While `1mol` of ice melts at `0^(@)C` and at constant pressure of `1atm, 1440 cal` of heat are absorbed by the system. The molar volume of ice and water are `0.0196` and `0.0180L` respectively. Calculate `DeltaH` and `DeltaU`.

Explanation

`DeltaH = 1440 cal`
`DeltaH = DeltaU +P(DeltaV)`
`[H_(2)O(s)] Ice rarr H_(2)O(l)`
`P DeltaV = (76 xx 13.6 xx 981)/(4.81 xx 10^(-7)) (18 -19.6) (CGS units)`
`= 0.0388cal`
Volume of ice `= 0.0196 L = 19.6 mL`
Volume of `H_(2)O = 0.0180 L = 18 mL`
`DeltaH = DeltaU + 0.0388`
`DeltaU = 1440 - 0.0388 = 1439.96 cal`