how many joules are required to evaporate 1 gram of boiling water?

q = mass x delta Hvap.

335 To calculate the amount of energy required to evaporate 1 gram of boiling water, we need to consider the specific heat capacity and the heat of vaporization of water.

1. Specific Heat Capacity: The specific heat capacity (C) of water is the amount of energy required to raise the temperature of water by 1 degree Celsius. For water, the specific heat capacity is approximately 4.18 Joules per gram per degree Celsius (J/g°C).

2. Heat of Vaporization: The heat of vaporization (ΔHvap) of water is the amount of energy required to convert 1 gram of liquid water to vapor at a constant temperature. The heat of vaporization of water is approximately 2260 Joules per gram (J/g).

To determine the energy required to evaporate 1 gram of boiling water, we need to consider two scenarios:

Scenario 1: Raising the temperature of water from 100°C to its boiling point (100°C):

To calculate the energy required for this, we use the specific heat capacity formula:
Energy = mass (grams) x specific heat capacity x temperature change
Energy = 1 gram x 4.18 J/g°C x (100°C - 100°C)
Energy = 0 Joules (No energy is required to raise the temperature to boiling point since it's already at the boiling point).

Scenario 2: Vaporizing the boiling water at its boiling point:

To calculate the energy required for vaporization, we use the heat of vaporization formula:
Energy = mass (grams) x heat of vaporization
Energy = 1 gram x 2260 J/g
Energy = 2260 Joules

Therefore, the amount of energy required to evaporate 1 gram of boiling water is 2260 Joules.