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Veronika [31]
3 years ago
5

A roast 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?
Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
4 0
(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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