Mathematics: Aroast turkey is taken from an oven when its temperature has reached 185 degrees f and is placed on a table in a room where the temperature is 75 degrees f. (a) if the temperature of the turkey is 150 degrees f after half an hour, what is the temperature after 45 minutes? (b) when will the turkey have cooled to 100 degrees f?

Aroast turkey is taken from an oven when its

Ответ:

raven681

14.03.2022

(a) Using Newton's Law of Cooling, $\dfrac{dT}{dt} = k(T - T_s)$ , we have $\dfrac{dT}{dt} = k(T - 75)$ where T is temperature after T minutes.
Separate by dividing both sides by T - 75 to get $\dfrac{dT}{T - 75} = k dt$ . Integrate both sides to get $\ln|T - 75| = kt + C$ .

Since T(0) = 185

, we solve for C:
$|185 - 75| = k(0) + C\ \Rightarrow\ C = \ln 110$
So we get $\ln|T - 75| = kt + \ln 110$ . Use T(30) = 150 to solve for k:
$\ln| 150 - 75 | = 30k + \ln 110\ \Rightarrow\ \ln 75 - \ln 110= 30k \Rightarrow \\ k= \frac{1}{30}\ln (75/110) = \frac{1}{30}\ln(15/22)$

So

$\ln|T - 75| = kt + \ln 110 \Rightarrow |T - 75| = e^{kt + \ln110} \Rightarrow \\ \\
|T - 75| = 110e^{kt} \Rightarrow T - 75 = \pm110e^{(1/30)\ln(15/22)t} \Rightarrow \\
T = 75 \pm110e^{(1/30)\ln(15/22)t}$

But choose Positive because T > 75. Temp of turkey can't go under.

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} \\
T(45) = 75 + 110e^{(1/30)\ln(15/22)(45)} = 136.929 \approx 137{}^{\circ}F$

(b)

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} = 75 + 110(15/22)^{t/30} \\
100 = 75 + 110(15/22)^{t/30} \\
25 = 110(15/22)^{t/30} 
\frac{25}{110} = (15/22)^{t/30} \\
\ln(25/110) / ln(15/22) = t/30 \\
t = 30\ln(25/110) / ln(15/22) \approx 116\ \mathrm{min}$

Dogs of the AMS.

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