determine the total heat required in KCal to convert 1.25g of ice at 0 degree celsius to water vapour at 100 degree celsius
The total heat required to convert 1.25 g of ice into vapor is a sum of fusion enthalpy, heat capacity and enthalpy of vaporization of water. Thus, enthalpy of fusion (to melt ice) is 335 J/g * 1.25 g = 419 J; heat capacity (to heat water to 100 C) 4.184 J/g/K * 1.25 g * 100 K = 523 J; enthalpy of vaporization (heat required for vaporization of boiled water) 2260 J/g * 1.25 g = 2825 J. The total heat is 419 J + 523 J + 2825 J = 3767 J. In kcal, 3767 J/4.184 J/cal = 900 cal or 0.9 kcal.