How much heat is released when 105 g of steam at 100.0°C is cooled to ice at - 15.0°C? The enthalpy of vaporization of water is 40.67 kJ/mol, the enthalpy of fusion for water is 6.01 kJ/mol, the molar heat capacity of liquid water is 75.4 J / (mol ⋅ °

c., and the molar heat capacity of ice is 36.4 J / (mol ⋅ °

c.

1) (Hvap) (moles of water) = 236.9783574kJ

(40.67) (105/18.02)

2) (change in temperature) (mass) (Cliquid) = 43.9345172kJ

(100) (105/18.02) (75.4) / 1000

3) (Hfus) (moles of water) = 35.01942286kJ

(6.01) (105/18.02)

4) (change in temperature) (mass) (Csolid) = 3.181465039kJ

(15) (105/18.02) (36.4) / 1000

Total released=319.1137625kJ