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Snezhnost [94]
3 years ago
11

106 grams of liquid water are in a cylinder with a piston maintaining 1 atm (101325 Pa) of pressure. It is exactly at the boilin

g 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?
Physics
1 answer:
xeze [42]3 years ago
4 0

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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