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Heat + Power Question!

  1. Dec 17, 2012 #1
    1. The problem statement, all variables and given/known data

    A microwave oven produces energy at a rate of P = 1219 W, all in the form of microwaves. When a cup of water is placed into this oven, 64.4% of the microwaves are absorbed by the water. If a cup containing mw = 212 grams of water (227 g equals 8 ounces) starts at temperature Tw = 19.6o C: find t, the time it will take the water to reach its boiling point.

    2. Relevant equations

    Q = mcΔT
    P = dW/dt

    3. The attempt at a solution

    Heat needed to raise 0.212 kg of water from 19.6 C to 100 C = (0.212)(4186)(100-19.6)
    Q = 71349.53 Joules

    Rate of Energy Absorption of Water = (64.4/100)(1219 Joules/sec) = 785 Joules/sec

    Since water is absorbing heat at 785 Joules per second, in order to absorb 71349.53 Joules altogether, we need:
    t = 785 J/s / 71349.53 J = 0.011 seconds.

    Am I approaching the problem incorrectly?

    Many thanks in advance!
     
  2. jcsd
  3. Dec 17, 2012 #2

    gneill

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    Staff: Mentor

    You're doing fine up to the last expression. Check how the units cancel.

    Also, does 0.011 seconds seem a reasonable amount of time to heat up that quantity of water in a microwave? :uhh:
     
  4. Dec 17, 2012 #3
    Thanks Gneill!

    Also, just curious: do you guys help out students (like myself) voluntarily? If so , that is awfully nice !
     
  5. Dec 17, 2012 #4

    gneill

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    Staff: Mentor

    You're welcome. Yes, it's a voluntary activity and we're happy to help!
     
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