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NEET PHYSICSThermodynamicsEasy

Question

When 1 kg1 \text{ kg} of ice at 0C0^{\circ}\text{C} melts into water at 0C0^{\circ}\text{C}, the resulting change in its entropy, taking the 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=qrevT\Delta S = \frac{q_{rev}}{T} where qrevq_{rev} is the heat absorbed reversibly and TT is the absolute temperature .

  1. Calculate Heat Absorbed (qq): The heat required to melt mass mm is given by q=m×Lfq = m \times L_f, where LfL_f is the latent heat of fusion. Given: m=1 kg=1000 gm = 1 \text{ kg} = 1000 \text{ g} and Lf=80 cal/gL_f = 80 \text{ cal/g}. q=1000 g×80 cal/g=80,000 calq = 1000 \text{ g} \times 80 \text{ cal/g} = 80,000 \text{ cal}

  2. Convert Temperature to Kelvin: T=0C=273 KT = 0^{\circ}\text{C} = 273 \text{ K}

  3. Calculate Entropy Change (ΔS\Delta S): ΔS=80,000 cal273 K293 cal/K\Delta S = \frac{80,000 \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 Thermodynamics. 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.

PHYSICSThermodynamicscirctextccirctextcresultingchangeentropy

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