Question

How much heat is required to boil $83.0{\text{ g}}$ of water at its boiling point?

Answer 1

When water is boiled at its boiling point i.e. at ${100^ \circ }{\text{C}}$ liquid water is converted to water vapour. Here, the phase change occurs from liquid to gas. To solve this we must use the equation that gives the relationship between heat required and the heat of vaporization of liquid water.

Complete solution:
We know that when water is boiled at its boiling point i.e. at ${100^ \circ }{\text{C}}$ liquid water is converted to water vapour. Here, the phase change occurs from liquid to gas.
During phase change, the temperature remains constant but heat transfer occurs. We are given that water is boiled at its boiling point. This suggests that liquid water at ${100^ \circ }{\text{C}}$ is converted to gaseous water i.e. water vapour at ${100^ \circ }{\text{C}}$.
Thus, the equation to calculate the heat required is as follows:
$Q = m\Delta {H_{{\text{vap}}}}$
Where $Q$ is the heat required,
$m$ is the mass,
$\Delta {H_{{\text{vap}}}}$ is the latent heat of vaporization.
The latent heat of vaporization is the heat required to change one mole of liquid substance to gaseous form under standard atmospheric pressure i.e. one atmosphere. The value of latent heat of vaporization at ${100^ \circ }{\text{C}}$ is $2256{\text{ kJ/kg}}$.
Substitute $83.0{\text{ g}} = 83.0 \times {10^{ - 3}}{\text{ kg}}$ for the mass, $2256{\text{ kJ/kg}}$ for the latent heat of vaporization. Thus,
$Q = 83.0 \times {10^{ - 3}}{\text{ kg}} \times 2256{\text{ kJ/kg}}$
$Q = 187.248{\text{ kJ}}$
Thus, the heat required to boil $83.0{\text{ g}}$ of water at its boiling point $187{\text{ kJ}}$.

Note: Remember that water boils at ${100^ \circ }{\text{C}}$. The latent heat of vaporization is the heat required to change one mole of liquid substance to gaseous form under standard atmospheric pressure i.e. one atmosphere. The value of latent heat of vaporization at ${100^ \circ }{\text{C}}$ is $2256{\text{ kJ/kg}}$.