calculate the enthalpy change on freezing 1 mol of water at 10degree Celsius to ice at -10 degree Celsius . delta(H)of fusion= 6.03 kJ/mol at 0 degree Celsius. Cp[water(liquid)]=75.3 J/mol/K Cp[water(solid)]=36.8 J/mol/K

Asked by Mahesh Padmanabh | 28th Jan, 2011, 10:43: PM

Expert Answer:

ΔH =Cp .ΔT   (for changing 1mol of liquid water from 10 -0 degree celcius)
      =75.3 x10 = .753kj
from (
0 to 10)  
36.8 x-10 =- .368kj
Total enthalpy change= .753 +(-6.03) + (-.368)
                               =-5.645kjmol-1 

Answered by  | 31st Jan, 2011, 09:50: AM

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