Question

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 `=3.36xx10^(5) J kg^(-1)`

Answer 1

Assuming that heat loss from the sides of the slab is negligible, the amount of heat flowing through the slab is
`Q=(kA(T_1-T_2)t)/(d)` ..(i)
If m is the mass of ice and L the latent heat of fusion, then
`Q=mL` ..(ii)
From Eqs. (i) and (ii) we have
`mL=(kA(T_1-T_2)t)/(d)`
or `k=(mLd)/(A(T_1-T_2)t)` ..(iii)
Given `m=4.8kg`,`d=10cm=0.1m`,`A=3600cm^2=0.36m^2`
`T_1=100^@C`,`T_2=0^2C` and `t=1h=(60xx60)s`
We know that `L=80 cal//g=80000cal//kg=80000xx4.2J//kg=3.36xx10^5J//kg`
Substituting these values in Eq. (iii) and solving, we get
`k=1.24J//s//m^@C` or `1.24W//m//K`