Question

Suppose a cup of coffee is at 100 degrees Celsius at time t = 0, it is at 70 degrees
at t = 10 minutes, and it is at 50 degrees at t = 20 minutes. Compute the ambient temperature.

So in the book I was given Newton's Heat equation.
dT/dt=k(A-T) where T is temperature of the coffee. t is time. A is the ambient temperature.
This is a seperable equation so I isolated k to one side with dt and dT/(A-T)
When I took the integral of both sides I got
ln(A-T)=kt+C
which is then
A=e^(kt+C)+T

How do I find ambient temperature or k or C? I am very confused and cannot get past this step.

Thanks!

Answer 1

I always like to subsume the e^C into a new C so the function is

Ce^(kt)+T

A(0) = 100
so, C + T = 100

A(10) = 70
so, Ce^10k+T = 70

also, use the 3rd point, and then you can solve for the constants.