Question

How many joules of energy are required to vaporize 423 grams of water at 100 degrees Celsius and 1 atm?

Answer 1

To vaporize 423 grams of water at 100 degrees Celsius and 1 atm, 954,750 J of energy are required.

Step-by-step explanation:

To calculate the amount of energy required to vaporize water at a specific temperature and pressure, we need to use the heat of vaporization. The heat of vaporization for water is approximately 2,250 J per gram. Since we have 423 grams of water, we can calculate the total energy required by multiplying the mass of water by the heat of vaporization:

Energy Required = mass of water * heat of vaporization

Energy Required = 423 g * 2,250 J/g = 954,750 J

Answer 2

955980 j

Step-by-step explanation:

Given data:

Mass of water = 423 g

Temperature = 100°C

Amount of energy required = ?

Solution:

For one gram of water heat of energy required to convert it into vapors is 2260 j/g.

Amount of heat required for 423 g.

423 g × 2260 j/g = 955980 j

The amount of energy released when one gram steam is condensed into water is -2260 j.

For 423 g water vapor to condensed energy released.

423 g × -2260 j/g = -955980 j