# I am trying to find out two unknown values from one equation

• May 21st 2008, 08:10 PM
SOS
I am trying to find out two unknown values from one equation
Does anyone know how to find two unknown values in the one equation it is for a year twelve maths b assignment. I am trying to find out what c and k are in this equation (by the way c and k are constants):
T=TS + ce^(-kt)
Where TS= Temperature of surroundings which equals 23 degrees
T= Temperature of the body at a time (t)
t= Time
c & k= constants
• May 21st 2008, 08:24 PM
CaptainBlack
Quote:

Originally Posted by SOS
Does anyone know how to find two unknown values in the one equation it is for a year twelve maths b assignment. I am trying to find out what c and k are in this equation (by the way c and k are constants):
T=TS + ce^(-kt)
Where TS= Temperature of surroundings which equals 23 degrees
T= Temperature of the body at a time (t)
t= Time
c & k= constants

You need data, say some times \$\displaystyle t_1, .. t_n\$, and the corresponding tempratures \$\displaystyle T(t_1), .. T(t_n)\$.

RonL
• May 21st 2008, 08:33 PM
SOS
I have Values for T and t but I don't know how to find the 2 unknowns in the 1 eqn
thanks ronL, in class we got the time and temperature which is:
t= 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
T= 93 75 62 51 45 41 37 35 33 31 30 29 28 27.8 27
But even if I substitute those numbers in I still have two unknowns left, and I don't know what to do after I substitute T and t in.
• May 21st 2008, 08:39 PM
Mathstud28
Quote:

Originally Posted by SOS
Does anyone know how to find two unknown values in the one equation it is for a year twelve maths b assignment. I am trying to find out what c and k are in this equation (by the way c and k are constants):
T=TS + ce^(-kt)
Where TS= Temperature of surroundings which equals 23 degrees
T= Temperature of the body at a time (t)
t= Time
c & k= constants

Check your problem again...these types of problems usually come with initial conditions such as

when t= y=
• May 21st 2008, 09:00 PM
SOS
The question in full
ok umm.. this is the question in full:
"Newton's Law of Cooling states that the rate of change of temperature of a body is proportional to the difference in temperature between the body and the temperature of its surrounds."
(I don't actually understand the above sentence is that what you mean by t= y=)
"The formula fo Newton's law of cooling is:
T=TS + ce^(-kt)
Where TS= Temperature of surroundings which equals 23 degrees

T= Temperature of the body at a time (t)
t= Time
c & k= constants"
I then have made a graph of time (t) verse temperature (T) when t=x and T=y
It then says to
"Determine how well Newton's Law of Cooling models your experimental data. Your solution should include the use of a spreadsheet and a comparative graph."
So I thought that it meant that I had to come up with values for the constants so that I could model the formula against the data I obtained in a graph.
• May 22nd 2008, 06:23 AM
CaptainBlack
Quote:

Originally Posted by SOS
thanks ronL, in class we got the time and temperature which is:
t= 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
T= 93 75 62 51 45 41 37 35 33 31 30 29 28 27.8 27
But even if I substitute those numbers in I still have two unknowns left, and I don't know what to do after I substitute T and t in.

You wish to fit a curve of the form:

\$\displaystyle T(t)=T_s+ce^{-kt}\$

to this data.

Rearrange:

\$\displaystyle (T(t)-T_s)=c e^{-kt}\$

Now take logs to get:

\$\displaystyle \log(T(t)-T_s)=-kt+\log(c)\$

so plot \$\displaystyle t\$ against \$\displaystyle u(t)=\log(T(t)-T_s)\$ , this should be a straight line and its slope will be \$\displaystyle -k\$, and the intercept \$\displaystyle \log(c)\$.

RonL