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106 grams of liquid water are in a cylinder with a piston maintaining 1 atm (101325 Pa) of pressure. It is exactly at the boiling point of water, 373.15 K. We then add heat to boil the water, converting it all to vapor. The molecular weight of water is 18 g/mol and the latent heat of vaporization is 2260 J/g. 1) How much heat is required to boil the water?

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Rau7star

Answer:

239.55 KJ

Explanation:

Given:

Mass 'm' = 106 g

Latent heat of vaporization'L'= 2260 J/g.

Molecular weight of water'M' = 18 g/mol

Pressure 'P' = 101325 Pa

Temperature 'T' = 373.15 K

Using the formula of phase change, in order to determine the amount of heat required, we have

Q = mL

Q = 106 x 2260

Q = 239560J = 239.55 KJ

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