Question

How much heat, in kJ, is required to convert 36.82 grams of water at 100°C to steam? (ΔHvap = 40.657 kJ/mol.)

Answer 1

Answer:

83.35 KJ

Explanation:

Step 1:

Data obtained from the question. This includes:

Mass of water = 36.82 g

Heat of vaporization (ΔHvap) =

40.657 kJ/mol.

Heat (Q) =?

Step 2:

Conversion of 36.82 g water to mole.

Molar Mass of H2O (M) = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O (m) = 36.82 g

Mole of H2O (n) =?

Number of mole = Mass/Molar Mass

Mole of H2O = 36.82/18

Mole of H2O = 2.05 mole

Step 3:

Determination of the heat required to convert 36.82g ( i.e 2.05 moles) of water at 100°C to steam. This is illustrated below:

Q = mΔHv

Q = nΔHv

Q = 2.05mol x 40.657 kJ/mol.

Q = 83.35 KJ