))/((4esigmaLT_(s)^(3)+K))`
Comparing with the given relation, proportianality constant `=(K)/(4esigmaLT_(s)^(3)+K)`