# Newton's rate equation, determine constant and re-use

I'm having trouble with the following homework question!

A container is full of liquid at 185C and is cooled by another fluid which was maintained at a temperature of 15C. The temperature of the liquid fell by 75C in 20 minutes. By using Newton's rate equation determine how long ti takes for the liquid in the container to falling temperature to 50C.

Provide an approximation for the time it takes the liquid to stabilize in temperature to 15C.

I tried to calculate this myself and determined the constant to be $k = 0.0290961$. I got this by using:

$$50-15=(110-15)e^{-20k}$$

I then used this constant to determine how long it would take for it to fall to 50 C and I got 34.3183 minutes. I then tried to use the Fourier version which is I think used for approximations:

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

This got me a number very close to 34 minutes again...

The question in the second part asks to provide an approximation for the time taken, but I realise now I can't do this because this equation would just tell you infinity (as that really is how long it would take for it to fall to that temperature). My friend used the Fourier equation in both cases and his answers make more sense to look at (45 minutes for the approximate time to 15 C), but in my notes the one I used is labelled as the rate equation.

What am I doing wrong? Here is my working in full:

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You are doing everything OK. It seems to me that the key is the word "approximation". You are right, mathematically the temperature will be 15 degrees only with $t\to\infty$. However, it is not difficult to calculate that after, e.g, 200 minutes, the temperature is 15.2, which is close enough to 15 to say that it takes approximately 200 minutes to stabilize at 15 degrees. – Artem Dec 15 '12 at 17:42
And of course you did not need to do it twice, because your two approaches are equivalent. – Artem Dec 15 '12 at 17:45
My friend got an answer of 45 minutes though, using this last equation I show (the dT/dt). So I think his trick is using an approximate form of the Newton equation rather than solving for something like 15.1 as the final temperature... but I thought with the accurate k from the Newton equation I should get a similar answer to him, but his k is 0.02 when mine is 0.029... The second one is supposed to be saying the time taken to get to approximate 15 C (from 185 C) but it comes out very close to my first answer... I haven't done the same thing twice. – deed02392 Dec 15 '12 at 17:47
The time constant is defined as $\tau=-1/k\approx 34.369$ min. In some applications a practical rule is to consider 5 time constants $5\tau \approx 172$ min, which corresponds to a final temperature of $T=15+(185-15)e^{-5}\approx 16.1$ C instead of 15C. – Américo Tavares Dec 15 '12 at 19:16

If you have the expression for the temperature $T$ at time $t$, then for your first question just solve $T(t)=50$ for $t$.
For the second question, well, $T(t)\to 15$ as $t\to\infty$, so I'm not sure what they want other than that observation.
Well he wants an approximation. So presumably there's a formula for determining this. I think that's what the Fourier equivalent is. But my friend use Fourier one to get k and I used Newton's law, and we got different ks. But his makes more sense. I just want to know why Newton's law which I figure is more accurate, gives me wrong looking times when his doesn't. – deed02392 Dec 15 '12 at 17:11
JohnD, I'm not sure why you've got an upvote because I don't think that helps me - can you expand what you mean by T(t). – deed02392 Dec 15 '12 at 17:23