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Heat Transfer and Power

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi, I'm tutoring an introductory Physics class at a Big 10 school, and we've stumbled across a problem that we can't solve. Well, I thought we could solve it, but we have to use a clumsy web program to enter our answers and we can't seem to solve this one.
    The problem is :
    You work as an attendant in a welcome booth in Death Valley. The outside temperature is 53oC. The booth has a surface area of 19m2, a thickness of 0.03m, and a thermal conductivity of 11 J/s*m*K (or W/m*k). You are supplied with an air conditioner with 3000W of power that works as an ideal Carnot refrigerator. What is the minimum temperature you can have in the booth?


    2. Relevant equations
    [tex]\Delta[/tex]Q/[tex]\Delta[/tex]t=k*A*[tex]\Delta[/tex]T/d
    OR
    W=k*A*[tex]\Delta[/tex]T/d

    3. The attempt at a solution
    This *seems* straight forward. I've been wracking my head over this and can't seem to come up with an acceptable, much less reasonable solution.
    Ideal carnot refrigerator means all power available to the air conditioner is available for cooling (the answer would be even more ridiculous if it wasn't)

    So, using the second equation given below, we can set our 3000W air conditioner equal to the rest of the equation...which is all given to us except the minimum temperature, which we are trying to find. It seems straightforward, but when everything is all plugged in and done, you get a [tex]\Delta[/tex]T of.430622 and thus a minimum temperature of 52.5694 oC

    I would hate to be that attendant! The computer program also doesn't like it either. I feel like there is a glaring error in my logic that I just can't pick up. This is due for them @ midnight on Wednesday, so I'll be thinking about it until then.

    Any help would be greatly appreciated :cool:
     
    Last edited: Nov 23, 2008
  2. jcsd
  3. Nov 23, 2008 #2
    I've been there - giving a student help and then a question coming up that you yourself can't answer!

    I'm no expert, but your figures appear to be right. If you think about it, you've essentially got a guy with a huge surface area - quantify 19 m^2 in your head - out in the desert. Additionally the thermal conductivity of 11 W/m/K is the equivalent of stainless steel. Insulation is about 0.04 W/m/K.

    So you've got a dude, in the desert, in a steel box which is an excellent conductor of heat. The heat from the outside is able to get into the box far quicker than you can shed it with your crappy 3kW cooler - even if it is an ideal refrigerator!

    I suppose you should look at who designed the question, I may well be wrong, but my sense is that it's a poorly designed question.
     
  4. Nov 23, 2008 #3
    That's as much as I thought.

    Thanks :)
    The trick now is getting the computer program to accept the answer...

    Ah, well, they will meet with their TAs/professors to see if they can get some more light shed on it!
     
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