Determine the amount of energy required to completely vaporize 250.0mL sample of water at room temperature ( assume 20 degrees).
To completely vaporize 250.0mL of water at room temperature (20 degrees Celsius), we first need to raise the temperature of the water to its boiling point and then provide enough energy to convert it from liquid to vapor.
1. Raise the temperature of the water from 20 degrees Celsius to its boiling point:
The specific heat capacity of water is 4.18 J/g°C. Since we have 250.0mL of water, and the density of water is approximately 1g/mL, the mass of the water is 250.0g.
The temperature change required is:
ΔT = 100°C (boiling point of water) - 20°C (initial temperature) = 80°C
The energy required to raise the temperature of the water is:
q1 = mcΔT
q1 = 250.0g * 4.18 J/g°C * 80°C
q1 = 83,600 J
2. Convert the liquid water to vapor:
The enthalpy of vaporization of water is approximately 2260 J/g.
The energy required to vaporize the water is:
q2 = m * Hv
q2 = 250.0g * 2260 J/g
q2 = 565,000 J
3. Total energy required:
Total energy = q1 + q2
Total energy = 83,600 J + 565,000 J
Total energy = 648,600 J
Therefore, it would require 648,600 Joules of energy to completely vaporize a 250.0mL sample of water at room temperature (20°C).