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Second Law of Thermodynamics-Convert Heat to Work

  1. Apr 10, 2012 #1
    A long time ago when I was in college there was a problem in the back of a chapter on thermodynamics in my freshman chemistry textbook:

    A newly designed ship travels the tropical seas by using the warm tropical seawater,converting the heat from the seawater to work, and then expelling ice cubes back into the sea after energy was extracted from the water.

    That is it. The question for this problem may have been to explain why this is impossible, why it is impractical or something like that. Anyway, the problem was not assigned and there was no discussion of it in class.

    Years later having forgotten most of my thermodynamics I still remember the problem.Is it impossible? Why?

    For ease of calculation let's say the temperature of the tropical seawater was 80 degrees F. After doing the work the ship expels water at 33 degrees Farenheit.
    I know the Second Law says that you can't convert heat into an eqivalent amount of work,not with 100% efficiency.But there are billions of gallons of tropical seawater in ocean. Even if the conversion is only 10% efficient, it should be enough to power the ship across the ocean, wouldn't it?
  2. jcsd
  3. Apr 10, 2012 #2
    I feel the amount of heat absorbed would be really less to convert it into work.
    Its similar to the problem where we cannot use the sound energy to generated by phone to charge itself again.
    And with that amount of heat, it would be possible to drive a paper boat (I am not saying sarcastically), weight would be a big matter.
    But yes, this topic is interesting. Using the water heat itself as energy. Let's look ahead for few more answers.
  4. Apr 10, 2012 #3


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    Entropy (in a closed system) cannot decrease. If a quantity of heat Q1 moves from a body at temperature T1 then the entropy that body loses is Q1/T1 (assuming it was not enough to change the temperature much). The same heat added to a body at T2 raises its entropy by Q1/T2. So, in the absence of any other associated entropy change, Q1/T2 > Q1/T1. I.e. T1 > T2. More generally, if only Q2 = Q1-W flows into the second body, W having been extracted to do work, Q2/T2 > Q1/T1. Since Q2 < Q1, this also requires T1 > T2, only more so.
    In the scenario given, T1 is warm surface water. Since heat is extracted from it until it forms ice, that heat must be flowing to some sink that's below freezing. So it might make sense if the fuel consists of liquid nitrogen.
  5. Apr 10, 2012 #4
    I never learnt this stuff very well but my first instinct is to suggest something along the lines of "the entropy of the seawater is maximal (it is in equilibrium), so the amount of useful work that can be done by it is zero".

    In other words, to convert heat in the water to work, what you're effectively asking for is for a small portion of the molecules to cease their random motion and begin all moving in the same direction (so they will freeze, and you can extract energy from their velocity). So you end up with an increase in useful energy.

    While this is permissible under conservation of energy it is not permissible under the 2nd law, because the entropy of such a state is lower than the entropy of water in equilibrium - there are fewer microstates corresponding to the macrostate of "water at 80 plus a few ice cubes moving around" than there are corresponding to "water at 80 in equilibrium".

    edit- too late! looks like someone beat me to it ^^ really nice answer
  6. Apr 10, 2012 #5


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    You cannot spontaneously cool your working fluid. You need a cold reservoir to discharge waste heat to. One might just as easily ask why you don't siphon water off the surface of the ocean, run it through a turbine, then discharge it 100m below the surface to generate electricity!

    I can't imagine this wasn't covered as it is basically the key concept of thermodynamics!
    Last edited: Apr 10, 2012
  7. Apr 10, 2012 #6
    If I look at this like a practical engineering problem.
    If the surface is 80 F, and the first Thermocline say below 30 feet is at 65 F.
    This Delta would give us a Carnot efficiency of 1-(291/299)=about 2.7%.
    The energy in the 8 degrees between 291K and 299K would be 9.1 Watt Hours
    per cubic meter of water.
    so 9.1 watt hours * 2.7%= .26 watt hours per cubic meter of water.
    This means you can only extract .00035 HP for each cubic meter of water.
    I think the work necessary to bring up a cubic meter of water from 30 feet would
    exceed the work generated.
  8. Apr 10, 2012 #7
    Is it just a matter of degrees? Suppose the temperature of the seawater is over a volcanic vent and the temperture is about 130 degrees F. On board the ship you have a "steam" turbine, but instead of using water to create steam you use an alcohol which boils at 110 degrees F. The hot tropical seawater now boils the alcohol which turns the turbine and powers the ship. The air in the ship is at 90 degrees F, the alcohol condenses and is recycled and heated again by the seawater to turn the turbine.
  9. Apr 11, 2012 #8


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    Sure, you can use any sustained temperature difference to derive useful work, but what about the expelled ice cubes? Either you have access to a heat sink below 0C or you're wasting effort producing ice.
  10. Apr 11, 2012 #9
    Years ago I read an article that proposed using the temperature difference between ocean layers to boil and condense ammonia the gas phase to be used in a turbine, does any one know if it got beyond the "back of a fag packet" stage?
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