The heat of vaporization of water at 100°C is 40.66 kJ/mol.Calculate the quantity of heat that is absorbed/released when 6.00 g of steam condenses to liquid water at 100°C.
A) 13.5 kJ of heat are absorbed.
B) 13.5 kJ of heat are released.
C) 122 kJ of heat are absorbed.
D) 122 kJ of heat are released.
E) 244 kJ of heat are released.

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, we use the formula: $ q = m \times \Delta H_{vap} $, where $ q $ is the heat absorbed or released, $ m $ is the mass of the substance in moles, and $ \Delta H_{vap} $ is the heat of vaporization. First, convert the mass of water to moles using its molar mass (18.015 g/mol for water): $ 6.00 \, \text{g} \times \frac{1 \, \text{mol}}{18.015 \, \text{g}} = 0.333 \, \text{mol} $. Then, calculate the heat released: $ 0.333 \, \text{mol} \times 40.66 \, \text{kJ/mol} = 13.54 \, \text{kJ} $. Therefore, 13.5 kJ of heat are released during the condensation of 6.00 g of steam at 100°C.

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