Suppose that a cup of boiling hot coffee (ie its temperature is $100^oC$) is left in a $20^oC$
room and sits until it has cooled to $60^oC$. The cup of coffee is then taken outside where
the temperature is $5^oC$. After 15 minutes, the coffee is now $25^oC$. How long has the
coffee been cooling in total?
Normally I would have some sort of idea where to start, but with this I have absolutely no idea how. Can someone give me some guidance on this?
Thanks.
Best Answer
Hint
Split the problem in two parts and use Newton'law of cooling which write $$T(t)=T_a+c e^{-kt}$$ For $t=0$, you have $T_{ini}=T_a+c$ which means that $c=T_{ini}-T_a$ and then $$T(t)=T_a+(T_{ini}-T_a) e^{-kt}$$ For the first situation, you are given $T_{ini}=100$ and $T_a=20$ and the coffee cools until its temperature reaches $T=60$; if you replace the numbers, the coffee was kept in the room for a time $t_1$ such that $e^{kt_1}=\frac{1}{2}$.
Now, for the second situation, you are given $T_{ini}=60$ and $T_a=5$ and it takes $15$ minutes in order $T=25$. So, $$25=60+(5-60)e^{-15k}$$ then $$k=\frac{1}{15} \log \left(\frac{11}{7}\right)$$ Back to the first situation, we then have $$e^{kt_1}=\frac{1}{2}$$ Replacing $k$ by the above value then gives $$t_1=\frac{15 \log (2)}{\log \left(\frac{11}{7}\right)} \simeq 23$$
So the coffee spent $23$ minutes indoors and $15$ minutes outdoors