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)`)

Answer

[`1.245 Wm^(-1) .^(@)C^(-1)]`

Explanation

Here, `A=0.36 m^(2), x=0.1 m,`
`T_(1)-T_(2)=100-0=100^(@)C, t=1 h=60xx60 s`
Mass of ice melted, `m=4.8 kg`
Latest heat of ice, `L=336xx10^(3) j kg^(-1)`
Heat required to melt the ice is
`a=mL=4.8xx336xx10^(3) j`
Now, `Q=(KA(T_(1)-T_(2))t)/(x)=mL`
or `K=(x mL)/(A(T_1)-T_(2))t=0.1xx(4.8)xx(336xx10^(3))/(0.36xx(100)xx(60xx60)) =1.245 Wm^(-1) .^(@)C^(-1)`