One end of a steel rod `(K=46Js^(-1)m^(-1)C^(-1))` of length `1.0m` is kept in ice at `0^(@)C` and the other end is kept in boiling water at `100^(@)C` . The area of cross section of the rod is `0.04cm^(2)` . Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice `=3.36xx10^(5)Jkg^(-1)` .

Answer

A

Explanation

`K=46 J//m-s-^(@)C, l=1 m,`
`A=0.04cm^(2)=4xx10^(-6)m^(2)`
`(Q)/(T)=(kA(T_1-T_2))/(l)`
`=(46xx4xx10^(6)xx(100-0))/(1)`
`=184xx(10^(-4))`
`mL==184xx(10^(-4))`
`m=(184xx10^(-4))`/L`
`m=(184xx10^(-4))/(3.36xx10^(5))`
`=5.5xx10^(-5)g.`