Question

Determine the amount of heat (in kJ) required to vaporize 1.05 kg of water at its boiling point. For water, ΔHvap = 40.7 kJ/mol (at 100∘C).

Answer 1

The amount of heat required to vaporize 1.05 kg of water at its boiling point is approximately 2372.23 kJ, calculated by converting the mass to moles and then multiplying by the heat of vaporization.

Step-by-step explanation:

To determine the amount of heat required to vaporize 1.05 kg of water at its boiling point, we first need to convert the mass of water to moles. The molar mass of water (H2O) is approximately 18.015 g/mol. Therefore:

• Convert 1.05 kg to grams: 1.05 kg × 1000 g/kg = 1050 g
• Calculate the number of moles: 1050 g ÷ 18.015 g/mol ≈ 58.29 moles
• Multiply the moles by the heat of vaporization (ΔHvap): 58.29 moles × 40.7 kJ/mol

Now, calculate the total heat (Q) required:

Q = 58.29 moles × 40.7 kJ/mol ≈ 2372.23 kJ.

Therefore, the amount of heat required to vaporize 1.05 kg of water at its boiling point is approximately 2372.23 kJ.