Calculate the total heat required to vaporize 1.5 kg of water at 100oC to water vapor at 100oC?
q = mass H2O x heat vaporization = ?
To calculate the total heat required to vaporize 1.5 kg of water at 100°C to water vapor at 100°C, we need to consider two steps:
1. Heating the water from 100°C to its boiling point:
The specific heat capacity of water is approximately 4.186 J/g°C.
To calculate the heat required to raise the temperature of the water, we use the formula:
Q1 = m * c * ΔT
where Q1 is the heat energy, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Plugging in the values:
Q1 = 1.5 kg * 4.186 J/g°C * (100°C - 100°C)
Q1 = 0 J
Since there is no temperature change during this step, no heat energy is required.
2. Vaporizing the water at the boiling point:
The heat required to vaporize a substance is known as the latent heat of vaporization. For water, it is approximately 2260 J/g.
To calculate the heat required to vaporize the water, we use the formula:
Q2 = m * L
where Q2 is the heat energy, m is the mass of the water, and L is the latent heat of vaporization.
Plugging in the values:
Q2 = 1.5 kg * 2260 J/g
Q2 = 3390 J
Therefore, the total heat required to vaporize 1.5 kg of water at 100°C to water vapor at 100°C is 3390 joules (J).