Calculate the amount of heat energy required to evaporate 27.5 g of water at 100.0ºC. (molar heat of vaporization of liquid water = 4.07 × 10⁴ J/mol)

Question

Quizplus · Accepted Answer

To find the amount of heat energy required to evaporate 27.5 g of water, we use the formula: $q = n \times \Delta H_{vap}$, where $n$ is the number of moles of water and $\Delta H_{vap}$ is the molar heat of vaporization. First, calculate the number of moles of water: $n = \frac{mass}{molar\ mass} = \frac{27.5\ g}{18.015\ g/mol} \approx 1.527\ mol$. Then, calculate the heat energy: $q = 1.527\ mol \times 4.07 \times 10^4\ J/mol \approx 62155.9\ J = 62.1\ kJ$.

Calculate the Amount of Heat Energy Required to Evaporate 27