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How many grams of steam at 100 °C would be required to raise the temperature
of 47.6 g solid benzene (CH) from 5.5 °C to 30.0 °C? Assume that heat is only
transferred from the steam (and not liquid water) and that the steam/water and
benzene are separated by a glass wall and do not mix. (The melting point of ben-
zene is 5.5 °C; AHus for benzene is 9.87 kJ/mol; specific heat for benzene is 1.63
J/g• °C; AH for steam at 100 °C is 40.7 kJ/mol.)
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Respuesta :

onyebuchinnaji

The mass of steam required to raise the temperature of water is 3.5 g.

The given parameters;

  • mass of the benzene, = 47.6
  • initial temperature of the benzene, = 5.5 ⁰C
  • final temperature of the benzene = 30 ⁰C

The molar mass of Benzene = 78.11 g/mol

The molar mass of water = 18 g/mol

The number of moles of the Benzene is calculated as follows;

[tex]n = \frac{47.6}{78.11} = 0.61 \ mole[/tex]

The mass of steam required is calculated as follows;

heat lost by steam = heat absorbed by benzene

[tex]\frac{m}{18} \times 40.7 \times 10^3 = 47.6(1.63)(30-5.5) \ + \ 0.61 \times 9.87 \times 10^3\\\\2261.11 m = 7921.61\\\\m = \frac{7921.61}{2261.11} \\\\m = 3.5 \ g[/tex]

Thus, the mass of steam required to raise the temperature of water is 3.5 g.

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