Create a mathematical model for when you put a freshly baked bread in the freezer. When you put the bread in the freezer the bread is $30^OC$ and the freezer has a constant temperature of $-20^OC$. Assume that the cooling of the bread follows Newton's Law of Cooling.
After 1 hour in the freezer the bread is $10^Oc$. What is the temperature after 2 hours?
Been asking a lot of these questions lately and I'm finally starting to get it, however I can't seem to get this one right.
Anyway, I start with putting up the model:
$\frac {dy}{dt} = -k(Y(t) - 20)$
Am I correct with using $-k$ since the temperature is cooling?
$ \frac {dy}{(Y(t)-20)} = -k dt$
$ln(Y(t)-20) = -kt + c $
$Y(t) - 20 = Ce^{-kt}$
Which finally gives me:
$Y(t) = Ce^{-kt} + 20$
The problem is that with this formula I get that after 0 hours when the temperature is $30^OC$ I get C = 10 and $Y(t) = 10e^{-kt} + 20$. When I'm trying to solve for k I get:
$10 = 10e^{-k} + 20$
$-10 = 10e^{-k}$
$-1= e^{-k}$
$-k = ln(-1)$ which obviously doesn't work, where am I doing something wrong here?
Yes, you should use "$-k$", but you should put $\frac{d T}{d t} = -k(T(t) - (-20^0$ C$))$, that is $\frac{d T}{d t} = -k(T(t) + 20^0$C$)$. The temperature difference between the body when at a temperature $T$ and the ambient temperature $-20^0$ C is $T-(-20^0$C$)) = T+20^0$C.