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Q.

A slab of stone of area 0.36 m2 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 56 minutes 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)

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a

2.05 J/m/s/°C

b

1.02 J/m/s/°C

c

1.33 J/m/s/°C

d

1.29 J/m/s/°C

answer is C.

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Detailed Solution

∆Q∆t=KA∆T∆x=mL∆t⇒K=mL∆xA∆T∆tK=4.8×3.36×105×0.156×60×100×0.36=1.33 J/m/s/oC
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