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Calculate the change in entropy when 1.00 kg of water at 100 ∘c is vaporized and converted to steam at 100 ∘c. Assume that the heat of vaporization of water is 2256×103j/kg.

Respuesta :

Answer:6.04 kJ/K

Explanation:

Given

mass of water [tex]m=1 kg[/tex]

Latent heat of vaporization is [tex]L=2256\times 10^{3} J/kg[/tex]

entropy at constant temperature is given by

[tex]\Delta S=\frac{Q}{T_0}[/tex]

[tex]Q=m\times L[/tex]

[tex]Q=1\times 2256\times 10^3[/tex]

[tex]\Delta s=\frac{2256\times 10^3}{100+273}[/tex]

[tex]\Delta s=6.04kJ/K[/tex]    

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