Question

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?

Answer 1

(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.