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Turn water to steam

  1. Jul 1, 2007 #1
    How much heat must be added to 120g of water at an initial temperature of 60C to
    heat it to the boiling point?
    completely convert the 100C water to steam?

    Specific heat capacity of water = 1.0cal/g-C
    It would take 120*1 to raise 120g of water up to 1C
    difference in temperature = 100C - 60C = 40C
    It would then take 120*40 to raise it to the boiling point
    120*40 = 4800cal

    specific heat of steam is 0.48 cal/g-C but what it the temperature of steam?
     
  2. jcsd
  3. Jul 1, 2007 #2

    Astronuc

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    Staff: Mentor

    Temperature of steam at 1 atm (0.101325 MPa, or 14.7 psia) is 100°C (212°F). Steam represents a phase change.

    Steam however can be superheated, but that would be in a vessel with higher steam pressure, otherwise the steam expands and cools.


    Remember, the problem increasing the water temperature, then transforming liquid water at 100°C to steam (water vapor) at 100°C.

    See
    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase2.html#c3
    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase.html
     
    Last edited: Jul 1, 2007
  4. Jul 1, 2007 #3
    But what do i use for the calculation of water to steam. Doesnt the system have to have some amount of heat added to convert it to steam.
     
  5. Jul 1, 2007 #4
    You need the latent heat of vapourization of water for the last part of the calculation, not the specific heat of steam ;)
     
  6. Jul 1, 2007 #5

    Astronuc

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    Staff: Mentor

    Look at the first link I provided.

    Converting liquid water to steam at 100°C (0.101325 MPa) requires a heat addition of 539 cal/g (2260 J/gm).
     
  7. Jul 1, 2007 #6
    The latent heat for boiling is 540 cal/g. That means that 1g requires 540 cal of heat to steam

    Thus, 120g requires 120 x 540 = 64800 cal to steam
     
  8. Jul 1, 2007 #7

    Astronuc

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    Staff: Mentor

    Correct.

    And then add the two heat requirements.
     
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