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ASHA 777 [7]
3 years ago
7

10 kg of liquid water is in a container maintained at atmospheric pressure, 101325 Pa. The water is initially at 373.15 K, the b

oiling point at that pressure.The latent heat of water -> water vapor is 2230 J/g. The molecular weight of water is 18 g.103 J of heat is added to the water.1)How much of the water turns to vapor?mass(vapor)=
Physics
1 answer:
rewona [7]3 years ago
5 0

Answer:

m=0.0462\ g of water is converted into vapour.

Explanation:

Given:

  • mass of water, m_w=10\ kg
  • pressure conditions, P=101325\ Pa
  • temperature conditions, T=373.15\ K
  • latent heat of vapourization of water, L=2230\ J.g^{-1}
  • amount of heat supplied to the water, 103\ J

<u>Now using the equation of heat considering latent heat only:</u>

<u>(</u>since water already at boiling point at atmospheric temperature<u>)</u>

Q=m.L

103=m\times 2230

m=0.0462\ g of water is converted into vapour.

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