10g ice at 0 degree celcius is added to 20g water at 90 degree celcius in a thermally insulated flask of negligible heat capacity. The heat of fusion of ice is 6kJ/mol. What is the final temperature and entropy of the system? (Cp of water = 75.42 J/mol)

Here, heat gained by ice = heat lost by warm water 
Or, Energy required to melt the ice Q= m xdH​fus = 10 x 6 KJ/mol = 60000J
Energy required to warm the melted ice Q2 =m x​ Cp xdT
​Here, dT = (Tf -00C )
So, Q2 = 10 x 75.42 x​(Tf -00C )
Now , ​energy to cool the liquid water Q3 = m xCp xdT = 20 x 75.42 J/mol x (900C-Tf)
​So, we have Q+ Q2 = Q3 
Or, 60000 J + ​10 x 75.42 x​(Tf -00C ) = ​20 x 75.42 J/mol x (900C-Tf
Solving we get, the final temperature = 34.270C
Hence, the entropy change dS = dSi + dSw
​The entropy change of ice =​

ΔSi= mi(HfusTfus+C lnTfinalTfus )= 10 ( 6000273 + 75.42 ln 307.27273)Or,ΔSi=10 ( 6000273 + 75.42 x 0.118) = 10 x 8429.5237 =308.77 J/K

The entropy change of water =
ΔSw= mw C lnTfinalTwater = 20 x 75.42 ln307.27363 = 20 x 75.42 x (-0.166)Or, ΔSw = -250. 4 J/K
Hence, 
ΔS = ΔSi + ΔSw = 308.77 J/K + (-250.4 J/K) = 58.37 J/K
 

  • -1
What are you looking for?