Answer :
The boiling point of a substance depends on the pressure exerted on it. In this case, the water is maintained at atmospheric pressure, which is approximately 101325 Pa. The boiling point of water at atmospheric pressure is 100 degrees Celsius or 373.15 Kelvin. Therefore, the initial temperature of the water is equal to its boiling point.
Given:
Mass of water (m) = 10 kg
Pressure (P) = 101325 Pa
Initial temperature (T) = 373.15 K
Since the water is already at its boiling point, it is in a state of equilibrium between the liquid and vapor phases. At this point, the water starts to vaporize or boil, converting from liquid to vapor.
To calculate the amount of energy required to convert the entire mass of water to vapor at its boiling point, we can use the formula:
Energy (Q) = mass (m) × latent heat of vaporization (L)
The latent heat of vaporization (L) for water at atmospheric pressure is approximately 2.26 × 10^6 J/kg.
Substituting the values into the formula, we get:
Q = 10 kg × 2.26 × 10^6 J/kg = 22.6 × 10^6 J
Therefore, the energy required to completely vaporize the 10 kg of water at its boiling point is 22.6 × 10^6 J.
It's important to note that during the phase change from liquid to vapor, the temperature of the water remains constant at the boiling point until all the water is converted to vapor. The energy supplied to the water is utilized to break the intermolecular bonds and convert the liquid water into gaseous water vapor.
Therefore, to completely vaporize 10 kg of liquid water at its boiling point of 373.15 K and atmospheric pressure, it would require an energy input of 22.6 × 10^6 J.
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