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NEET PHYSICSThermal Properties of MatterMedium

Question

When 1 kg1\text{ kg} of ice at 0C0^{\circ}\text{C} melts to water at 0C0^{\circ}\text{C}, the resulting change in its entropy, taking latent heat of ice to be 80 cal/g80\text{ cal/g}, is

A

8×104 cal/K8 \times 10^4\text{ cal/K}

B

80 cal/K80\text{ cal/K}

C

293 cal/K293\text{ cal/K}

D

273 cal/K273\text{ cal/K}

Step-by-Step Solution

The change in entropy (ΔS\Delta S) during a phase transition at a constant temperature is given by the formula ΔS=QT\Delta S = \frac{Q}{T} [1], where QQ is the heat absorbed and TT is the absolute temperature. Given: Mass of ice, m=1 kg=1000 gm = 1\text{ kg} = 1000\text{ g} Latent heat of fusion of ice, L=80 cal/gL = 80\text{ cal/g} Absolute temperature, T=0C=273 KT = 0^{\circ}\text{C} = 273\text{ K} The total heat absorbed to melt the ice is Q=m×L=1000 g×80 cal/g=80000 calQ = m \times L = 1000\text{ g} \times 80\text{ cal/g} = 80000\text{ cal}. Now, calculating the change in entropy: ΔS=80000 cal273 K293 cal/K\Delta S = \frac{80000\text{ cal}}{273\text{ K}} \approx 293\text{ cal/K}.

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.

PHYSICSThermal Properties of Mattercirctextccirctextcresultingchangeentropy

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