100 g ice at 0°C is put in water in bucket at 50°C .on complete melting change in entropy will be( assuming no change in temperature of water)?
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Change of entropy of ice
S1 = ΔQ / T = mL / T
= 80 × 100 / (0+273)
= 8 × 10^3 / 273 cal/K
Change of entropy of water
S2 = ΔQ / T = mL / T
= 80 × 100 / (273+50)
= 8 × 10^3 / 323 cal/K
Total change of entropy
S2 - S1 = 8 × 10^3 / 323 - 8 × 10^3 / 273
= - 4.5 cal/K
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