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Thread: Transposition of Formulae - Can someone check my answers

  1. #1
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    Transposition of Formulae - Can someone check my answers

    Hi,

    I've been asked the following question:

    The temperature of an electrical device at a time 't' is given by the function:

    $\displaystyle {T}_{2} = {T}_{1}(1-{e}^{\frac{-t}{\tau}}) $

    Where $\displaystyle {T}_{1} $ is the temperature at time t = 0 and $\displaystyle \tau $ is a constant

    Find a) $\displaystyle {T}_{1} $ when $\displaystyle {T}_{2} $ is 60 degrees Celsius, t = 50 seconds and $\displaystyle \tau $ = 90 seconds

    b) The time t, for $\displaystyle {T}_{2} $ to be $\displaystyle \frac{1}{3} $ the value of $\displaystyle {T}_{1} $


    I believe the answer for (a) to be:

    $\displaystyle \frac{60}{1-{e}^{\frac{-50}{90}}} = {T}_{1} = 140.763593 $

    However for question (b) I've gotten:

    $\displaystyle 46.92119767 = 140.763593(1-{e}^{\frac{-t}{90}})$

    $\displaystyle \frac{46.92119767}{140.763593} = 1-{e}^{\frac{-t}{90}}$

    $\displaystyle (-1)\frac{46.92119767}{140.763593} = (-1)1-{e}^{\frac{-t}{90}}$

    $\displaystyle -\frac{46.92119767}{140.763593} = -1+{e}^{\frac{-t}{90}}$

    $\displaystyle -\frac{46.92119767}{140.763593} + 1 = -1+1+{e}^{\frac{-t}{90}}$

    $\displaystyle -\frac{46.92119767}{140.763593} + 1 = {e}^{\frac{-t}{90}}$

    $\displaystyle ln(-\frac{46.92119767}{140.763593} + 1) = ln({e}^{\frac{-t}{90}})$

    $\displaystyle ln(-\frac{46.92119767}{140.763593} + 1) = \frac{-t}{90}$

    $\displaystyle ln(-\frac{46.92119767}{140.763593} + 1)(90) = -t$

    $\displaystyle (-1)ln(-\frac{46.92119767}{140.763593} + 1)(90) = (-1)-t$

    $\displaystyle (-1)ln(-\frac{46.92119767}{140.763593} + 1)(90) = t$

    $\displaystyle t = 36.49185973 $ Seconds

    Can someone please let me know if I've gotten these correct?

    Thanks.
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  2. #2
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    They look fine, though your LaTeXing needs work...
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  3. #3
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    Brill thanks.

    Yeh my LaTeXing kinda sucks.. my handwritings even worse.
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  4. #4
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    You don't need to use the "$\displaystyle T_1$" from the first problem. When $\displaystyle T_2= \frac{1}{3}T_1$, the equation becomes $\displaystyle \frac{1}{3}T_1= T_1(1- e^{-t/\tau})$. Divide both sides by $\displaystyle T_1$: $\displaystyle \frac{1}{3}= 1- e^{-t/90}$ so $\displaystyle e^{-t/90}= \frac{2}{3}$. $\displaystyle -t/90= ln(2/3)$, $\displaystyle t= -90 ln(2/3)$.
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