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

The quantity of heat required convert 36.82 grams of water at 100°C to steam is 83.09 KJ

From the question,

We are to determine the quantity of heat required to convert 36.82 g of water at 100°C to steam.

Using the formula

Q = nΔHvap

Where

Q is the quantity of heat

n is the number of moles

and ΔHvap is the heat of vaporization

First, we will determine the number of moles of water present

Mass of water present = 36.82 g

Using the formula

Molar mass of water =  18.01528 g/mol

Number of moles of water present =

Number of moles of water present = 2.0438 moles

Now, using the formula

Q = nΔHvap

Q = 2.0438 × 40.657

Q = 83.09 KJ

Hence, the quantity of heat required convert 36.82 grams of water at 100°C to steam is 83.09 KJ

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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