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Effective heat andThermometers / Rate heat is lost

  1. Aug 14, 2011 #1
    1. The problem statement, all variables and given/known data
    I am a bit confused on these two questions in my textbook, I have the answers to them as well.

    The answer the thermometer question is that the thermometer bulb should be covered in a wet cloth.

    The answer for the 2nd question is 10.1 MW m^-2.


    2. Relevant equations
    How could a thermometer be modified to allow it to measure the effective temperature (windchill factor)?

    Calculate the rate at which thermal energy could be lost from a 3mm thick copper pipe which is carrying water at 100 C, if the surrounding air is at 20 C.



    3. The attempt at a solution
    For the 2nd question, I have tried to use the formula Q/t =( kA deltaT)/L where K=the materials conductivity, A is the surface area, L is the thickness in metres and delta T is the change in temperature.

    I know that delta T=80 C and the L=0.003 and the thermal conductivity of copper is 380 Wm^-1K^-1. But I am having problems finding the A value.
     
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  3. Aug 14, 2011 #2

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    Hi again paperdoll! :smile:

    What do you know about "windchill factor"?

    A 3 mm thick copper pipe is mathematically a cylinder with diameter 3 mm.
    Do you know what the surface of a cylinder is?
     
  4. Aug 14, 2011 #3
    Hi, I Like Serena :)
    I know the SA formula for a cylinder is SA= 2 pi r 2 + 2 pi r h except I can't find the "h" value in the question.

    erm I know that windchill factor is how the cold air feels on the skin which isn't exactly a measure of temperature, but heat right? I'm not too sure will try and read it again.
     
  5. Aug 14, 2011 #4

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    Good! :smile:

    Since we're talking about a copper pipe of undetermined length, we'll just take a piece out of it of length "h".
    This piece will "bleed" heat through the metal of the pipe.
    However the 2 endings won't "bleed" heat, because they are connected to the rest of the pipe.

    So if we take a piece of pipe of length "h", what is the surface through which heat is transmitted?

    Can you fill that in in your formula (including the "h")?


    Well, in practice, if the wind blows past a human being, the moisture on his skin will vaporize faster than it otherwise would.
    Because of this effect it feels colder than it actually is.

    To measure the temperature including the windchill factor, the method is to use a wet sphere, keep it in the wind, and measure the temperature of that.
     
  6. Aug 14, 2011 #5
    I'm trying to think...if its asking for surface area, will I need to include the inside of the pipe as well as the outside? without the the 2 endings of the pipe, the SA would be = 2 pi r h for one side, and double that to include the inside/outside of the pipe. The "h" value is still unknown though :confused:

    So does that mean that to measure windchill factor, we are actually measuring the rate at which the moisture is being vaporized off an object? rather than the actual temperature itself? I asked a classmate and she said that thermometers could only measure temperature (average kinetic energy of the body) but in the case of windchill, its the heat (transfer) which we are looking to measure.
     
  7. Aug 14, 2011 #6

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    Which surface area would be relevant?
    The inside of the copper pipe is 100 oC and copper is a good conductor for heat...

    Forget your reservations about "h" for now and substitute the area (including "h") in your formula.
    If you did it right, you'll see that something funny happens to "h". :wink:


    We're measuring a combination of the two.

    If there is no wind, you'll simply measure the temperature.
    The more wind there is, the more moisture is vaporized, and the lower the temperature becomes, which is still what you measure.

    After a while there will be an equilibrium where heat absorbed will be equal to heat lost due to vaporization.
    At that time you're still actually measuring a temperature (so you're not measuring the heat transfer).
     
    Last edited: Aug 14, 2011
  8. Aug 15, 2011 #7
    I think I understand the thermometer now :)
    but I think I will ask my physics teacher tomorrow about the other
     
  9. Aug 15, 2011 #8

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    I just checked out your formula which I didn't do before.

    It gives the heat transferred per unit of time by a specific area of the pipe.
    However, it is not specified what this area is.
    The formula does have the thickness of the pipe in it, which I didn't notice before.

    You have Q/t = kA(deltaT)/L.

    Since you do not know the area A, you can only calculate how much heat is transferred per unit of area.

    In other words, you can calculate Q/(A t) = k(deltaT)/L.
     
  10. Aug 15, 2011 #9
    I just filled out that and got the answer. Must keep that in mind the next time I come across a question without area. :smile:
    Thanks again I Like Serena :redface:
     
  11. Aug 15, 2011 #10

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    Till next time! :approve:
     
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