AIPMT MAINS 2012 Physics Thermal Conductivity MCQ Question
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)
1.24 J/m/s/°C
1.29 J/m/s/°C
2.05 J/m/s/°C
1.02 J/m/s/°C
Correct Answer
Detailed Explanation
The thermal conductivity can be calculated using the formula Q = kA(T₁-T₂)t/d, where Q is the heat transferred, A is the area, T₁ and T₂ are the temperatures, t is the time, and d is the thickness. Solving for k gives the correct answer.
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