Question

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)?

Answer 1

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