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Efficiency Problem!

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine the amount of time it will take a 1000 W electric kettle to heat up 1.0 L of water to boiling point if the water starts at 10 °C and the kettle is made from 400 g of iron. The transfer of heat energy from the kettle to the water is 63%.


    2. Relevant equations
    P = W/Δt
    Q = mcΔT

    3. The attempt at a solution
    Okay so what i was thinking was find 63% of the 1000 W electric kettle and then that is the power used to bring the water to a boil so I did -

    Pfor water = 1000W x 0.63
    = 630 W

    then I would use this power in the formula P = W/Δt to find time so I did -

    Δt = (mwater x cwater x (Tf of water - Ti of water) ) / 630 W

    = (1kg x 4.2E3 J/kgC x (100 C - 10 C )) / 630 W
    = 600s ||or|| 10 min

    apparently this answer is wrong because I am suppose to incorporate the mass of the iron kettle but I don't know how. Please help!
     
  2. jcsd
  3. Dec 5, 2013 #2

    CWatters

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    You have the correct equation for the energy needed to heat the water mwater x cwater x (Tf of water - Ti of water). Write similar equation for the energy needed to heat the iron. What laws do you know about energy?

    PS: I believe they are saying that the 63% figure applies between the kettle and the water not the element and the water.
     
    Last edited: Dec 5, 2013
  4. Dec 5, 2013 #3
    Δt = (mwater x cwater x (Tf of water - Ti of water) + miron x ciron x(Tf iron - Ti iron)) / 630 W

    Is ^ this correct then. I know that energy cant be created or destroyed and that it is only transferred from one form to another.
     
  5. Dec 8, 2013 #4

    CWatters

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    The more I think about this problem the more I think it's badly specified and your first answer based on 63% of 1000W could be justified.
     
  6. Dec 8, 2013 #5

    haruspex

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    I agree. Based on the masses and specific heats, you'd think 96% of the heat would go into the water (did I get that right?). If only 63% went that way then the rest has been radiated/convected away to the room. This leaves the OP calculation as the only route to the answer. The mass and composition of the kettle are irrelevant.
     
  7. Dec 8, 2013 #6
    Okay so then the 600 s answer is right then because the only thing affecting the heating up of the water is the amount of power the kettle has and not what the kettle is made from. And since the kettle is only 63% efficient then only 630W is going towards heating the water.
     
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