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Homework Help: Ideal carnot engine

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data
    An ideal Carnot engine operates between a heat reservoir and a block of ice of mass M. An external energy source maintains the reservoir at a constant temperature [itex]T_{h}[/itex]. At time t=0 the ice is at its melting point [itex]T_{0}[/itex], but it is insulated from everything except the engine, so it is free to change state and temperature. The engine is operated in such a way that it extracts heat from the reservoir at a constant rate [itex]P_{h}[/itex].
    (a) Find an expression for the time [itex]t_{1}[/itex] at which all the ice is melted, in terms of the quantities given and any other thermodynamic parameters.
    (b) Find an expression for the mechanical power output of the engine as a function of time for times [itex]t > t_{1}[/itex].

    2. Relevant equations
    [tex]Q=ML_{f}[/tex]
    [tex]W=Q_{h}-Q_{c}[/tex]
    [tex]\frac{dQ_{h}}{dt}=\frac{dQ_{c}}{dt} + \frac{dW}{dt}[/tex]
    3. The attempt at a solution
    First of all I assumed
    [tex]\frac{dW}{dt}=0[/tex]
    so
    [tex]\frac{dQ_{h}}{dt}=\frac{dQ_{c}}{dt}= P_{h}[/tex]

    [tex]\int dQ_{h} = P_{h}\int dt = P_{h}t_{1} = ML_{f}[/tex]
    so
    [tex]t_{1} = \frac{ML_{f}}{P_{h}}[/tex]

    but somehow to answer manages to squeeze out

    [tex]t_{1} = \frac{ML_{f}T_{h}}{P_{h}T_{0}}[/tex]

    Whats the deal here? That answer doesn't seem to make much sense since it appears to be saying the hotter the heat reservoir, the longer it takes to melt the ice. I would have thought it to be the other way around.

    And for part (b) I am not sure where to start.
     
    Last edited: Feb 10, 2010
  2. jcsd
  3. Feb 10, 2010 #2

    Mapes

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    Where do you get [itex]dW/dt=0[/itex]? Is the engine not producing power? It certainly should be, if it's perfectly efficient and there's a nonzero temperature difference between the two reservoirs.
     
  4. Feb 10, 2010 #3
    I took it to mean W = constant, so d/dt W = 0
    Still not sure how to proceed
     
  5. Feb 10, 2010 #4

    Mapes

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    But W isn't constant, the total work you extract increases every second that the engine operates.

    Think entropy: how much comes in from the hot reservoir? How much leaves with work? How much goes to the cold reservoir? How much is created? That's the key to analyzing Carnot engines.
     
  6. Feb 15, 2010 #5
    I'm not really sure about the entropy of any of them
    perhaps
    [tex]dS_{h}=\frac{dQ_{h}}{T_{h}}[/tex]
    and maybe
    [tex]dS_{c}=\frac{ML_{f}}{T_{c}}[/tex]
    No idea about how much entropy leaves as work.
     
  7. Feb 15, 2010 #6

    Mapes

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    Great; although you're mixing differential and non-differential terms in the second equation, it's the right idea.

    There is no entropy transfer associated with work.
     
  8. Feb 15, 2010 #7
    I was just writing random equations not sure how this fits into the overall scheme of things
     
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