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posted by  jbar1628 on 11/3/2009 2:02:56 PM  |  status: Closed  |  Earned Karma: 44

Help with this problem

Course Textbook Chapter Problem Needs by
Differential Equations N/A N/A N/A 11/4/2009 at 12:00:00 PM
Question Details:

 

It takes a room thermometer
where the temperature F is 90 and is carried outside where the temperature
air is F 0. After three quarters of a minute the thermometer
set F 60.
_____

a) What is the thermometer reading in t = 1 ?

b) How long does the thermometer in reach C ° 0 ?

c) Which is the temperature when t -> ∞?
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Oracle
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posted by anonymousxyz on 11/3/2009 5:56:52 PM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:


T( t ) = T_{Env} + (T_0 - T_{Env})e^{-kt}


T_{Env} = 0^\circ \text{ F}, \qquad T_0 = 90^\circ \text{ F}


T( t ) = 0 + (90 - 0)e^{-kt} = 90e^{-kt}


\text{(a)}


\text{Find \textit{k}:}


\begin{align*} & T( 0.75 ) = 60^\circ \text{ F} &\Rightarrow\\\\& 90e^{-0.75k} = 60 &\Rightarrow\\\\& e^{-0.75k} = \frac{2}{3} &\Rightarrow\\\\& \ln e^{-0.75k} = \ln \left(\frac{2}{3}\right) &\Rightarrow\\\\& {-0.75k} \ln e = \ln \left(\frac{2}{3}\right) &\Rightarrow\\\\& {-0.75k} = \ln \left(\frac{2}{3}\right) &\Rightarrow\\\\& k = \left(\!{-\frac{4}{3}}\right) \ln \left(\frac{2}{3}\right) \end{align*}


T( t ) = 90e^{-\left(\!{-\frac{4}{3}}\right) \ln \left(\frac{2}{3}\right)t} = 90\left(e^{\ln \left( \frac{2}{3} \right)}\right)^{\frac{4t}{3}} = 90\left\left( \frac{2}{3} \right)}^{\frac{4t}{3}}


T( 1 ) = 90\left\left( \frac{2}{3} \right)}^{\frac{4}{3}} = 52.4148279^\circ \text{ F} \approx 52.415^\circ \text{ F}


\text{(b)}


\begin{align*} & T( t ) = 0^\circ \text{ C} = 32^\circ \text{ F} &\Rightarrow\\\\& 90 \left( \frac{2}{3} \right)^{\frac{4t}{3}} = 32 &\Rightarrow\\\\& \left( \frac{2}{3} \right)^{\frac{4t}{3}} = \frac{32}{90} &\Rightarrow\\\\& \ln \left( \frac{2}{3} \right)^{\frac{4t}{3}} = \ln \left( \frac{32}{90} \right) &\Rightarrow\\\\& \left(\frac{4t}{3}\right) \ln \left( \frac{2}{3} \right) = \ln \left( \frac{32}{90} \right) &\Rightarrow\\\\& \end{align*}


t = \frac{ 3 \ln \left( \frac{32}{90} \right)} { 4\ln \left( \frac{2}{3} \right) } = 1.91275478 \text{ min } \approx 1.913 \text{ min }


\begin{align*} &\text{(c)} \\\\& T( t ) = 90e^{-\left(\!{-\frac{4}{3}}\right) \ln \left(\frac{2}{3}\right)t} = 90e^{-0.540620144t} \\\\& \lim_{t\rightarrow \infty }T( t ) &=\\\\& \lim_{t\rightarrow \infty }90e^{-0.540620144t}&=\\\\& \lim_{t\rightarrow \infty } \frac{90}{e^{0.540620144t}}&=\\\\& \frac{90}{e^\infty}&=\\\\& \frac{90}{\infty} &=\\\\& 0^\circ\text{ F } \end{align*}
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