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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 one hour 4.8 kg of ice is melted. The thermal conductivity of slab is: [Given latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹]

1. 1.24 J/m/s/°C
2. 1.29 J/m/s/°C
3. 2.05 J/m/s/°C
4. 1.02 J/m/s/°C

Correct answer: 1.29 J/m/s/°C

Solution

The heat conducted through the slab is used to melt the ice. Using the formula for heat conduction, Q = kAΔTt/d, and equating it to the heat required to melt the ice, Q = mL, we solve for k. Substituting values: m = 4.8 kg, L = 3.36 × 10⁵ J/kg, A = 0.36 m², ΔT = 100°C, t = 3600 s, d = 0.1 m, we get k = 1.29 J/m/s/°C.

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