To find the mass of liquid water required to absorb a certain amount of energy upon boiling, we can use the concept of specific heat of vaporization. The specific heat of vaporization is the amount of energy needed to convert one kilogram of a substance from a liquid to a gas at constant temperature and pressure.
The question states that we need to absorb 5.23x10^4 kJ of energy (which is 52300 kJ) on boiling. We will assume that the specific heat of vaporization for water at its boiling point is 2260 kJ/kg, which is a commonly cited value for water at 100°C (or 373.15 Kelvin).
The formula to calculate the mass (m) of water required to absorb this energy (Q) during the phase change is:
\[ Q = m \times L \]
where:
- \( Q \) is the energy required (in kJ),
- \( m \) is the mass of the water (in kg),
- \( L \) is the specific heat of vaporization (in kJ/kg).
Rearranging the formula to solve for \( m \), we get:
\[ m = \frac{Q}{L} \]
Now, we plug in the values:
\[ Q = 52300 \, \text{kJ} \]
\[ L = 2260 \, \text{kJ/kg} \]
So,
\[ m = \frac{52300 \, \text{kJ}}{2260 \, \text{kJ/kg}} \]
\[ m = \frac{52300}{2260} \, \text{kg} \]
Now we perform the division to calculate the mass:
\[ m \approx 23.14159292 \, \text{kg} \]
Therefore, approximately 23.14 kg of liquid water is required to absorb 52300 kJ of energy on boiling.
