1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook