One end of a brass rod of length 2.0 m and cross section `1cm^2` is kept in stream at `100^@C` and the other end in ice at `0^@C`. The lateral suface of the rod is covered by heat insulator. Determine the amount of ice melting per minute. Thermal conductivity of brass is `110 W//m-K` and specific latent heat of fusion of ice is `80 cal//g`.

Explanation

Let us find the amount of heat absorbed by ice per minute,
`A=1xx10^-4m^2,Deltat=60s,K=110 W//mK`,
Using `DeltaQ=KADeltaTDeltat//L` we have
`DeltaQ=(110)(1xx10^-4)((100)(60))/(2)=33J`
Now, if m be the mass of ice melting per minute, then, from `Delta=mL`,
`m=(DeltaQ)/(L)=(33J)/(80xx4.2 J//g)=0.098g`