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NEET PHYSICSEasy

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}

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