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?

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

Since [tex]T(0) = 185[/tex], we solve for C:
[tex]|185 - 75| = k(0) + C\ \Rightarrow\ C = \ln 110[/tex]
So we get [tex]\ln|T - 75| = kt + \ln 110[/tex]. Use T(30) = 150 to solve for k:
[tex]\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)[/tex]

So

[tex]\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} [/tex]

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

[tex]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[/tex]

(b)

[tex]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}[/tex]

Dogs of the AMS.