# Thread: Law of cooling problem

1. ## Law of cooling problem

I am going to use Uo for u nought

A cup of coffee has a temperature of Uo when freshly poured. Then it has cooled to 60 C after 1 min and to 40 C after 2 min. Also the room temperature is 20 C. The temperature u(t) of the coffee at time t changes at a rate proportional to the difference of its temperature and the room temperature: du/dt = -k (u-20) for a constant k>0

Treating k and Uo as fixed parameters, calculate the constant k

I have done similar problems but the initial temperature was given so then I could use seperable equations. Not sure how to go about this I keep running into two variables without a system to solve.

Any help would be greatly appreciated.

2. Originally Posted by billbarber
I am going to use Uo for u nought

A cup of coffee has a temperature of Uo when freshly poured. Then it has cooled to 60 C after 1 min and to 40 C after 2 min. Also the room temperature is 20 C. The temperature u(t) of the coffee at time t changes at a rate proportional to the difference of its temperature and the room temperature: du/dt = -k (u-20) for a constant k>0

Treating k and Uo as fixed parameters, calculate the constant k

I have done similar problems but the initial temperature was given so then I could use seperable equations. Not sure how to go about this I keep running into two variables without a system to solve.

Any help would be greatly appreciated.
You don't need initial conditions. You have the conditions...

u(1) = 60 and
u(2) = 40.

Can you post up what you've done.

From a quick look I got $u(t) = Ce^{-kt} + 20$.

3. Yeah. I got that, but then using initial conditions I would have:

60 = ce^-k + 20 and 40 = ce^-2k + 20

I am still left with c and a k in both of them. If I was given initial temp I would lose the e^k and just be left with the c. That's where my confusion is.

4. Originally Posted by billbarber
Yeah. I got that, but then using initial conditions I would have:

60 = ce^-k + 20 and 40 = ce^-2k + 20

I am still left with c and a k in both of them. If I was given initial temp I would lose the e^k and just be left with the c. That's where my confusion is.
It's just two equations two unknowns.

$Ce^{-k} = 40$
$Ce^{-2k} = 20$

$Ce^{-k} = 40$
$2Ce^{-2k} = 40$

$e^{-k} = 2e^{-2k}$

=> $\frac{e^{-2k}}{e^{-k}} = \frac{1}{2}$

=> $e^{-k} = \frac{1}{2}$

=> $-k = \ln(2^{-1})$

=> $-k = -\ln(2)$

=> $k = \ln(2)$

5. Can't believe I didn't see that one......

Thanks.

6. Originally Posted by billbarber
Can't believe I didn't see that one......

Thanks.
Lol no worries. Earlier I forgot how to differentiate $x^n$. Too much maths around exam time...