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Liono4ka [1.6K]
3 years ago
13

When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2.20×106J/kg2.20×10

Physics
1 answer:
Grace [21]3 years ago
5 0

Answer:

Explanation:

a ) When 1 kg water is boiled at constant pressure of 1  atm , its volume increases by following volume

(.824 - .001 )m³

.823 m³

work done by steam  = increase in volume x pressure

.823 x 10⁵ J

Heat added

=  latent heat of vaporization x mass

= 2260000 J x 1

= 22.6 x 10⁵ J

Increase in internal energy of gas

= heat added - work done by gas

= (22.6 - .823) x 10⁵ J

= 21.777 x 10⁵ J .

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