Newton's law of cooling and thermometer

1. Nov 24, 2009

chrisdk

1. The problem statement, all variables and given/known data
Hi, I need some help in solving Newton's law of cooling problem.

Use Newton’s law of cooling to determine the
reading on a thermometer 5 min after it is taken from an oven
at 72 ◦ C to the outdoors where the temperature is 20 ◦ C, if
the reading dropped to 48 ◦ C after one min.

2. Relevant equations

$$\frac{dT}{dt}=k(T-R)$$

3. The attempt at a solution

Last edited: Nov 24, 2009
2. Nov 24, 2009

Staff: Mentor

What have you tried to do? You have to make an attempt before anyone will give you any help - them's the rules.

3. Nov 24, 2009

chrisdk

OK, we let T be the temperature of the object. Then, by Newton's Law of Cooling, $$\frac{du}{dt}=k(u-20)$$, for some constant k. Let $$y=T-20$$. Then $$\frac{dy}{dt}=\frac{du}{dt}=ky$$. Thus we are now using the formula $$y=y_{0}e^{kt}$$. Since T is initially 72 ◦ C , $$y_{0}=72-20=52$$. So, $$y=52e^{kt}$$.
This is what I can do, so I'm asking you for some further help.

Last edited: Nov 24, 2009
4. Nov 24, 2009

Staff: Mentor

You have thrown in an extra variable that you don't need.

Then, by Newton's Law of Cooling, $$\frac{dT}{dt}=k(T-20)$$, for some constant k. Let $$u=T-20$$. Then $$\frac{du}{dt}=ku$$, and the solution to this differential equation is $$u=u_{0}e^{kt}$$. Since T is initially 72 ◦ C , $$u_{0}=72-20=52$$. So, $$u=52e^{kt}$$.

Reverting to T (we really want to know the temperature), we have T - 20 = 52ekt, or T(t) = 20 + 52ekt.

You're given that T(1) = 48, so use this fact to solve for k. After you have k, evaluate T(5) and you're done.

5. Nov 24, 2009

chrisdk

this helped me a lot, thank you! :)