Respuesta :

Answer:

[tex]Q=977216.256\ J=977.216\ kJ[/tex]

Explanation:

Given:

  • mass of  steam, [tex]m=384\ g[/tex]
  • temperature of steam, [tex]T_{is}=100^{\circ}C[/tex]
  • temperature of resultant water, [tex]T_{fw}=31^{\circ}C[/tex]

We have,

  • latent heat of vapourization of water, [tex]L=2256\ J.g^{-1}[/tex]
  • specific heat capacity of water, [tex]c=4.186\ J.g^{-1}[/tex]

When we cool the steam of 100°C then firstly it loses its latent heat to convert into water of 100°C and the further cools the water.

Now the heat removed from steam to achieve the final state of water:

[tex]\rm Q=latent\ heat\ of\ vapourization+sensible\ heat\ of\ water[/tex]

[tex]Q=m(L+c.\Delta T)[/tex]

[tex]Q=384(2256+4.186\times (100-31))[/tex]

[tex]Q=977216.256\ J=977.216\ kJ[/tex]

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