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NEET PHYSICSThermal Properties of MatterMedium

Question

A slab of stone of area of 0.36 m20.36\text{ m}^2 and thickness 0.1 m0.1\text{ m} is exposed on the lower surface to steam at 100C100^{\circ}\text{C}. A block of ice at 0C0^{\circ}\text{C} rests on the upper surface of the slab. In one hour 4.8 kg4.8\text{ kg} of ice is melted. The thermal conductivity of slab is : (Given latent heat of fusion of ice =3.36×105 J kg1=3.36 \times 10^5\text{ J kg}^{-1})

A

1.24 J/m/s/C1.24\text{ J/m/s/}^{\circ}\text{C}

B

1.29 J/m/s/C1.29\text{ J/m/s/}^{\circ}\text{C}

C

2.05 J/m/s/C2.05\text{ J/m/s/}^{\circ}\text{C}

D

1.02 J/m/s/C1.02\text{ J/m/s/}^{\circ}\text{C}

Step-by-Step Solution

Heat transferred through the stone slab is used to melt the ice. According to the law of thermal conduction, the heat transferred QQ is given by: Q=KA(ΔT)tdQ = \frac{KA(\Delta T)t}{d} where KK is thermal conductivity, AA is area, ΔT\Delta T is temperature difference, tt is time, and dd is thickness. Also, heat required to melt the ice is Q=mLfQ = mL_f, where mm is mass and LfL_f is latent heat of fusion. Equating both expressions for QQ: KA(ΔT)td=mLf\frac{KA(\Delta T)t}{d} = mL_f K=mLfdA(ΔT)tK = \frac{m L_f d}{A(\Delta T)t} Substitute the given values: m=4.8 kgm = 4.8\text{ kg}, Lf=3.36×105 J/kgL_f = 3.36 \times 10^5\text{ J/kg}, d=0.1 md = 0.1\text{ m}, A=0.36 m2A = 0.36\text{ m}^2, ΔT=100C0C=100C\Delta T = 100^{\circ}\text{C} - 0^{\circ}\text{C} = 100^{\circ}\text{C}, t=1 hour=3600 st = 1\text{ hour} = 3600\text{ s}. K=4.8×3.36×105×0.10.36×100×3600K = \frac{4.8 \times 3.36 \times 10^5 \times 0.1}{0.36 \times 100 \times 3600} K=161280129600=1.24 J/m/s/CK = \frac{161280}{129600} = 1.24\text{ J/m/s/}^{\circ}\text{C}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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