How Does Newton's Law of Cooling Apply to a Roast Turkey?

[Tineeyyy]

Newtons Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between object and it's surroundings. A roast turkey is taken from the oven when it's temperature has reached 185 F and is placed on a table in a room where temperature is 75 F.
1. If the temperature of the turkey is 150 F after half an hour, what is the temperature after 45 minutes?
2. When will the turkey have cooled to 100 F?

 

[DaalChawal]

Assuming $T_1$ and $T_s$ are the temperature of the turkey initially and the temperature of the surroundings respectively. $T_2$ is the temperature of the turkey after some time $t$ then integrating the above expression we get $\Delta T = \Delta T_o e^{-k \Delta T}$ , where $\Delta T_o$ is the initial temperature difference between the turkey and the surroundings.
Now as per the questions put the values and solve.

 

[HOI]

Let the temperature of the turkey 't' minutes after taking it out of the oven be 'T(t)'. Saying that the rate of cooling is proportional to difference between the temperature of the turkey and the temperature of the room means that $\frac{dT}{dt}= kT$ for some constant k. Then $\frac{dT}{T}= kdt$. Integrating, $ln(T)= kt+ c$. Solving for T, $T(t)= Ce^{kt}$. We have to determine the two constants, C and k.

When t= 0, the turkey has just been taken out of the oven, $T(0)= Ce^0= C= 185$. 30 minutes later, $T(30)= Ce^{30k}= 185e^{30k}=150$. $e^{30k}= \frac{150}{185}= \frac{30}{37}$ so $e^{kt}= (e^{30k})^{t/30}= \left(\frac{30}{37}\right)^{t/30}$. $T(t)= 185\left(\frac{30}{37}\right)^{t/30}$.

So when t= 30, $T(30)= 185\left(\frac{30}{37}\right)^{30/30}= 185\left(\frac{30}{37}\right)$.

The turkey will be at 100 F when $185\left(\frac{30}{37}\right)^{t/30}= 100$.