A block of ice at `0^(@)C` rest on the upper surface of the slab of stone of area `3600cm^(2)` and thickness of `10cm`. The slab is exposed on the lower surface to steam at `100^(@)C`. If `4800g` of ice is melted in one hour, then calculate the thermal conductivity of stone.
(Given the latent heat of fusion of ice `=80cal//g`)

Options

A. `K=2.96xx10^(-3)cal//cms.^(@)C`
B. `K=1.96xx10^(3)cal//cm s.^(@)C`
C. `K=0.96xx10^(3) cal//cm s.^(@)C`
D. none of the above

Explanation

(a) We know `theta=(KADeltaT)/(Deltax)t`
Quantity of heat flowing through the stone in one hour
`=4800xx80=384xx10^(3)`cal
`K=(theta)/txx(Deltax)/(DeltaT)1/A=(384xx10^(3))/3600xx10/100xx1/3600`
`K=2.96xx10^(3)`cal/cm `s.^(@)C`