How Do I Graph Logarithms with Negative Rate of Cooling?

[Peter G.]

I am finished collecting data for my experiment and now I am processing it.

I have a graph of temperature against time from which I will determine, after processing, whether or not rate of cooling is proportional to the (excess temperature)^5/4

I've produced my tables, calculated uncertainties and etc. but when I decided to go log my rate of cooling, I noted that it is negative. How am I supposed to do a graph of the log of excessive temperature against the log of the rate of cooling if my gradient for the rate of cooling was negative?

Thanks,
Peter G.

EDIT: Anyone has any ideas? I am happy to explain it more clearly if needed, give more information etc... I am really confused and I am doing everything within my reach to try and figure this out but I am helpless!

 

[I like Serena]

Actually, the rate of "heating" is negative.
The rate of "cooling" is positive, so you can take a log.

 

[Peter G.]

Hi, thanks!

But would you mind explaining? Because in my graph, the gradient is negative. When the temperature is high, the rate of cooling is faster in it.

 

[I like Serena]

Hi, thanks!

But would you mind explaining? Because in my graph, the gradient is negative. When the temperature is high, the rate of cooling is faster in it.

The rate of heating is the "increase in temperature per unit of time".
The rate of cooling is the "decrease in temperature per unit of time".
So the rate of cooling is the opposite of the rate of heating.

Your gradient is the rate of heating, so you need to apply a minus sign.

At a high excess temperature the rate of cooling is indeed higher than at a low temperature.
This is exactly what you need to model, that is, the rate of cooling versus the excess temperature.
Your graph shows the temperature versus the time.

If you would make a separate column for the excess heat, and another separate column for the rate of cooling, you can make a new graph from that.

If you furthermore take the log of the rate of cooling, and graph that versus the excess heat, you should find a slope that according to your problem statement might be 5/4.

 

[Peter G.]

Thanks a lot for your help

I did all the calculations with several of my attempts and the best line I got was of gradient of 1.73 as an oppose to 1.25, which, I think is not that far off

Peter G.