In order to calculate the enthalpy change for H2O at 5∘C to ice at −5∘C, we need to calculate the enthalpy change of all the transformation involved in the process.
(a) Energy change of 1mol,H2O(l), at 5∘C→1mol,H2O(l),0∘C
(b) Energy change of 1mol,H2O(l) at 0∘C→1mol,H2O(s)(ice),0∘C
(c) Energy change of 1mol, Ice (s), at 0∘C→1mol, Ice (s),−5∘C
Total ΔH
=CP[H2O(l)]ΔT+ΔH freezing +CP[H2O(s)]ΔT
=(75.3Jmol−1K−1)(−5)K+(−6×103Jmol−1)+(36.8Jmol−1K−1)(−5)K
ΔH=−6.56kJmol−1 (exothermic process)
So, ΔH=6.56kJmol−1.