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Homework Help: Fundamentals of Engineering Problem

  1. Apr 5, 2005 #1
    Q. Water is boiled at 1 atm pressure in a coffee maker equipped with an immersion type electric heating element. The coffee maker initially contains 1kg of water. Once boiling started, it is observed that half of the water in the coffee maker evaporated in 25 minutes. If the heat loss from the coffee maker is negligible, the power rating of the heating element is :

    Is there a formula to find this.

    I can't find any formula, or any problem close to this one in my text book.

    Any insights on how to get started.

    Something to do with ideal gas law, may be:

    PV = nRT
  2. jcsd
  3. Apr 5, 2005 #2


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    First, look up the latent heat of vaporization for water (2260 kJ/kg).

    You have the mass of water that was boiled off. Multiply the two and that will give you the energy used to evaporate that much water (theoretically).

    Next you have the amount of time it took to evaporate that much water.......What is the deifinition of power? Energy per unit of time...
  4. Apr 11, 2005 #3
    Mass of water is = 1 kg

    Latent heat of vaporization = 2260 KJ /kg

    Q = Lv * M = 2260 KJ /kg * 1 kg = 2260 KJ


    Power = Energy / Unit time = 2260 KJ / 25 minutes
    = 90.4 ,,,, is this correct...
  5. Apr 11, 2005 #4


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    Watts (or kilowats) are not in j/min. Also check as to how much water actually boiled.
  6. Apr 11, 2005 #5
    1/2 of it is boiled so,

    Q = Lv * m

    = 2260 KJ/kg * 1/2 kg

    = 1130

    Power = 1130 / 25 minutes = 45.2 Watts , Is this right

    or may be 45.2 / 1000 = 0.045 Kw
  7. Apr 11, 2005 #6


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    You are still using minutes in your calculation of power. Power is in J/s. Convert 25 minutes into seconds and then divide 1130 by that number. You should end up with approximately .75 kW or about 750 W.
    Last edited: Apr 11, 2005
  8. Apr 12, 2005 #7
    Oh, yes you are right, should have converted to seconds,

    Anyway thanks, for pointing it out.
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