# Newton's Law of Cooling Help

I have a question about Newton's Law of cooling. Basically I understand that the equation,
http://album6.snapandshare.com/3936/45466/853596.jpg [Broken]
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...

http://album6.snapandshare.com/3936/45466/853597.jpg [Broken]

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?

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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 $$T_{0}$$

arildno
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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.

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 $$T_{0}$$

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

arildno said:
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.

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?

arildno
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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 $A=T_{i}-T_{0}$

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

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!!

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!!

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

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?