A cup of coffee with cooling constant k = .09 min^-1 is placed in a room at tempreture 20 degrees C.
How fast is the coffee cooling(in degrees per minute) when its tempreture is T = 80 Degrees C?
Is this just a straightforward application of newtons cooling law where y = 80?
the question goes on to ask to use linear approximation to estimate the change in tempreture over the next 6s when T = 80 degrees C?
Im not too sure how to solve this one, is it just using the linear approximation formula?
and lastly: if the coffee is served at 90 degrees C how long wil it take to reach an optimal drinking tempreture of 65 degrees C?
Is this solved by using the solution to the cooling forumula, the one that is y(t) = b + Ce^(kt) and then just solving for t or do i have the wrong idea?
any help is appreciated, thanks
Best Answer
In order not to break truth-in-solving regulations, let me first state that the Newton Law of Cooling is not a good mathematical model for the cooling of coffee. (There are good models, but they are more complicated.) But let's hold our noses and go on.
Let $y$ be the temperature at time $t$. Then $$\frac{dy}{dt}=-k(y-20).$$ This gives you, without any work, the answer to the first question. Just put $y=80$.
For the approximate change, use the linear approximation. This essentially assumes that over the next $6$ seconds (which is $\frac{1}{10}$ minutes, we need to use that), the rate of change remains exactly what it was at temperature $80$. So multiply the first answer by $\frac{1}{10}$.
or the third question, let $t=0$ when the hot coffee is served. Then by stuff you know, $$y=20+Ce^{-kt},$$ where you can determine $C$ from the fact that $y(0)=90$.
Then set $y=65$ and use logarithms to solve for $t$.