The enthalpy of vaporization of water is 40.79 kJ/mol. This means that it takes 40.79 kJ of energy to vaporize one mole of water.

We can use this information to calculate the amount of heat needed to vaporize 24.40 grams of water. First, we need to calculate the number of moles of water in 24.40 grams:

$$24.40 \text{ g H}_2\text{O} \times \frac{1 \text{ mol H}_2\text{O}}{18.02 \text{ g H}_2\text{O}} = 1.354 \text{ mol H}_2\text{O}$$

Next, we can use the enthalpy of vaporization to calculate the amount of heat needed to vaporize this amount of water:

$$1.354 \text{ mol H}_2\text{O} \times 40.79 \text{ kJ/mol} = 55.14 \text{ kJ}$$

Therefore, it takes 55.14 kJ of heat to completely vaporize 24.40 grams of water at its boiling point.