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

Answer

A

Explanation

Rate of heat given by steam = Rate of heat taken by ice
`implies (dQ)/(dt)=(KA(100-0))/(Deltax)=m(dL)/(dt)`
`implies (K xx100 xx 0.36)/(0.1)=(4.8 xx 3.36 xx 10^5)/(60 xx 60)`
`K=1.24 J s^(-1) m^(-1) C^(-1)`