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Rate of heat flow by conduction question

  1. Aug 7, 2010 #1
    1. The problem statement, all variables and given/known data

    a bushwalker wears clothing that is 2.0cm thick with a surface area of 1.9m^2. The material has k = 0.042 WM^-1K^-1. Her skin temperature is 33 degrees C and the ambient temp. is 0 degrees C.

    Calculate rate of heat loss through her clothing while she is sitting

    2. Relevant equations

    Rate = (kA(T1-T2))/L

    where K = conductivity, A = surface area, T1 and T2 = termperature at two faces and L = thickness.

    3. The attempt at a solution
    This is a fairly straightforward problem, however I am unsure as to whether to convert the degrees Celsius units given as T1 and T2 values into kelvins, which gives me and answer of -960W (is it even possible to have negative Watts?) or to leave them and cancel the kelvin units of K with them in order to obtain an answer in watts.
     
  2. jcsd
  3. Aug 7, 2010 #2
    If you let T1=ambient temp. and T2=skin temp., then the heat lost rate should be negative, as that means the heat flow goes from the skin to the surrounding, not the other way. If you exchange T1 and T2 (T1=skin temp. and T2=ambient temp.), then the heat lost rate should be positive for the same reason.

    EDIT: In the first case, the heat lost rate you calculate = the rate of the heat lost from the surrounding to the skin. Heat actually goes from the skin to the surrounding, so that your result is negative makes sense.
     
  4. Aug 7, 2010 #3
    ok, so I did the right thing in converting the degrees C into kelvins? and -960W is a realistic ballpark figure?
     
  5. Aug 7, 2010 #4
    You can convert the temp. into any unit. It doesn't matter, as eventually you still have to take the different T1-T2. The different doesn't depend on whether its unit is K or C.
    In practice, nobody says something has negative power. But in the physics context, negative power does make sense. It points out that the heat flow should be the way around. Heat Q can be either positive or negative, then why can't its rate be negative? :wink:
     
  6. Aug 7, 2010 #5
    I see. Thanks a lot! :)
     
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