Anonymous
Anonymous asked in Science & MathematicsChemistry · 4 months ago

how much heat necessary to boil 40.0g of H2O at 100 degrees C, then raise the temp of H2O vapor from 100.0 to 110.0 degrees C. ?

The specific heat of water vapor = 2.080 J/g degrees C. The heat of vaporization of water = 2,257 J/g.

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  • 4 months ago

    Heat to vaporize the water:

    q = (40.0 g ) X 2257 J/g = 9.03X10^4 J

    Heat to raise temp of steam:

    q = m c (T2-T1)

    q = 40.0 g (2.080 J/gC) (110.0-100.0) = 823 J

    Total = 9.11X10^4 J = 91.1 kJ

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