Calculate the enthalpy change of freezing of `1.0` mol of water at `10^(@)C` to ice at `-10^(@)C, Delta_(fus)H=6.03 kJ mol^(-1)` at `0^(@)C`.
`C_(P)[H_(2)O(l)]=75.3 J mol^(-1) K^(-1)`
`C_(P)[H_(2)O(s)]=36.8 J mol^(-1) K^(-1)`

Explanation

(i) Heat change required to lower the temperature of one mol of water from `10.0^(@)C` to `0^(@)C`-
`{:("Formula",,,,DeltaH_(1)=nCpDeltaT),("Substitution",,,,DeltaH_(1)=1.0xx75.3xx10=-753J mol^(-1)):}`
(ii) Heat change required to convert `1` mol of `H_(2)O(l)` at `)^(@)C` to `H_(2)O(s)` at `0^(@)C`
`because DeltaH_(2)=DeltaH_("fusion")`
`:. DeltaH_(2)= -6.03KJ mol^(-1)`
(iii) Heat change required to change `1` mole of ice from `0^(@)C` to `-10.0^(@)C`
`{:("Formula",,,,DeltaH_(3)=nCpDeltaT),("Substitution",,,,Delta H_(3)= -36.8xx10xx1=-368J mol^(-1)):}`
`ul("Total heat change")=DeltaH_(1)+DeltaH_(2)+DeltaH_(3)`
`=(-0.753-6.03-0.368)KJ mol^(-1)`
`:.` Total enthalpy `= -7.151 KJ mol^(-1)`