Physics, 22.04.2020 03:29 zekrader18

When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2.20×106J/kg2.20×10
6
J/kg and the boiling point is 120∘C120
∘
C. At this pressure, 1.00 kg of water has a volume of 1.00×10−3m31.00×10
−3
m
3
, and 1.00 kg of steam has a volume of 0.824m30.824m
3
. (a) Compute the work done when 1.00 kg of steam is formed at this temperature. (b) Compute the increase in internal energy of the water.

Answers: 3

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