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Homework Help: Cooling Curve

  1. May 19, 2008 #1
    When a liquid is cooled, the difference between the tempurature of the liquid and the surrondings is measured. The results are:


    t(min) 10 20 30 40
    T(°C) 60.7 36.8 22.3 13.5


    Where t= time from the start of the cooling
    T= tempurature difference

    Q(i) Process the above data and show that the cooling curve has the form:

    T=Ae^(-kt)

    I'm not sure how to do this. I have gone through the text book and there are no similar questions to this. The only one close has a different form.
    I tried getting a value for A using a computer programme, but it said that there is no algebraic solution. I also tried getting a value for -k that worked but i'm not sure how to transform this data to show a cooling curve.

    b) Find the values of A and k.

    Thanks for any help.
     
  2. jcsd
  3. May 19, 2008 #2
    If the data obeys the equation T = Ae-kt, what happens if you take the natural logarithm of both sides?
     
  4. May 23, 2008 #3
    Wouldn't this transform it into a straight line ?
     
  5. May 24, 2008 #4

    Defennder

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    Homework Helper

    Yes it would. And that makes it a whole lot easier to obtain the value of A and k graphically. Plot the resulting straight line graph with appropriate axes. Then from the gradient and y-intercept, you could get the values you need after some algebraic manipulation.
     
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