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Entropy - Problem

  1. Nov 27, 2004 #1
    Hi! I have a problem that I simply can't solve. I was hoping you could push me in the right direction. English isn't my native language, so please excuse errors relating to that.

    A pump fetches water from the bottom of a sea where the water has a temperature of 4C (277K). It transfers this water into a heat carrier (direct translation from swedish - don't know the actual term) in a house where the temperature is 58C (331K). The electrical power of the pump is 1.00kW and when it's running it adds a heat power of 3.55kW to the heat carrier. Approximate how much the total entropy increases per time unit due to the pump while running. The excess heat is emitted at 20C (293K).

    Again, I apologize for any linguistic errors. And I don't expect anyone to solve this for me; I just really, really need some pointers.

    Thanks in advance,
  2. jcsd
  3. Nov 27, 2004 #2


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    You should plot some scheme and attach it here. I don't know what is a heat carrier, and I don't understand well the problem itself.

    Some scheme will help us to help you too.
  4. Nov 27, 2004 #3

    Andrew Mason

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    The house is at a temperature of 58C?! Something seems odd here.

    My guess is that this is a simple heat pump. The heat is transferred from the sea water to the house via a heat pump that cools the sea water and heats fluid in a heat exchanger ('heat carrier'?) that is then pumped through radiators in the house.

    The change in entropy is given by
    [tex]\triangle S = \triangle Q/T_c[/tex]
    So the rate of change of Entropy per unit time is:
    [tex]\triangle S/\triangle t = \frac{\triangle Q/\triangle t}{T_c}[/tex]

  5. Nov 28, 2004 #4
    Thanks for responding! Your interpretation appears to be correct.

    I agree with your solution, but I have now been instructed to use a Carnot-process to solve it, which really doesn't help me. Any ideas?

    Oh, and just a question regarding your formula:
    What does [tex]\Delta Q[/tex] and [tex]T_c[/tex] represent?

    Thanks again,
  6. Nov 28, 2004 #5

    Andrew Mason

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    The Carnot process is an ideal. But it does not apply here. For a heat pump, the efficiency of the Carnot cycle for a heat pump would be:
    [tex]T_H / (T_H - T_C) = 293/16 = 18.3[/tex]
    Here, the heat transferred is 3.55 kJ for each kJ of energy added.
    [tex]\Delta Q[/tex] is the heat transferred and
    [tex]T_c[/tex] is the temperature of the cold reservoir.
    You are given 3.55 kW as the heat transfer rate, which is [itex]\Delta Q/\Delta T[/itex]

    Last edited: Nov 28, 2004
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