Calculate change of entropy of universe:

5Kg of ice at 100C is kept in atmosphere that is at 30⁰C. 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_iC_idT + M_iL_i + m_wC_wdT

= 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 - 10⁰C to 0⁰C + Entropy change as ice melts at 0⁰C to water at 0⁰C + Entropy change of water as it gets heated from 0⁰C to 30⁰C

 

= 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

 

Posted Date: 10/16/2012 6:42:50 AM | Location : United States

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