The temperature in a room is 21 degrees Celsius. A thermometer which has been kept in it is placed outside. After 5 minutes the thermometer reading is 16 Celsius. Five minutes later, it is 13 Celsius. Find the outside temperature.
The temperature in a room is 21 degrees Celsius. A thermometer which has been kept in it is placed outside. After 5 minutes the thermometer reading is 16 Celsius. Five minutes later, it is 13 Celsius. Find the outside temperature.
This is related to Newton's Law of Cooling, I presume. (See part way down the page of this link.)
We have that
$\displaystyle T(t) = T_{env} + (T(0) - T_{env})e^{-t/t_0}$
where $\displaystyle T(0)$ is the temperature of the body (in this case the thermometer), $\displaystyle T_{env}$ is the temperature of the environment, and $\displaystyle t_0$ is the "time constant," which we don't know yet.
We know that $\displaystyle T(0) = 21~^oC$ and that at $\displaystyle t = 5~min$, $\displaystyle T(5) = 16~^oC$, and at $\displaystyle t = 10~min$, $\displaystyle T(10) = 13~^oC$.
So:
$\displaystyle T(5) = 16 = T_{env} + (21 - T_{env})e^{-5/t_0}$
and
$\displaystyle T(10) = 13 = T_{env} + (21 - T_{env})e^{-10/t_0}$
Two equations and two unknowns. We want $\displaystyle T_{env}$ and don't care about the time constant, so what I recommend you do is, instead of trying to solve for $\displaystyle t_0$ directly, solve the top equation for $\displaystyle e^{-5/t_0}$. Then note that $\displaystyle e^{-10/t_0} = \left ( e^{-5/t_0} \right )^2$ and use that for your substitution. You'll be left with an equation for $\displaystyle T_{env}$.
Sorry, I gotta run.
-Dan