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Does Q = enthalpy in Constant Temperature?

  1. Mar 18, 2014 #1
    1. The problem statement, all variables and given/known data
    In a constant temperature process, where heat is added - Does the enthalpy after the heat is added equal the initial enthalpy plus the Q value?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 18, 2014 #2
    For an ideal gas, if the temperature is constant, the change in enthalpy is zero.
     
  4. Mar 18, 2014 #3
    For my particular question I have steam with a given pressure and dryness fraction. Really not sure what to do here
     
  5. Mar 18, 2014 #4
    Enthalpy is a measure of the total energy of a system (relative to some selected "zero point", we have no theory which allows us to determine the absolute energy of any system). In Classical Physics Energy is conserved. It is neither lost nor gained in any isolated system. In Relativity, if you include Mass, Momentum and stress in your definition of Energy, which is the way it works, then Energy is conserved (locally). Meaning the Laws of Thermodynamics are locally obeyed. So, you need to engrave in your soul that unlike vampires, and Magneto, energy can be neither created nor destroyed. Heat is the flow of energy. If energy is flowing into a system, you shouldn't need to ask if enthalpy is increasing by *EXACTLY* the same amount. Of course it is.
    Dryness fraction is an engineering term, and until 30 seconds ago, I'd never heard of it.
    Try: http://www.engineeringtoolbox.com/wet-steam-quality-d_426.html
    or
    https://en.wikipedia.org/wiki/Enthalpy–entropy_chart
    it seems to be a simple problem of adding components of enthalpy together (computed separately).
     
  6. Mar 18, 2014 #5
    No problem. If you have saturated vapor and saturated liquid in equilibrium at a given quality, temperature, and pressure, and you add or remove heat so that the quality changes, then Q = ΔH (provided you don't add or remove enough heat to take you out of the two phase region). Also, ΔS=Q/T.

    Chet
     
  7. Mar 18, 2014 #6
    Ok makes sense so far just wasn't sure.

    It also asks for work later in the question.

    (these are only for a few marks aswell which is why it's frustrating. The harder questions I have a good idea how they go)

    I appreciate the help buddy
     
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