The lower surface of a slab of stone of face-area `3600cm^(2)` and thickness `10cm` is exposed to steam at `100^(@)C` . A block of ice at `0^(@)C` rests on the upper surface of the slab. `4.8g` of ice melts in one hour. Calculate the thermal conductivity of the stone. Latent heat of fusion of ice `=3.36 xx 10^(5) J kg^(-1)`

Explanation

The amount of heat trsnsferred though the slab to the ice in one hour is
`Q=(4.8xx10^(-3)kg)xx(3.36xx10^(5)J kg^(-1)` . `=4.8xx336J` . Using the equation `Q=(KA(theta_(1)-theta_(2))t)/(x)` , `4.8xx336J=(k(3600cm^(2))(100^(@)C)xx(3600s))/(10cm)` . Or, `K=1.24xx10^(-3)Wm^(-1)`^(@)C^(-1)` .