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 Q1 = m xdHfus = 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 Q1 + 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 =
The entropy change of water =
Hence,
Or, Energy required to melt the ice Q1 = m xdHfus = 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 Q1 + 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 =
The entropy change of water =
Hence,