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

Respuesta :

Answer:

[tex]m=0.0462\ g[/tex] of water is converted into vapour.

Explanation:

Given:

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

Now using the equation of heat considering latent heat only:

(since water already at boiling point at atmospheric temperature)

[tex]Q=m.L[/tex]

[tex]103=m\times 2230[/tex]

[tex]m=0.0462\ g[/tex] of water is converted into vapour.

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