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Math Help - Newtons Law of Cooling

  1. #1
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    Exclamation Newtons Law of Cooling

    Hi all,

    Ive been trying to work this question out on my own for over a week now and cant seem to crack it. I hope its in the right Sub forum and im really sorry if its not.

    A hot metal body is allowed to cool in still air and the difference in temperature, θ (Celsius) between the body and its surroundings is recorded at, t (seconds) from the beginning of the cooling period.

    its gives me the following:

    (t) θ
    40.5 60
    96.0 50
    137.0 40
    210.5 30
    327.0 20
    488.5 10

    The question asks me to verify graphically the relationship between θ and t is of the form: θ=θoe-kt (-kt is to the power of θoe) Hence estimate the values of the constants θo and k. Then using this relationship compile a table of values of θ for values of t between 0 and 600 at intervals of 60.

    I have taken the natural log (ln) of each of the temperatures in the table above and plotted them in the graph. I drew a line of best fit and found the y intercept, and took the natural exponential of it which is:

    e4.25 = 70.11

    I think 70.11 is the initial temperature at time 0. so θo= 70.11?

    Is k the gradient of the line? if so how do i calculate the values of θ?


    All help is greatly appreciated, it would be amazing to get this out of the way!
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  2. #2
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    Newton's law says that \theta = \theta_0e^{-kt}. Take natural logarithms of that equation: \ln\theta = \ln\theta_0 - kt. So what you should do is to graph the logarithm of theta against t. The graph should then look like a straight line, with gradient –k and intercept \ln\theta_0. I haven't tried to check the numbers, but it looks as though that is what you have done. If so, you're right!

    Once you have the values for \theta_0 and k, you can plug them into the original formula \theta = \theta_0e^{-kt}, and use it to find \theta for any given value of t.
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  3. #3
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    Ok i think im right so far. All i need now is to estimate the values of the constants θo and k.

    I have worked out

    θo = 70.11 and k = -0.004

    Now to compile a table of values of θ for values of t between 0 and 600 at intervals of 60.


    (t) θ
    0 70.11
    60 ?
    120 ?
    180 ?
    240 ?
    300 ?

    And so on...

    do i still use ?

    I cant figure out how to do this?!
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  4. #4
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    Thanks Opalg! But, plugging them straight into the formula is the bit im having trouble with! Would you be able to explain or show and example please?
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  5. #5
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    Once you've done as Opalg suggested, you'll need to fit a straight line to the data using linear regression (least squares fit) or some such estimator. What do you get when you do that?
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    Sorry i dont know what that means! Im just not so confident about changing the subject of the formula to t.
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  7. #7
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    Quote Originally Posted by skyhook2 View Post
    Ok i think im right so far. All i need now is to estimate the values of the constants θo and k.

    I have worked out

    θo = 70.11 and k = -0.004

    Now to compile a table of values of θ for values of t between 0 and 600 at intervals of 60.


    (t) θ
    0 70.11
    60 ?
    120 ?
    180 ?
    240 ?
    300 ?

    And so on...

    do i still use ?

    I cant figure out how to do this?!
    If \theta_0 = 70.11 and k=-0.004, then your formula is \theta = 70.11e^{-0.004t}. For example, when t = 120, \theta is given by

    \theta = 70.11e^{-0.004\times120} = 70.11e^{-0.48} \approx 70.11\times0.62\approx43.4.

    But there's something wrong with this! It says that after 120 seconds, the temperature difference between the hot object and its surroundings (43.4 degrees) is greater than it was at t=60 seconds, namely 40.5 degrees according to the given data. In other words, the object has started to heat up again!

    Once you see that this can't be correct, you should look back to check the values that you got for \theta_0 and k. The value of \theta_0 is the temperature difference at time 0, which should obviously be greater than its value at any later time. But you are told that \theta = 488.5 when t=10. So the value \theta_0 = 70.11 is very wrong. I should think it would be about ten times that.
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  8. #8
    A Plied Mathematician
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    Here's what I got - I think it looks pretty reasonable.

    Newtons Law of Cooling-newtons-law-cooling-fit.jpg
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  9. #9
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    opalg, if you look back to the first part it gives me the following (you have theta and t the wrong way round :

    (t) ----------- θ
    40.5 ----------- 60
    96.0 ----------- 50
    137.0 ----------- 40
    210.5 ----------- 30
    327.0 ----------- 20
    488.5 ----------- 10

    I have worked out the gradient wrong!
    I think it is -0.2.
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  10. #10
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    Quote Originally Posted by skyhook2 View Post
    you have theta and t the wrong way round
    That would explain it!
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