The heat of vaporization of water at $100^{\circ} \mathrm{C}$ is $40.66 \mathrm{~kJ} / \mathrm{mol}$. Calculate the quantity of heat that is absorbed/released when $9.00 \mathrm{~g}$ of steam condenses to liquid water at $100^{\circ} \mathrm{C}$.
$20.3 \mathrm{~kJ}$ of heat are released.
$81.3 \mathrm{~kJ}$ of heat are absorbed.
$81.3 \mathrm{k}$ of heat are released.
$20.3 \mathrm{~kJ}$ of heat are absorbed.

Question

The heat of vaporization of water at $100^{\circ} \mathrm{C}$ is $40.66 \mathrm{~kJ} / \mathrm{mol}$. Calculate the quantity of heat that is absorbed/released when $9.00 \mathrm{~g}$ of steam condenses to liquid water at $100^{\circ} \mathrm{C}$.
$20.3 \mathrm{~kJ}$ of heat are released.
$81.3 \mathrm{~kJ}$ of heat are absorbed.
$81.3 \mathrm{k}$ of heat are released.
$20.3 \mathrm{~kJ}$ of heat are absorbed.

Accepted Answer

***B*** ∻Key Concept∻ ⚹Heat of vaporization and mass-to-mole conversion⚹ ∻Explanation∻ ⚹To find the heat absorbed or released during the condensation of steam, you need to convert the mass of water to moles using its molar mass (approximately $18.02 \mathrm{~g/mol}$ for water), and then multiply by the heat of vaporization. The calculation is as follows: $\frac{9.00 \mathrm{~g}}{18.02 \mathrm{~g/mol}} \times 40.66 \mathrm{~kJ/mol} = 20.3 \mathrm{~kJ}$. Since condensation is the opposite of vaporization, the heat is released, not absorbed. However, the question asks for the heat absorbed, which is the same magnitude but opposite in sign, hence $20.3 \mathrm{~kJ}$ of heat are absorbed.⚹◊What is the heat of vaporization of water at $100^{\circ}$C typically?⍭ Generate me a similar question◊

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