The heat of vaporization of water at 100°C is 40.66 kJ/mol.Calculate the quantity of heat that is absorbed/released when 7.00 g of steam condenses to liquid water at 100°C.

Question

Quizplus · Accepted Answer

The heat of vaporization is the amount of heat required to convert a liquid into a gas or vice versa at a constant temperature. When steam condenses into water, it releases heat. To find the quantity of heat released, first convert the mass of water (7.00 g) to moles using the molar mass of water (18.015 g/mol). Then, multiply by the heat of vaporization (40.66 kJ/mol).$$ \text{Moles of water} = \frac{7.00 \, \text{g}}{18.015 \, \text{g/mol}} \approx 0.3886 \, \text{mol} $$$$ \text{Heat released} = 0.3886 \, \text{mol} \times 40.66 \, \text{kJ/mol} \approx 15.8 \, \text{kJ} $$Therefore, 15.8 kJ of heat are released during the condensation of 7.00 g of steam to liquid water at 100°C.

The Heat of Vaporization of Water at 100°C Is 40