A slab of stone of area 3600 `cm^(2)` and thickness 10 cm is exposed on the lower surface to steam at `100^(@)C`. A block of ice at `0^(@)C` rests on upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of the stone In `js^(-1) m^(-1) k^(-1)` is (latent heat of ice `= 3.36 xx 10^(5) j//kg`)

Options

A. 12
B. 10.5
C. 1.02
D. 1.24

Answer

d

Explanation

Given area of stab of stone (A) = `3600 cm^(2)`
`= 3600 xx 10^(-4) m^(2)`
Thickness of stone stab = 10 cm `= 10 xx 10^(2) m`
Temperature difference `(Delta theta) = 100^(@)C`
we know that,
`(Q)/(t) = (KA Delta theta)/(I)`
`implies (4.8 xx 3.36 xx 10^(5))/(60 xx 60) = (K xx 3600 xx 10^(-4) xx 100)/(10 xx 10^(-2))`
`implies K = (4.8 xx 3.36 xx 10^(5) xx 10 xx 10^(-2))/(3600 xx 10^(-4) xx 100 xx 60 xx 60)`
`implies K = (161280)/(129600)`
`implies K = 1.24 js^(-1) m^(-1) K^(-1)`