Calcualte the enthalpy change on freezing of 1.0 mole of water at `10.0^(@)C` to ice at `-10^(@)`C. `Delta_(fs)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 Jmol^(-1)K^(-1)`

Explanation

Total `DeltaH = ` ( 1 mol water at `10^(@)C rarr ` 1 mol of waterat `0^(@)C) + (` 1 mol water at `0^(@)C rarr 1 `mol ice at `0^(@)C ) +(` 1 mol ice at `0^(@) C rarr ` 1 mol ice at `- 10^(@)C) ` `(DeltaT = T_(2) - T_(1))`
`=C_(p) [H_(2)O(l) ]xx DeltaT + DeltaH _("freezing") +C_(p)[H_(2)O (s) ] xx DeltaT `
`= ( 75.3J K^(-1) mol^(-1) ) ( 0- 10) K( - 6.03 kJ mol^(-1)) + ( 36.8 J K^(-1) mol^(-1) ) ( - 10 K) ( DeltaH_("freezing" ) =-DeltaH_("fusion") )`
`=-753 J mol^(-1) - 603 kJ mol^(-1) - 368 J mol^(-1)`
`= - 0.753 kJ mol^(-1) - 6.03 kJ mol^(-1) -0.368kJ mol^(-1) kJ mol^(-1) = -7.151kJ mol^(-1)`
Note `:` Directly also, as in each step , heat is evolved,each step will have a negative sign with `DeltaH`.