How much heat is necessary to vaporize 100 g of water at 100 °C to form steam at 100 °C?
Don't you read your posts to see if they have been answered? I answered your post the first time.
To determine how much heat is necessary to vaporize 100 g of water at 100 °C to form steam at 100 °C, we need to consider the concept of heat transfer and the specific heat capacity of water.
First, it's important to note that at 100 °C, water is already at its boiling point, so in this case, we are dealing with a phase change from liquid water to steam. To undergo this phase change, the water needs to absorb a specific amount of heat energy known as the heat of vaporization.
The heat of vaporization for water is approximately 2,260 kilojoules per kilogram (kJ/kg). To find the amount of heat energy required to vaporize 100 g of water, we need to convert the mass to kilograms.
Step 1: Convert mass to kilograms:
100 g = 0.1 kg (since 1 kg = 1000 g)
Step 2: Calculate the heat energy required:
Heat energy (Q) = mass (m) × heat of vaporization (H)
Q = 0.1 kg × 2,260 kJ/kg
Q ≈ 226 kJ (rounded to the nearest whole number)
Therefore, approximately 226 kilojoules of heat energy is necessary to vaporize 100 g of water at 100 °C to form steam at 100 °C.