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 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)
A) 1.24 J/m/s/°C
B) 1.29 J/m/s/°C
C) 2.05 J/m/s/°C
D) 1.02 J/m/s/°C
Explanation
[Solution from source] Heat rate = mL/t = (4.8 × 3.36 × 10⁵)/3600 = 448 W. Setting KAΔT/d = 448 gives K × 0.36 × 100/0.1 = 360K = 448, so K ≈ 1.24 J m⁻¹ s⁻¹ °C⁻¹. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.