nemired9

2021-12-31

When a body is removed from an oven, its temperature is measured at ${230}^{\circ }F$. Four minutes later its temperature is ${120}^{\circ }F$. If the room temperature is found to be ${70}^{\circ }F$, how long will it take for the chicken to cool off to a temperature of ${90}^{\circ }F$. What is the steady state temperature of the body?

lovagwb

According to question given that
When a body is removed from an oven measured temperature $={23}^{\circ }F$
4 minutes later temperature is $={120}^{\circ }F$
Room temperature $={70}^{\circ }F$
Apply the Newtons

Mason Hall

$T-{T}_{s}=\left({T}_{0}-{T}_{s}\right){e}^{-kt}$
${T}_{0}={210}^{\circ }F$
${T}_{s}={50}^{\circ }F$
$T={100}^{\circ }F$
$T-T=\left({T}_{0}-{T}_{s}\right){e}^{-kt}$
$100-50=\left(210-50\right){e}^{-k×4}$
$\frac{5}{16}={e}^{-4k}$
$\mathrm{ln}\left(\frac{5}{16}\right)=-4k$
$k=\frac{\mathrm{ln}\left(\frac{5}{16}\right)}{-4}=0.291$.
Find t hv $t=70$
$70-50=\left(210-50\right){e}^{-0.291×t}$
$\frac{2}{16}={e}^{-0.291t}$
$\mathrm{ln}\left(\frac{1}{8}\right\}\right)=-0.291t$
$t=7.15min$.

Vasquez

We have two constants to solve for, but if we plug in what we know we can find them.
At $t=0:210=C{e}^{0}+70⇒C=140$
At $t=30:140=140{e}^{-k\cdot 30}+70⇒{e}^{k\cdot 30}=\frac{1}{2}$
$⇒k=-\frac{1}{30}\mathrm{ln}\left(\frac{1}{2}\right)=\frac{1}{30}\mathrm{ln}\left(2\right)$
So we now have the full function of T with respect to t
$T=140{e}^{-\frac{1}{30}\mathrm{ln}\left(2\right)t}+70$
Now to finish the problem we solve for t when T=100
$\begin{array}{}100=140{e}^{-\frac{1}{30}\mathrm{ln}\left(2\right)t}+70\\ \frac{30}{140}={e}^{-\frac{1}{30}\mathrm{ln}\left(2\right)t}\\ \mathrm{ln}\left(\frac{3}{14}\right)=-\frac{1}{30}\mathrm{ln}\left(2\right)t\\ t=30\frac{\mathrm{ln}\left(\frac{14}{3}\right)}{\mathrm{ln}\left(2\right)}\\ t=30\cdot 2.22239=66.67177\end{array}$

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