A slab of stone of area of `0.36 m^(2)` and thickness `0.1 m` is exposed on the lower surface to steam at `100^(@)C`. A block of ice at `0^(@)C` rests on the upper surface of the slab. In one hour `4.8 kg` of ice is melted. The thermal conductivity of slab is
(Given latent heat of fusion of ice `= 3.63 xx 10^(5) J kg^(-1)`)

Options

A. `1.29J//m//s//^(@)C`
B. `2.05J//m//s//^(@)C`
C. `1.02J//m//s//^(@)C`
D. `1.24J//m//s//^(@)C`

Answer

D

Explanation

`(dQ)/(dt)=(KA)/(L)(T_(1)-T_(2))`
`Q=(KA)/(L)(T_(1)-T_(2))t`
`Q=mL_(f)`
`(KA)/(L)(T_(1)-T_(2)t=mL_(f)`
`K=(mL_(f)(L))/(A(T_(1)-T_(2))t)`
`K=(4.8xx3.36xx10^(5)xx0.1)/(0.36xx100xx3600)J//m//s//^(@)C`
`=(4.8xx3.36)/(0.36xx36)`
`1.24J//m//s//^(@)C`