Question

Calculate the amount of heat (in kilojoules) required to vaporize 2.20 kg of water at its boiling point (ΔHvap = 40.7 kJ/mol at 100∘C).

Answer 1

The amount of heat required to vaporize 2.20 kg of water at its boiling point is 4963.2 kilojoules, calculated by multiplying the mass of the water by the latent heat of vaporization (Q = m × Lv).

Step-by-step explanation:

To calculate the amount of heat required to vaporize 2.20 kg of water at its boiling point, we can use the formula Q = m × Lv where 'm' is the mass of water in kilograms and 'Lv' is the latent heat of vaporization. Given that the latent heat of vaporization of water is 2256 kJ/kg at 100°C, we can plug in our values:
Q = 2.20 kg × 2256 kJ/kg
This calculation will give us the heat in kilojoules needed to vaporize the water. Performing the calculation:
Q = 2.20 kg × 2256 kJ/kg = 4963.2 kJ

Therefore, 4963.2 kilojoules of heat is required to vaporize 2.20 kg of water at its boiling point.