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Can you measure watts through heat?

  1. Dec 12, 2006 #1
    I am in grade 11 physics and haven't yet learned about thermodynamics or anything dealing with heat really.

    I am wondering basically if you can reliably measure the power difference between two settings on a stove through temperature. i.e use a thermometer, measure the temp of the element at high, compare to temp at medium high. Could you tell the difference in power consumption? My guess is that you can, but I need to know for sure.
     
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  3. Dec 12, 2006 #2

    russ_watters

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    The difficulty is in knowing the heat transfer rate from the element. Since you have conduction, convection, and radiation, and all are temperature dependent (and not necessarily in the same way), it would not be easy to do.

    So the trick would be to find a way to remove nearly all of the heat from the element to something easily measurable. Perhaps a large insulated pot of water, for example.
     
  4. Dec 12, 2006 #3

    Danger

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    Would the efficiency/resistance/whatever of the element at different temperatures enter into the matter?
     
  5. Dec 12, 2006 #4
    So theoretically this would be plausible (albeit impractical) by building a well insulated pot and sticking a thermometer in the top to measure how long it takes to boil a precise amount of water at a precise temperature. Then set up a ratio between the two results. Would this give you an reasonable ratio for power consumption?
     
    Last edited: Dec 12, 2006
  6. Dec 12, 2006 #5
    I think that Danger's post would be my greatest worry, that the element's efficiency fluctuates with the temp setting.
     
  7. Dec 12, 2006 #6

    russ_watters

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    No. All heating elements are 100% efficient.
     
  8. Dec 12, 2006 #7

    russ_watters

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    Yes. And not just a ratio - you could calculate the actual amount of energy/power and compare it to what you measure with an ammeter/wattmeter.
     
  9. Dec 12, 2006 #8
    great, thanks!
     
  10. Dec 13, 2006 #9
    Actually I was thinking about this and it should have become apparent to me sooner that since 1 joule of energy raises 1ml of water 1^C, if I had 1250ml of water in an insulated pot, it would take 93750J of energy to heat it from 25^C to boiling. And of course I forgot but am now reminded that 1J/s=1watt. So I would just divide 93750J by however many seconds it took and I could get the watts.

    I guess this scenario will reinforce the power formula in my mind, heh.
     
    Last edited: Dec 13, 2006
  11. Dec 13, 2006 #10

    Danger

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    Please clarify, Russ. I thought that nothing was 100% efficient aside from superconductivity (from which work can't be extracted). Doesn't the resistance of a conductor change with the temperature?
     
  12. Dec 13, 2006 #11
    Resistors are 100% efficient because all the energy in must come out in one form or another. (Radiation, convection, and conduction). The energy is not being partitioned to places like chemical reactions or sound which would result in less than 100% energy conversion.
     
    Last edited: Dec 13, 2006
  13. Dec 13, 2006 #12

    You can do this very easily with one temperature measurement. You can measure the power going in using a multimeter, and compare that to the value you get using your thermometer measurement, and see how close they are.

    Now, when you say reliably, keep in mind that in heat transfer, that means you are 40-50% off in your answer.
     
  14. Dec 13, 2006 #13

    Danger

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    Thanks for that clarification, Cyrus. I had a different impression of what Russ meant. Although the element itself might be 100% efficient, that is not transferred to the vessel. A lot of the heat still goes into the air rather than into the vessel. If that were not the case, you could comfortably rest your hand on the heating element beside the vessel. (Which, I suppose, you can do with inductive cooktops.)
     
  15. Dec 13, 2006 #14

    russ_watters

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    True - the element itself is 100% efficient at making heat, but you won't be able to capture all of it (though you can get pretty close).
     
  16. Dec 13, 2006 #15

    russ_watters

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    ....keeping in mind that all energy lost to inefficiency (in most instances I can think of) is in the form of heat. In a car, for example, you may get 35% of your energy sent to the wheels while losing 65% through the engine block and radiator. The total is 100%. In the example in this thread, it just so happens that all of what we are looking for is heat and it is the only form of energy transferred. No steam engine here.

    You can even go a step further, though: at the end of the day, your car is right back where it started, so the total amount of mechanical work done is zero. All of the energy created by the engine eventually became lost to heat.
     
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