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Percent Efficieny Problem

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data

    The heating element in a kettle has a resistance of 10 ohms when it is plugged into a 110V outlet. When heating water, only 75% of the energy produced is actually absorbed by the water. How much heat energy does the water absorb if it takes 4 minutes to boil the water?



    2. Relevant equations

    V = IR
    P = VI
    Percent Efficiency = Useful Energy Output / Total Electrical Output



    3. The attempt at a solution

    1. Finding Energy Produced by Kettle

    I = V / R
    I = 11a

    P = VI
    =110 x 11
    =1210 w

    290 400 j (1210 x 60 x 4) of energy produced in 4 minutes
    290 400j x 0.75
    =217 800j / 3600
    =60.5 watt hours is absorb by the water.

    However, my teacher wrote this on the board:

    E = V x I x t
    =110 x 11 x 240 (4 minutes in seconds)
    =80 2/3 watt hours (output)

    % Efficiency = Useful Energy Output / Total Electrical Input
    75% = 80 2/3 / Total Electrical Energy Input
    E = 80 2/3 divided by 0.75
    E absorbed by water = 108 watt hours

    His answer confuses me... how could the energy absorbed by the water be more then the energy output?

    Please help, have a "quest" on this tomorrow.
     
  2. jcsd
  3. Mar 8, 2009 #2

    Delphi51

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    Homework Helper

    Teachers do make mistakes! He should have multiplied by .75 instead of dividing.
     
  4. Mar 8, 2009 #3

    lanedance

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    Homework Helper

    I think you make a fair point & I agree with your way

    the energy absorbed by the water should not be more then the energy output...
     
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