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NEET PHYSICSMedium

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}.

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