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Homework Help: Change in entropy

  1. Feb 24, 2005 #1
    I'm stuck on a entropy problem and have no idea how to solve it!!

    Heat Q flows spontaneously from a reservoir at 394 K into a reservoir that has a lower temerature T. Because of the spontaneous flow, thirty percent of Q is rendered unavailable for work when a Carnot engine operates between the reservoir at temperature T and reservoir at 248 K. Find the temperature T.

    First of all... I don't even understand the problem. Does the 394 K reservoir flow into the hot reservoir, with temperature T? And then the water flows from that reservoir through a Carnot engine into the cold reservoir, which is temperature 248 K?

    I know the change in entropy equals the heat (Q) divided by the temperature (T). I also know that the work unavailable (W) equals the temp of the coldest reservoir multiplied by the change of entropy of the universe.

    I would appreciate any help!! Thanks
  2. jcsd
  3. Feb 24, 2005 #2


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    I'm not sure that you even need to consider entropy, enthalpy or energy. I think somewhere there is an equation for a Carnot Engine's efficiecy as a function of high and low temperature. So, if you have one temperature, that being your Tlow (248) and efficiency. For 100% efficiency in a Carnot Engine, all heat is converted to work, so I would assume the problem states that efficiecny = 0.7. If this is how it is, then plug it in and get your Tresevior. Although, if this was the case, then I'm not sure why they gave you the first temperature.....oh well, that's just my two cents.
  4. Feb 24, 2005 #3
    Here is my working,
    [tex]Efficiency = \frac{output work}{input heat}[/tex]
    thus Efficiency = 0.7
    For Carnot engine,
    [tex]Efficiency = 1 - \frac{T_c}{T_h}[/tex]
    The [tex]T_c[/tex] is the T that we want to find. Why we can't put it as a denominator? Well, the first info tells us that it's not possible.
    As for your question,
    Since the heat is flowing spontaneously, it is always flowing in the manner from high temperature to low temperature.
  5. Feb 25, 2005 #4

    Andrew Mason

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    I am not sure about the question either. It seems to be an entropy question. The part about the Carnot engine doesn't seem to make any sense. Perhaps Clausius2 can figure it out.

    If the spontaneous flow of heat from R1 (first reservoir) to R2 (at T) caused 30% of the energy to be unavailable for work from R1 to R2 (which is not what the question says):

    [tex]\Delta ST = .3Q[/tex]
    [tex]QT/T - QT/T_H = .3Q \rightarrow T = .7T_H = 276K[/tex]

    But I don't see how that relates to the third reservoir.

    The Carnot cycle operating between R2 and R3 would have an efficiency of:

    [tex]\eta = \frac{T - T_C}{T} = \frac{28}{248} = .11[/tex]

    (i.e 89% of the energy is unavailable for work) but that seems to be irrelevant???

  6. Feb 26, 2005 #5


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    I'm with Minger. The first temperature doesn't make me any sense. With the data of Carnot engine the problem is closed.
  7. Feb 26, 2005 #6
    I think the question is like this.
    First the Q is transfered from the 394K reservoir to the reservoir with temperature of T. This Q will then become the 100% of the input energy that works between the T and the 248K reservoir. When we want to find T, there are 2 possibility for T, either it is a numerator or denominator. So the first info is there to tell us that the T we find cannot be exceeding 394K. That limits it to be the denominator.
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