How much heat is released when 115 g of steam at 100.0°C is cooled to ice at -15.0°C?

Question

How much heat is released when 115 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).

Would someone be willing to show me the steps? I'm a little lost to this problem

Roger the Mole · Answer

Since all the given constants are in moles, convert the mass of water to moles:
(115 g) / (18.01532 g H2O/mol) = 6.3835 mol H2O

(40.67 kJ/mol) x (6.3835 mol) = 259.617 kJ from condensing the steam

(75.4 J/(mol⋅°C)) x (6.3835 mol) x (100 - 0) °C = 48132 J = 48.131 kJ from cooling the water to 0°C

(6.01 kJ/mol) x (6.3835 mol) = 38.365 kJ from freezing the water

(36.4 J/(mol⋅°C)) x (6.3835 mol) x (0 - (-15.0)) °C = 3485 J = 3.485 kJ from cooling the ice to -15°C

259.617 kJ + 48.131 kJ + 38.365 kJ + 3.485 kJ = 350 kJ total

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How much heat is released when 115 g of steam at 100.0°C is cooled to ice at -15.0°C?

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