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Homework Help: Heat capacitance

  1. Mar 21, 2010 #1
    1. The problem statement, all variables and given/known data

    When water is boilded under a pressure of 2 atm, the heat of vaporization is 2.2 x 10^6 J/kg and the boiling point is 120 degree C. At this pressure, 1 kg of water has a volume of 10^-3 cubic meter and 1 kg of steam has a volume of 0.824 cubic meter.

    a.) Compute the work done when 1 kg of steam is formed at this temperature.
    b.) Compute the increase in the internal energy of the water.

    2. Relevant equations

    boiling point = 393 K
    number of moles = 1000 / 18.0

    first law of thermo: change in internal energy = energy input - work done by system

    3. The attempt at a solution

    a.) W = p(V2-V1) since it is isobaric
    = 2x10^5 (0.824 - 10^-3)
    = 1.646 x 10^5 J (work done by water)

    (Is this correct?)

    b.) pV = nRT
    2x10^5x10^-3 = (1000/18.0) x 8.31 x T
    T = 0.43 K

    therefore: change in T = 393 - 0.43
    = 392.57 K

    (Don't I need to know the heat capacitance to work out the heat used to increase the temperature of the water to the boiling point? If I find this, then I can simply add this to the heat of vaporization to get the total heat input. Then, from first law of thermodynamics, I can find the change in internal energy.)
  2. jcsd
  3. Mar 21, 2010 #2
    [tex]W_{b}= \int Pdv[/tex]

    where pressure is constant and [tex]dv[/tex] is the change in volume. The volume goes from .01m^3 to .824m^3.

    This will give you the work done BY the system if it was in a piston cylinder device.

    To computer the work required to vaporize 1kg of water, simply multiply the the heat of vaporization by the amount of water.

    [tex]2.2e6 kJ/kg * 1kg = 2200 kJ[/tex]

    Using your own formula for the first law of thermo:

    [tex]\Delta h = 2200 kJ - 164.6 kJ = 2035 kJ[/tex]
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