how much heat is needed to vaporize a 400.0 g of liquid water at 310.0 K?
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To find out how much heat is needed to vaporize a given mass of liquid water, we need to use the concept of heat of vaporization.
The heat of vaporization is the amount of heat energy required to convert a given substance from its liquid phase to its vapor phase, at a constant temperature and pressure.
In the case of water, the heat of vaporization is approximately 40.7 kJ/mol.
To calculate the heat required, we'll go through these steps:
1. Convert the given mass of water from grams to moles.
- The molar mass of water (H2O) is approximately 18.02 g/mol.
- Divide the given mass (400.0 g) by the molar mass to get the number of moles:
Moles = Mass / Molar mass = 400.0 g / 18.02 g/mol
2. Calculate the heat required using the molar quantity.
- Multiply the number of moles by the heat of vaporization:
Heat = Moles × Heat of vaporization
Let's perform the calculations:
1. Moles = 400.0 g / 18.02 g/mol ≈ 22.206 mol
2. Heat = 22.206 mol × 40.7 kJ/mol ≈ 903.58 kJ
Therefore, approximately 903.58 kJ of heat is needed to vaporize 400.0 g of liquid water at 310.0 K.