The heat of vaporization of water at 100 °C is 40.66 kJ mol^-1.Calculate the quantity of heat that is absorbed/released when 9.00 g of steam condenses to liquid water at 100 °C. A)20.3 kJ of heat are absorbed. B)20.3 kJ of heat are released. C)81.3 kJ of heat are absorbed. D)81.3 kJ of heat are released.

Question

Quizplus · Accepted Answer

The heat of vaporization is the amount of heat required to convert 1 mole of a substance from liquid to gas at its boiling point. When steam condenses to liquid water, it releases heat. The molar mass of water is approximately 18.015 g/mol, so 9.00 g of water is 9.00 g / 18.015 g/mol = 0.5 mol. The quantity of heat released when 0.5 mol of steam condenses is 0.5 mol * 40.66 kJ/mol = 20.33 kJ. Therefore, 20.3 kJ of heat are released.

The Heat of Vaporization of Water at 100 °C Is