A cylindrical rod of 50 cm length and having `1 cm^(2)` cross sectional area isused as a conducting material between an ice bath at` 0^(@)C` and a vacuum chamber at `27^(@)C` as shown in figure. The end of rod which is inside the vacuum chamber behaves like a black body and is at temperature `17^(@)C` in steady state. Find the thermal conductivity of the material of rod and rate at which ice is melting in the ice bath. Given that latent heat of fusion of ice is `3.35 xx l0^(5) J//kg`..

Explanation

It is given that the system is in steady state. This means that any part of rod is not absorbing any heat. So heat absorbed by the end of the rod which is in vacuum chamber by radiation is fully conducted to the ice bath through the rod. Thus we have
Rate of that conduction through the rod =
Rate of heat absorption by radiation for the vacuum chamber
or (kA(T_(B)-T_(A)))/(.L)`
`= sigmaA(T_(vc)^(4)-T_(B)^(4))`
or (k(17^(@)C - 0^(@)C)/(0.5)` ` = 5.67 xx 10^(-8) [(300)^(4)-(290)^(4)]`
or `k = (5.67 xx 10^(-8)[(300)^(4) - (290)^(4)] xx 0.5)/(17)`
`= 1.713 W//m^(@)C`
Using this value of k we can find the rate of heat obtained by the ice bath as
`(dQ)/(dt) =(kA(T_(A)-T_(B)))/(l)`
`= (1.713 xx 1xx 10^(-4) xx17)/(0.5)`
`= 5.82 xx 10^(-3) J//s`
This heat is used to melt the ice in ice bath. If m mass of ice is being melted per second, then we have
`(dQ)/(dt) = mL`
or `5.82 xx 10^(-3) = m xx 3.35 xx 10^(5)`
or `m=1.74 xx 10^(-8) kg//s`.