Calculate change of entropy of universe:
5Kg of ice at 100C is kept in atmosphere that is at 30^{0}C. Calculate change of entropy of universe when it melts and comes into thermal equilibrium with atmosphere. Take latent heat of fusion as 335KJ/kg and specific Heat of ice is half of that of water.
Sol: Mass of ice, m = 5Kg
Temperature of ice = -100C = 263K Temperature of atmosphere = 300C = 303K
Heat absorbed by ice from the atmosphere = Heat in solid phase + latent heat + heat in liquid phase
= m_{i}C_{i}dT + M_{i}L_{i }+ m_{w}C_{w}dT
= 5 × 4.187/2 ( 0 + 10) + 5 × 335 + 5 × 4.187 × ( 30 - 0)
= 104.675 + 1675 + 628.05
Q = 2407.725KJ
Entropy change of atmosphere ( ? s)atm = - Q/T = -2407.725/303
(?s)_{atm} = - 7.946KJ/k
Entropy change of ice(?s)_{ice}
= Entropy change as ice gets heated from - 10^{0}C to 0^{0}C + Entropy change as ice melts at 0^{0}C to water at 0^{0}C + Entropy change of water as it gets heated from 0^{0}C to 30^{0}C
= 5 x 1.7409 = 8.705KJ
Entropy of universe = Entropy change of atmosphere (?s)_{atm} + Entropy change of ice( ? s)_{ice}
= - 7.946KJ/k + 8.705KJ
= 0.7605329KJ/kg .......ANS