How much heat is necessary to vaporize 100 g of water at 100 °C to form steam at 100 °C?
what would the final answer be in because we cant just put the answer on the work.
To determine the amount of heat necessary to vaporize water, we need to use the concept of heat of vaporization. The heat of vaporization is the amount of energy required to convert a substance from its liquid phase to its gaseous phase at a constant temperature.
For water at its boiling point of 100 °C, the heat of vaporization is approximately 40.7 kJ/mol. To calculate the amount of heat needed to vaporize a specific mass of water, we need to use the molar mass of water, which is approximately 18.015 g/mol.
Now, let's calculate the amount of heat needed to vaporize 100 g of water at 100 °C:
1. Convert the mass of water (100 g) to moles:
Number of moles = mass / molar mass
= 100 g / 18.015 g/mol
≈ 5.550 mol
2. Multiply the number of moles by the heat of vaporization:
Heat = Number of moles × Heat of vaporization
= 5.550 mol × 40.7 kJ/mol
≈ 225.405 kJ
Therefore, approximately 225.405 kJ of heat are required to vaporize 100 g of water at 100 °C to form steam at 100 °C.