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Equation used to find the efficiency of a kettle

  1. Feb 17, 2015 #1
    So, I'm doing this coursework right now, and I know that this equation is what's needed to get to the efficiency but I have only a little idea why.

    I'm given this formula:
    https://s.yimg.com/hd/answers/i/669504ce234c41af99c351e0c28eea10_A.png?a=answers&mr=0&x=1424239496&s=8a1c1557912955a6446f8f8e90ff7a57 [Broken]

    It's Efficiency = (mass (volume of water) x water capacity x change in temperature)/ (voltage x current x time taken)

    Is there any theory connected to this? Or other equations that led to this equation?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Feb 17, 2015 #2

    SteamKing

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    Perhaps you should look up the definition of 'efficiency'.

    Also, don't post duplicate threads in different forums. That's a violation of the rules of PF.
     
  4. Feb 18, 2015 #3
    Already have, but am I right to assume that:

    mass x water capacity x change in temperature = energy output
    and the voltage x current x time taken = energy input?
     
  5. Feb 18, 2015 #4

    SteamKing

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    That's correct.
     
  6. Feb 18, 2015 #5

    DEvens

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    I think C has to be the heat capacity of water, not the "water capacity" whatever that may be.

    The heat capacity of water will be in units of energy per change in temperature per unit mass. For example, Joules per degree C per kilogram.
     
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