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

[ineedhelpnow]

Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the objects and its surroundings. Suppose that a roast turkey is taken from an oven when its temperature has reached $200^oF$ and is placed on a table in a room where the temperature is $85^oF$. If $u(t)$ is the temperature of the turkey after t minutes, then Newton's Law of Cooling implies that $\frac{du}{dt}=k(u-85)$. This could be solved as a separable differential equation. Another method is to make the change of variable $y=u-85$. If the temperature of the turkey is $170^oF$ after half an hour, what is the temperature after $10$ minutes.

Confused. Please help.

 

[MarkFL]

Newton's law of Cooling states that the time rate of change of the temperature $T$ of an object is proportional to the difference between the ambient temperature $M$ and the temperature of the object. Stated mathematically, this is:

\(\displaystyle \frac{dT}{dt}=-k(T-M)\) where \(\displaystyle T(0)=T_0,\,0<k\in\mathbb{R}\) and \(\displaystyle T>M\).

The ODE is separable and may be written:

\(\displaystyle \frac{1}{T-M}\,dT=-k\,dt\)

Integrating, using the boundaries, and dummy variables of integration, we find:

\(\displaystyle \int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t \,dv\)

Can you proceed?

 

[ineedhelpnow]

i kind of feel like i know how to do it but at the same time I am really confused and i can't think that straight right now. do you mind going through the problem?

 

[MarkFL]

There is another way to solve the ODE associated with the IVP, and that is to write it in standard linear form:

\(\displaystyle \frac{dT}{dt}+kT=kM\)

Compute the integrating factor:

\(\displaystyle \mu(t)=e^{k\int\,dt}=e^{kt}\)

Multiply through by this factor:

\(\displaystyle e^{kt}\frac{dT}{dt}+ke^{kt}T=kMe^{kt}\)

Observing that the left side is now the derivative of the product of the integrating factor and the dependent variable, we may now write:

\(\displaystyle \frac{d}{dt}\left(e^{kt}T \right)=kMe^{kt}\)

Integrate with respect to $t$:

 

[ineedhelpnow]

im doing it as a seperable equation. so i integrated both sides and i have ln(u-85)+C=kt. what do i do now?

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and when it says what is the temperature after 10 minutes, does it mean another ten minutes (so 40 min altogether) or does it mean if it was 170 after half an hour what was it after 10 min?

 

[MarkFL]

I would write the solution to the general ODE as:

\(\displaystyle \ln(u-85)=-kt+C\)

simply because I like the heat transfer coefficient ($k$) to be non-negative. I would next translate from logarithmic to exponential form, and then use the two given points on the curve:

\(\displaystyle u(0)=200\)

\(\displaystyle u(30)=170\)

to determine the two parameters, $C$ and $k$. Then, once you have the solution to the IVP, evaluate:

\(\displaystyle u(10)\).

 

[ineedhelpnow]

Figured it out. Thanks.