# Thread: Find an accurate rate of decay

1. ## Find an accurate rate of decay

Hello everyone!
I have one question.
Usually when we have a set of data to find the rate of decay,
T = T0 - Tm e^(-kt) + Tm
T0 - Initial temperature
Tm - Surrounding temperature (At 20'C)
k - Constant value
T - Temperature

For eg. when time = 1, T = 90 (Knowing that t= 0 T=100)
So, we will write
90 = 100 - 20 e^(-k*1) + 20
90-20 = 80 e^(-k)
7/8 = e^(-k)
Take natural logs on both sides,
ln 7/8 = -k
k = -0.133531

And because we have t=1,2,3,4,5,6,7,8....n
So I found all the k values and averaged them to find a mean value
for k.

However, I was thinking just by averaging the rate of decay and find
an averaged constant value to find the final function that models
the real data (Experimented data) is not accurate enough.

So I wish to seek for some help on this. Can you please tell me
another way to find a good 'k' value that has higher accuracy to
model the data?

2. use a spreadsheet to do an exponential regression

3. Yep! However, that is one of the ways I used to model the data.
The k value here Im trying to find is for the algebraic method, not by using any technologies. Thank you anyways

4. ## Line of best fit

Hello Modernised
Originally Posted by Modernized
Hello everyone!
I have one question.
Usually when we have a set of data to find the rate of decay,
T = T0 - Tm e^(-kt) + Tm
T0 - Initial temperature
Tm - Surrounding temperature (At 20'C)
k - Constant value
T - Temperature

For eg. when time = 1, T = 90 (Knowing that t= 0 T=100)
So, we will write
90 = 100 - 20 e^(-k*1) + 20
90-20 = 80 e^(-k)
7/8 = e^(-k)
Take natural logs on both sides,
ln 7/8 = -k
k = -0.133531

And because we have t=1,2,3,4,5,6,7,8....n
So I found all the k values and averaged them to find a mean value
for k.

However, I was thinking just by averaging the rate of decay and find
an averaged constant value to find the final function that models
the real data (Experimented data) is not accurate enough.

So I wish to seek for some help on this. Can you please tell me
another way to find a good 'k' value that has higher accuracy to
model the data?
If I understand you correctly, your question is: given a set of experimental values of $\displaystyle t$ and $\displaystyle T$, and a function of the form $\displaystyle T = Ae^{-kt}+B$, how do I find the best value of $\displaystyle k$?

The problem with the method you've described - that of calculating a value of $\displaystyle k$ for each pair of values of $\displaystyle t$ and $\displaystyle T$ and then finding the mean of the answers - is that it will give equal weight to all the experimental data, without rejecting any values that are (through experimental error, perhaps) inconsistent with the majority of the others.

A better method would probably be to manipulate the equation much in the way you have already done, like this:

$\displaystyle T = Ae^{-kt} +B$

$\displaystyle \Rightarrow \ln(T-B) = \ln(Ae^{-kt}) = \ln(A)-kt$

and then plot values of $\displaystyle \ln(T-B)$ against $\displaystyle t$, the graph of which will (theoretically) be a straight line with gradient $\displaystyle -k$. You'll then be able to plot the best straight line through these points, and reject any data that are clearly in error.

5. But...
If I sub in all the data points.. I will still end up with lots of different
k values..

wouldnt I?

so at the end, I will still need to average them.. does that help?

6. Originally Posted by Modernized
But...
If I sub in all the data points.. I will still end up with lots of different
k values..

wouldnt I?

so at the end, I will still need to average them.. does that help?
No - you just plot a single straight line - the best line you can through all the points - find its gradient, and that's your single value of k.

7. Can you please show me how to actually do it? (I understand the concept)
But its just that.. it doesnt make sense for me to find all the k values and average it at the end..
lets say,
t=1, T=96
t=3, T=93
t=6, T=87
t=9, T=84

Can you please show me how to find a good single k value for this?
T=(To-Tm) e^(-kt) + Tm
With this model!

Thank you so much.. Im so confused - -

8. Originally Posted by Modernized
Can you please show me how to actually do it? (I understand the concept)
But its just that.. it doesnt make sense for me to find all the k values and average it at the end..
lets say,
t=1, T=96
t=3, T=93
t=6, T=87
t=9, T=84

Can you please show me how to find a good single k value for this?
T=(To-Tm) e^(-kt) + Tm
With this model!

Thank you so much.. Im so confused - -
Read this: Regression: Fitting Functions to Data

No - you just plot a single straight line - the best line you can through all the points - find its gradient, and that's your single value of k.

Isn't that just simply using the LinReg function on the calculator to find an expression for the linear function.. and it gives you the m (gradient) value..?

But Im trying to do a different thing. With all the different T and t values!
I want to find a single k value without having to go thru,
eg,
77e^-kt = 90
sub t = 2
then find k
and do this again and again for all other datas then average all the k value.

I want to find the k value more accurate by one method.. can you please please explain ? please.. Im really confused and frustrated! Argh

10. Hello Modernized
Originally Posted by Modernized
Isn't that just simply using the LinReg function on the calculator to find an expression for the linear function.. and it gives you the m (gradient) value..?

But Im trying to do a different thing. With all the different T and t values!
I want to find a single k value without having to go thru,
eg,
77e^-kt = 90
sub t = 2
then find k
and do this again and again for all other datas then average all the k value.

I want to find the k value more accurate by one method.. can you please please explain ? please.. Im really confused and frustrated! Argh
Mr F has referred you to an article that explains what you need to do.

I've done it 'manually' using your data (with $\displaystyle T_m=20$), by plotting $\displaystyle \ln(T-T_m)$ against $\displaystyle t$, as I suggested. The graph is shown, with the line indicating what I estimate to be the line of best fit.

Since the equation of this line is $\displaystyle \ln(T-T_m)= \ln(T_0-T_m) -kt$, its gradient gives a (single!) estimate of the value of $\displaystyle -k$ (and, of course, the intercept gives the value of $\displaystyle \ln(T_0-T_m)$).

The gradient, then, is approx $\displaystyle \frac{4.355-4.1}{0-11}=-0.023$, giving the value of $\displaystyle k$ as $\displaystyle 0.023$.