Calculate the amount of heat (in kJ) required to convert 62.6 g of water to steam at 100°C. (The molar heat of vaporization of water is 40.79 kJ/mol.)

The amount of heat (in kJ) required to convert 62.6 g of water to steam at 100°C is 141.86kJ

The formula for calculating the required amount of heat is expressed according to the equation;

Given the following parameters

Substitute the given parameters into the formula to have:

Hence the amount of heat (in kJ) required to convert 62.6 g of water to steam at 100°C is 141.86kJ

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The amount of heat (in kJ) required to convert 62.6 g of water to steam at 100°C is 141.74 kJ

Explanation:

The latent heat of vaporization is the heat required to transform a giving mass of a liquid completely to vapor at its boiling point thus

The mass of water = 62.6 g

The latent heat of vaporization of water = 40.79 kJ/mol

The quantity of heat = Q = LH·m  where  Q = required heat energy,

However the molar mass of water = 18.01528 g/mol

Therefore 62.6 g of water contains (62.6 g)/(18.01528 g/mol) = 3.47 moles

∴ Q = 3.47 moles × 40.79 kJ/mol = 141.74 kJ

The required amount of heat to convert 62.6 g of water to steam at 100 °C  is 141.74 kJ

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