What is the significance of the constant A in Newton's Law of Cooling equation?

[prace]

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

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?

 

[courtrigrad]

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]

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.

 

[prace]

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?

 

[prace]

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]

I don't get your question!
Let's start with the diff.eq, with an assigned initial temperature T_{i}=T(0)[/tex], and an ambient temperature T_{0}[/tex]<br /> We have the diff.eq:<br /> \frac{dT}{dt}=k(T-T_{0}), T(0)=T_{i}<br /> Introduce the new variable: <br /> y(t)=T(t)-T_{0}\to\frac{dy}{dt}=\frac{dT}{dt}, y(0)=T_{i}-T_{0}<br /> Thus, we have the diff.eq problem:<br /> y(t)=ky, y(0)=T_{i}-T_{0}\to{y}(t)=(T_{i}-T_{0})e^{kt}<br /> <br /> Thus, solving for T(t), we get:<br /> T(t)=T_{0}+(T_{i}-T_{0})e^{kt}<br /> or more obscurely:<br /> T(t)=T_{0}+Ae^{kt}<br /> where A=T_{i}-T_{0}

 

[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!

 

[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!

 

[arildno]

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

 

[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?