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Question

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0C to ice at 10.0C
fusH=6.03 KJ mol1 at 0C
Cp[H2O(1)]=75.3Jmol1K1
Cp[H2O(s)]=36.8Jmol1K1

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Solution

The process is represented below.
Liquid(100C)ΔH1−−Liquid(00C)ΔH2−−solid(00C)ΔH3−−Solid(100C
ΔH1=nCP[H2O(l)]×ΔT=1×75.3×10=753J/mol=0.753kJ/mol
ΔH2=n(ΔfusH0)=1×6.03=6.03kJ/mol
ΔH3=nCP[H2O(s)]×ΔT=1×36.8×10=368J/mol=0.368kJ/mol
ΔH1+ΔH2+ΔH3=0.7536.030.368=5645kJ/mol

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