Temperature Problem

• Jan 30th 2011, 03:10 PM
VonNemo19
Temperature Problem
Hi. I'm doing my homework problem set and I came across this one:

A small metal bar, whose initial temperature is $20^\circ{C}$, is dropped into a large container of boiling water. How long will it take the bar to reach $90^\circ{C}$ if it is known that its temperature increases $2^\circ{C}$ in $1$ second? How long will it take the bar to reach $98^\circ{C}$?

So, I'm sure that we use Newton's cooling law which states that the rate at which the temperature of an object increases or decreases is proportional to the difference between the object and the surrounding temperature, or

$\frac{dT}{dt}=k(T-T_m)\text{ where }k<0$.

So, from the problem, I can infer that (1) $T(0)=20$; (2) $T(1)=22$; and (3) $T_m=100$ (because this is the boiling point of water).

Newton's differential equation is seperable, so...

$\int\frac{dT}{T-100}=k\int{dt}$ implies that $T(t)=ce^{kt}+100$

Now, $T(0)=20\Rightarrow{ce}^{k(0)}+100=20\Rightarrow{c} =-80$

To find $k$ we use the fact that $T(1)=22$. So, ${-80e}^{k(1)}+100=22$.

All I want to know is if I've done anything wrong so far. So?
• Jan 30th 2011, 03:14 PM
dwsmith
Quote:

Originally Posted by VonNemo19
Hi. I'm doing my homework problem set and I came across this one:

A small metal bar, whose initial temperature is $20^\circ{C}$, is dropped into a large container of boiling water. How long will it take the bar to reach $90^\circ{C}$ if it is known that its temperature increases $2^\circ{C}$ in $1$ second? How long will it take the bar to reach $98^\circ{C}$?

So, I'm sure that we use Newton's cooling law which states that the rate at which the temperature of an object increases or decreases is proportional to the difference between the object and the surrounding temperature, or

$\frac{dT}{dt}=k(T-T_m)\text{ where }k<0$.

So, from the problem, I can infer that (1) $T(0)=20$; (2) $T(1)=22$; and (3) $T_m=100$ (because this is the boiling point of water).

Newton's differential equation is seperable, so...

$\int\frac{dT}{T-100}=k\int{dt}$ implies that $T(t)=ce^{kt}+100$

Now, $T(0)=20\Rightarrow{ce}^{k(0)}+100=20\Rightarrow{c} =-80$

To find $k$ we use the fact that $T(1)=22$. So, ${-80e}^{k(1)}+100=22$.

All I want to know is if I've done anything wrong so far. So?

Looks good. An easy way to test it is once you have the DE take the limit as t goes to infinity. In this case, the value better be 100.
• Jan 30th 2011, 03:31 PM
VonNemo19
Quote:

Originally Posted by dwsmith
Looks good. An easy way to test it is once you have the DE take the limit as t goes to infinity. In this case, the value better be 100.

Oh, Ok. That's a good way to look at it. So, you mean to say that once I've found a a function $y=\phi$ which is a solution to the DE, to take $\displaystyle\lim_{x\to\infty}\phi(x)$?
• Jan 30th 2011, 03:33 PM
dwsmith
Quote:

Originally Posted by VonNemo19
Oh, Ok. That's a good way to look at it. So, you mean to say that once I've found a a function $y=\phi$ which is a solution to the DE, to take $\displaystyle\lim_{x\to\infty}\phi(x)$?

Since your equation is T(t), I would label it as $\displaystyle\lim_{t\to\infty}T(t)$
• Jan 30th 2011, 03:42 PM
zzzoak
You said that:

it is known that its temperature increases http://www.mathhelpforum.com/math-he...2a4b60fde6.png in http://www.mathhelpforum.com/math-he...9a6f75849b.png second.

Why then you need DE ?
• Jan 30th 2011, 03:53 PM
VonNemo19
Quote:

Originally Posted by zzzoak
You said that:

it is known that its temperature increases http://www.mathhelpforum.com/math-he...2a4b60fde6.png in http://www.mathhelpforum.com/math-he...9a6f75849b.png second.

Why then you need DE ?

I thought that at first, too, but then I realized that just because it was observed that the temp went up 2 degrees in the first second didn't mean that this increase would be constant. The facts that this was a problem in cooling/heating and that I just read the section about Newton's DE gave it away too. ;)
• Jan 30th 2011, 04:07 PM
zzzoak
If it is http://www.mathhelpforum.com/math-he...2a4b60fde6.png in the first second then I was wrong. Sorry.You need this DE.