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

Answer

B

Explanation

`Q = (KA (DeltaT)t)/(Delta x) = m L`
`K = (mL Delta x)/(A (Delta T) t) rArr (4.8 xx 3.36 xx 10^5 xx 0.1)/(0.36 xx 100 xx 3600)`
`K = 56/45 = 1.24 J//s //""^@C`