The heat of vaporization of water (also called the enthalpy of vaporization) is the heat absorbed when liquid water becomes water vapor at constant temperature and pressure. It is denoted \( \Delta H_{\text{vap}} \) and is positive for vaporization because the process is endothermic.
Key value for water (normal boiling point)
At the normal boiling point (\(100^\circ\text{C}\), \(1\,\text{atm}\)), a widely used textbook value is:
\[ \Delta H_{\text{vap}}(\text{H}_2\text{O},\,100^\circ\text{C}) \approx 40.65\,\text{kJ/mol} \]
Unit conversion that is often requested
Using the molar mass of water \(M(\text{H}_2\text{O}) = 18.015\,\text{g/mol}\), the same heat of vaporization of water can be written per gram:
\[ \Delta H_{\text{vap}} \approx \frac{40.65\times 10^3\,\text{J/mol}}{18.015\,\text{g/mol}} \approx 2256\,\text{J/g} \]
This is the familiar “latent heat of vaporization” near \(100^\circ\text{C}\) (about \(2.256\,\text{kJ/g}\)).
How to use the heat of vaporization of water in calculations
For vaporization at the boiling point (temperature constant during the phase change), the heat required is:
\[ q = n\,\Delta H_{\text{vap}} \]
If mass is given, convert to moles with \(n = \frac{m}{M}\). If a per-gram value is used instead, the same idea becomes \(q = m\,L_v\), where \(L_v\) is in \(\text{J/g}\).
Visualization: where \(\Delta H_{\text{vap}}\) appears on a heating curve
Worked example using the heat of vaporization of water
Problem. How much heat is needed to vaporize \(25.0\,\text{g}\) of water at \(100^\circ\text{C}\) and \(1\,\text{atm}\)?
Step 1: Convert grams of water to moles.
\[ n = \frac{m}{M} = \frac{25.0\,\text{g}}{18.015\,\text{g/mol}} \approx 1.387\,\text{mol} \]
Step 2: Apply \(q = n\Delta H_{\text{vap}}\).
\[ q = (1.387\,\text{mol})(40.65\,\text{kJ/mol}) \approx 56.4\,\text{kJ} \]
Common pitfall
The heat of vaporization of water applies only to the phase change at constant temperature. If water starts below \(100^\circ\text{C}\), heating the liquid to \(100^\circ\text{C}\) requires an additional step using \(q = m c \Delta T\), and only then does \(q = n\Delta H_{\text{vap}}\) apply.
Quick practice (with answers)
Assume water is already at \(100^\circ\text{C}\) and vaporizes at \(1\,\text{atm}\) so that \( \Delta H_{\text{vap}} \approx 40.65\,\text{kJ/mol} \).
| # | Prompt | Answer (final) |
|---|---|---|
| 1 | Find \(q\) to vaporize \(10.0\,\text{g}\) of water. | \(q \approx 22.6\,\text{kJ}\) |
| 2 | How many moles can be vaporized by \(81.3\,\text{kJ}\) of heat? | \(n = 2.00\,\text{mol}\) |
| 3 | What mass of water can be vaporized by \(50.0\,\text{kJ}\)? | \(m \approx 22.1\,\text{g}\) |
| 4 | Compute \(q\) for \(0.750\,\text{mol}\) of water vaporized. | \(q \approx 30.5\,\text{kJ}\) |
Check: each answer comes from \(q = n\Delta H_{\text{vap}}\) and \(n = \frac{m}{18.015\,\text{g/mol}}\), using the heat of vaporization of water at \(100^\circ\text{C}\).