The heat necessary for water vaporization is determined by the equation:
\[
Q = n \Delta H_{\text{vap}}
\]
Here,
- \( n = 2 \, \text{mol} \) represents the mole quantity.
- \( \Delta H_{\text{vap}} = 40.79 \, \text{kJ/mol} \) is the enthalpy of vaporization.
The calculation for the heat is as follows:
\[
Q = 2 \times 40.79 = 81.58 \, \text{kJ}
\]
Therefore, 81.58 kJ of heat is needed to vaporize 2 moles of water.