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Newton's Law of Cooling Flawed?

  1. Feb 11, 2010 #1
    Newton's Law of Cooling (not the formal definition):

    (change in time) = -ln ((Tf - S)/(Ti - S)) / k

    Tf = final temperature
    Ti = initial temperature
    S = temperature of environment
    k = heat transfer coefficient

    Say that you wanted to cool something (such as a person) to a negative temperature (Tf would be negative) and the temperature of the environment was positive. This would mean that you would have to take the -ln (-#). Obviously, you can't do this. In another situation, say the surrounding temperature was the same as the final temperature. This would mean that you would have to take the -ln (0) which can't be done. How can I apply this formula to these situations?
     
  2. jcsd
  3. Feb 11, 2010 #2

    Mapes

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    Hi MHrtz, welcome to PF. Perhaps that's an indication that it's not possible to cool something by warming it? :smile:

    For the other part: conduction can only bring an object's temperature asymptotically close to the surrounding environment's. In other words, you can get arbitrarily close to the ambient temperature, but in theory it would take infinite time ([itex]-\ln 0[/itex]) to reach it. Does this help answer your question?
     
  4. Feb 11, 2010 #3
    what you actually compute here is the time that an object takes to cool from [itex] T_i [/itex] to
    [itex] T_f [/itex]. The fact that you get no answer in the first case is to be expected because the object wil never cool to a temperature below the environment.
    You get no answer in the second case, because you try to compute the time that your object reaches [itex] T_f [/itex], but your object is at [itex] T_f [/itex] all the time.
     
  5. Feb 11, 2010 #4
    Ok, I see what you mean. I guess I was too focused on the formula itself rather than what it implied. Thank you for the help.
     
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