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Heat transfer problem

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data


    Estimate how long it should take to bring a cup of water to boiling temperature in a typical 600 watt microwave oven, assuming all energy ends up in the water(assume any reasonable initial temperature for the water.) What can you conclude about the amount of heat added to the water?

    2. Relevant equations

    P=W/t
    possibly Q=W=m*c*delta(T)

    3. The attempt at a solution

    P=600 watts
    W=Q=m*c*delta(T)
    c=4.2 J/kg*1
    m=.250 grams ( I guessed what the mass of the cup of water would be)
    W=(.250)(4.2)(73 K)=76.7 Joules
    t=W/P=76.7 joules/600 watts

    W/P=t
    T_initial=300 K

    T_final=373 K
    c=4.2
    delta(T)=T_final-T_initial=373 K - 300 K= 73 K
    not sure if I should just guesstimate the mass of the cup of water.

    W=(.250)(4.2)(73 K)=76.7 Joules
    t=W/P=76.7 joules/600 watts

    Boiling the cup of water in the cup involves work not heat. Energy must be leaving the system(water in this case) to be described as heat. If the cup of water was cooling off, then I could describe that process as heat, not work.
     
  2. jcsd
  3. Jan 31, 2009 #2

    Redbelly98

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    It looks like you have the right idea. However, the mass of water will be a lot more than 0.250 g.

    Can you come up with a better estimate for m? Perhaps first you should estimate the volume first.
     
  4. Jan 31, 2009 #3
    I actually meant to write the mass of the cup was supposed to be .250 kg , not .250 grams. Is my explanation for why there is no heat involved in the process correct?
     
  5. Jan 31, 2009 #4

    Redbelly98

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    Well, no. There is heat involved. The microwaves are a means of adding heat to the water. In fact, the sole function of a microwave oven is to heat water.

    Doing work would involve a change in volume, or some kind of displacement, since
    W = P ΔV .or. F d cosθ​

    I guess on a microscopic level, heating something is equivalent to doing work on many individual molecules all at once. But we don't really consider that as work in a practical sense.
     
  6. Jan 31, 2009 #5

    Mapes

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    This isn't true; any transfer of thermal energy is described as heat. The water is being heated through irradiation; its thermal energy increases.
     
  7. Jan 31, 2009 #6
    Heat is the spontanteous flow of energy from one object to another and the amount of energy that enter a system . When spontaneous energy is involved, energy usually flows from a higher state of energy to a lower state of energy. hot objects have higher thermal energy than cold objects do. When we rub our hands together and they get hotter, heat is not involved. When cup of water at room temperature temperature goes up , its because of work not heat.
     
  8. Jan 31, 2009 #7

    Mapes

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    This also describes work.

    You can keep asserting this, but those responding to your question disagree. More importantly, in the water/cup system you can't identify any significant examples of work, whether it's force + change in distance, pressure + change in volume (other than routine thermal expansion), electric charge + change in voltage, etc. I invite you to try.

    The microwaves carry entropy, which is the defining characteristic of heat transfer.

    During thermal expansion and boiling, the water and gas does do work on the atmosphere; is this what you meant?
     
  9. Jan 31, 2009 #8
     
  10. Jan 31, 2009 #9

    Mapes

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    OK, I think I see where the problem lies. You're concluding that the process of water getting hotter in a microwave over can't be classified as heating because the physical microwave oven is cooler. But your textbook example of heat transfer from a hotter to a colder body assumes that the bodies are each in thermodynamic equilibrium (i.e., they each have a single, well-defined temperature). A microwave oven is not in thermodynamic equilibrium. And I stand behind my earlier points that entropy transfer is associated with heat, not work, and that none of the standard generalized force-displacement pairs can be connected to water getting warmer in a microwave oven.

    Of course it's spontaneous above 100°C! Seriously, I think you should consider your arguments more carefully.
     
  11. Jan 31, 2009 #10

    Redbelly98

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    Radiation is a well-known, and scientifically accepted, means of transferring heat.
    http://en.wikipedia.org/wiki/Heat_transfer#Radiation

    What if the water is in a black cup and gets heated by direct sunlight?

    Even if we accepted your definition of heat transfer, and your arguments that heating is not taking place, you have not shown that doing work is taking place either.
     
  12. Jan 31, 2009 #11
    Work , according to Scroeder is the transfer of energy into or out of a system.(Scroeder, p.19 )
    Energy of a system will increase and therefore its temperature will increase. The temperature of the cup of water reaching its boiling pt. Therefore its temperature changes, therefore there is work being done on the cup. I don't know , maybe Scroeder did not defined heat properly .

    You do make a good point since radiation is a type of heat transfer , and microwaves are part of the EM spectrum and excited the water molecules inside the microwave.
     
  13. Jan 31, 2009 #12

    Mapes

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    No, these are not good definitions, because heat is also the transfer of energy into or out of a system. Jeez, even Wikipedia is more precise. A good distinguishing characteristic is whether entropy is transferred (heat) or a mechanical variable (displacement, volume, charge, magnetization, strain, etc.) (work).
     
  14. Jan 31, 2009 #13
    Well take it up with the author of my thermal physics textbook not me. I didn't right these definitions for heat. We are at a point in the class where we have not covered entropy. But from what I can remember what definition of entropy is from another class, entropy of a system depends on the number of arrangements that system can have or its multiplicity.
     
  15. Jan 31, 2009 #14

    Redbelly98

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    Defining work as "the transfer of energy" would mean it is not mutually exclusive of adding/removing heat, and that heat transfer is one way to do work.

    But from what I remember of thermodynamics, heat transfer and doing work are separate, distinguishable processes for changing the energy content of a system:
    ΔE = W + Q​
    gives the change in energy of a system, where W is work done on the system, and Q is the heat added to it.
     
  16. Jan 31, 2009 #15

    Redbelly98

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    At this point, it's probably best to wait and see what your professor says about all this.
     
  17. Jan 31, 2009 #16

    Redbelly98

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    Back to the original question ...

    That's okay. However, note that c=4.2 J/g/K, not 4.2 J/kg/K as you wrote before.
     
  18. Jan 31, 2009 #17
    My estimate came out to be near 10 minutes. Is that a reasonable time?
     
  19. Jan 31, 2009 #18

    Redbelly98

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    10 minutes sounds a little long. Can you show your calculation?
     
  20. Jan 31, 2009 #19
    yes. t=Q/P=(250)*373*(4.2)/600 =652 seconds= almost 11 minutes.
     
  21. Feb 1, 2009 #20

    Mapes

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    Earlier you calculated the temperature difference as 73 K, not 373 K.
     
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