Final answer:
The energy required to vaporize 400.0 g of water at its boiling temperature is calculated using the heat of vaporization and is 902.56 kJ. This does not match the provided options, so the answer is D. None of the above.
Step-by-step explanation:
To calculate the amount of energy required to vaporize 400.0 g of water at its boiling point, we can use the given heat of vaporization of water, which is 2,256.4 kJ/kg. Since we have 400.0 g (or 0.400 kg) of water, we use the formula Q = mLv, where Q is the energy in kilojoules, m is the mass in kilograms, and Lv is the latent heat of vaporization in kJ/kg.
Q = 0.400 kg × 2,256.4 kJ/kg = 902.56 kJ.
The heat of vaporization of water is the amount of energy needed to convert a given amount of water from its liquid state to its gaseous state at the same temperature. It is given as 2,256.4 kJ/kg. To find the energy required to vaporize 400.0 g of water at its boiling temperature, we can use the equation: Qv = mLv. Substituting the given values, we get: Qv = (0.4 kg) x (2,256.4 kJ/kg) = 902.56 kJ.
Therefore, the energy required to vaporize 400.0 g of water at its boiling temperature is 902.56 kJ. This does not match any of the options provided, so the correct answer is D. None of the above.