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A slab of stone of area of 0.36 m^(2) an...

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)`)

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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)`..(4.11)
If m is the mass of ice and L the latent heat of fusion , then
`Q=mL`..(4.12)
Equating (4.11) and (4.12), we have
`mL= (kA(T_(1)-T_(2))t)/(d)`
or `k = (mLd)/(A(T_(1)-T_(2))t)`...(4.13)
Givenm= 4.8 kg,
d=10 cm =0.1m,
`A=3600cm^(2) = 0.36 m^(2)`
`T_(1)= 100^(@)C, T_(2)=0^(@)C and t=1 hour = (60 xx 60)s`.
We know that `L=80cal g^(-1) = 80,000 cal kg^(-1)`
`=80,000 xx 4.2 J kg^(-1) = 3.36 xx 10^(5) j kg^(-1)`
Substituting these values in equation (4.13) and solving , we get
`k= 1.24 j s^(-1)m^(-1)``""^(@)C^(-1) or 1.24 Wm^(-1)K^(-1)`
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PHYSICS GALAXY - ASHISH ARORA-HEAT TRANSFER -Illustrative Example 4.4
  1. A slab of stone of area of 0.36 m^(2) and thickness 0.1 m is exposed o...

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