A slab of stone of area of 0.36 m² and thickness 0.1 m is exposed on the lower surface to steam at 100^oC. A block of ice at 0^o 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 = 3.36 x 10⁵ J Kg^-1)

1.24 J/m/s/^oC

1.29 J/m/s^oC

2.05 J/m/s^o C

1.02 J/m/s^oC

Solution

1.24 J/m/s/^oC

δQ over δt space equals space fraction numerator K A over denominator L end fraction left parenthesis T subscript 1 minus T subscript 2 right parenthesis Q space equals fraction numerator K A over denominator L end fraction space left parenthesis T subscript 1 minus T subscript 2 right parenthesis t Q space equals m L subscript f fraction numerator K A over denominator L end fraction left parenthesis T subscript 1 minus T subscript 2 right parenthesis space t space equals m L subscript f K space equals fraction numerator m L subscript f L over denominator A left parenthesis T subscript 1 minus T subscript 2 right parenthesis t end fraction K equals fraction numerator 4.8 space x 3.36 space x space 10 to the power of 5 space x 0.1 over denominator 0.36 space x space 100 space x space 3600 end fraction space equals space fraction numerator 4.8 space x space 3.36 over denominator 0.36 space x space 36 end fraction space equals space 1.24 space J divided by m divided by s divided by C presuperscript o

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