Newton's Law of Cooling Help

1. Oct 28, 2006

prace

I have a question about Newton's Law of cooling. Basically I understand that the equation,

Comes from the DE, dT/dt = K(T-To)

Using this, I am to solve this problem:

A thermometer is taken from an inside room to the outside, where the air temperature is 5 °F. After 1 minute, the thermometer reads 55 °F, and after 5 minutes the reading is 30 °F. What is the initial temperature?

So to start, I solved for e^k...

So now that I have e^k, what do I do? My guess is that A is the initial Temperature? But I am not sure and my text does not really explain it too well. So, basically, I guess I am asking, what is the constant A in the general formula mean? And if it is not the initial temperature, or initial condition, then what can I do next with this problem?

2. Oct 28, 2006

Rewrite it as $$y(t) = y_{0}e^{kt}$$ where $$y = T - 5$$.

So $$y(t) = y_{0}(\frac{1}{2})^{\frac{1}{4}t}$$

$$y(1) = y_{0}(\frac{1}{2})^{\frac{1}{4}}$$

Solve for $$y_{0}$$ and then get $$T_{0}$$

3. Oct 28, 2006

arildno

You now have:
$$A=50*(\frac{1}{2})^{-\frac{1}{4}}$$

The initial temperature is now found by computing T(0)

As for what A is, it is the DIFFERENCE between the initial temperature and the ambient temperature.

4. Oct 28, 2006

prace

So what you are saying here is that $$y_{0}$$ in your equation is $$T_{0}$$, which is the initial temperature?

5. Oct 28, 2006

prace

I don't think I am understanding this at all here... Sorry to put you through this, but, if A is the difference between the initial temperature and the ambient temperature, what is the variable for the initial temperature if $$T_{0}$$ is not the initial temperature, but the ambient temperature that arises as time gets very large or goes to infinity?

6. Oct 28, 2006

arildno

Let's start with the diff.eq, with an assigned initial temperature $T_{i}=T(0)[/tex], and an ambient temperature [itex]T_{0}[/tex] We have the diff.eq: $$\frac{dT}{dt}=k(T-T_{0}), T(0)=T_{i}$ Introduce the new variable: [tex]y(t)=T(t)-T_{0}\to\frac{dy}{dt}=\frac{dT}{dt}, y(0)=T_{i}-T_{0}$$
Thus, we have the diff.eq problem:
$$y(t)=ky, y(0)=T_{i}-T_{0}\to{y}(t)=(T_{i}-T_{0})e^{kt}$$

Thus, solving for T(t), we get:
$$T(t)=T_{0}+(T_{i}-T_{0})e^{kt}$$
or more obscurely:
$$T(t)=T_{0}+Ae^{kt}$$
where $A=T_{i}-T_{0}$

7. Oct 28, 2006

prace

$$T(t)=T_{0}+(T_{i}-T_{0})e^{kt}$$

Wow... This really made it clear here. Sorry for the obscure questions, but you really nailed it for me here. I am going to try a few problems in my text and see how they work out. Thanks again!!

8. Oct 29, 2006

prace

Ok, so I worked it out and I got ~ 64.5°. If anyone has the time, would you mind checking this for me as I don't have the answer to this in my text. Thanks!!

9. Oct 29, 2006

arildno

I haven't worked it out, but:
Start having confidence in yourself!
I'm sure you managed it all right.

10. Feb 19, 2010

Sarah12345

I don't understand how we find k in problems like this where no initial temperature is given.
Do you have to compare the temps at t=1 and t=5?