Respuesta :
The heat change will be
Moles of water = mass / Molar mass = 65/ 18 = 3.61 mol
specific heat of ice =2.09J /g C
specific heat of water = 4.184 J/g C
Specific heat of vapour= 2.01 /g C
Heat of fusion = 3.33X10⁵ J /kg = 333 J /g
Heat of vaporization = 2.26 X10⁶J/kg = 2260J/g
Q1 = heat change when vapours get cooled to 373.15 K
Q2 = heat change when vapours get converted to liquid water
Q3 = heat change when liquid water cools to 273.15 K
Q4= heat change when liquid water freezes to ice
Q5= heat change when ice cools from 273.15K to 139 K
Q1= mass of water X specific heat of vapours X change in temperature
Q1 = 65 X 2.01 /g C X (421-373.15) = 6251.60 J = 6.252 kJ
Q2 = heat of vaporization X mass = 2260 X 65 = 146900 = 146.9 kJ
Q3 = mass X specific heat of water X change in temperature =
Q3 = 65 X 4.184 X (373.15-273.15) = 65 X 4.184 X 100 = 27196 J = 27.196kJ
Q4 = heat of fusion X mass =333X65 = 21645 J = 21.645 kJ
Q5 = mass X specific heat of ice X change in temperature
Q5 = 65 X 2.09 X (273.15-139) = 18224.3 J = 18.224 kJ
Total energy = 6.252 +146.9+27.196+ 21.645+ 18.224 = 220.217
As this is energy released so it will be expressed in negative
-220.217
from the given options the correct answer will be -219.4 kJ
The answer is little different as the reference values of specific heats or enthalpy may vary.
