Cooling Curve Analysis: t vs. T Difference

[Stacyg]

When a liquid is cooled, the difference between the temperature of the liquid and the surrondings is measured. The results are:


t(min) 10 20 30 40
T(°C) 60.7 36.8 22.3 13.5


Where t= time from the start of the cooling
T= temperature difference

Q(i) Process the above data and show that the cooling curve has the form:

T=Ae^(-kt)

I'm not sure how to do this. I have gone through the textbook and there are no similar questions to this. The only one close has a different form.
I tried getting a value for A using a computer programme, but it said that there is no algebraic solution. I also tried getting a value for -k that worked but I'm not sure how to transform this data to show a cooling curve.

b) Find the values of A and k.

Thanks for any help.

 

[Tedjn]

If the data obeys the equation T = Ae^-kt, what happens if you take the natural logarithm of both sides?

 

[Stacyg]

Wouldn't this transform it into a straight line ?

 

[Defennder]

Yes it would. And that makes it a whole lot easier to obtain the value of A and k graphically. Plot the resulting straight line graph with appropriate axes. Then from the gradient and y-intercept, you could get the values you need after some algebraic manipulation.