One end of a rod of length `20cm` is inserted in a furnace at `800K` The sides of the rod are coverd with an insulating meterial and the other end emits radiation like a black boy The temperature of this end is `750K` in the steady state The temperature of the surrounding air is `300K` Assuming radiation to be the only imporatant mode of energy transfer between the surrounding and the open end of the rod, find the thermal conductivity of the rod Stefan's constant `sigma = 6.0 xx 10^(8) W m^(-2) K^(-4)` .

Explanation

According to the equation.

Energy received per second at end (1)
`Q_(1) = (KA(T_(1)-T))/(L)`.............(i)
Energy radiated per second at end(2)
`Q_(2) = (sigmaA(T^(4)-T_(0)^(4))`...................(ii)
In steady state
`Q_(1) = Q_(2)`
`(KA(T_(1)-T_(2))/(L) = (sigmaa(T_(4)-T_(0))/(T_(1)-T)`
`K=(sigmaL(T_(4)-T_(0)^(4))/(T_(1)-T)`
`=6 xx(6 xx 1-^(-8) xx 0.2[(600)^(4)-(300)^(4)]/(800-600)`
`=(1.2 xx 10^(-8) xx 6^(4)-3^(4) xx 10^(8))/((200)`
`=(6 xx10^(-3) (1296 -81)) = 7.29 `W//m^(@)C`