How much heat is lost when changing 65 g of water vapor (H2O) at 421 K to ice at 139 K?

-146.9 kJ

-204.6 kJ

-48 kJ

-219.4 kJ

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.