[Math] Solve this question involving temperatures

applicationscalculusderivativesproblem solving

So I am given 2 formulas:
$$ \frac{dT}{dt}=-k(T_t-T_s)$$
Where $\frac{dT}{dt}$ rate at which the object's temperature is changing

$T(t)$ is the temperature of the object at time $t$

$T(s)$ is the constant surrounding temperature

$k$ is some real number to be found


Scenario: Suppose that a cup of soup cooled from 90◦C to 60◦C after 10 minutes in a room whose temperature was 20◦C.
How much longer would the soup take to cool to 35◦C

Please show all working so that I may learn. Thanks in advance.

Best Answer

Notice, we have the following equation $$T(t)-T(s)=(T(0)-T(s))e^{-kt}$$

$\color{red}{\text{Given condition}}$: It takes $t=10\ minutes $ for temperature fall $\color{blue}{90^\circ \ C \to 60^\circ \ C}$

Setting the corresponding values, final temperature $T(t)=60^\circ\ C$ , initial temperature $T(0)=90^\circ\ C$ & surrounding/room temperature $T(s)=20^\circ\ C$ we get $$60-20=(90-20)e^{-k(10)}$$

$$e^{-10k}=\frac{40}{70}=\frac{4}{7}$$ $$-10k=\ln\left(\frac{4}{7}\right)$$ $$k=-\frac{1}{10}\ln\left(\frac{4}{7}\right)\tag 1$$

Now, for temperature fall $\color{red}{90^\circ \ C \to 35^\circ \ C}$ , setting $T(t)=35^\circ \ C$, $T(0)=90^\circ\ C$ & $T(s)=20^\circ\ C$ we get $$35-20=(90-20)e^{-kt}$$ $$e^{-kt}=\frac{15}{70}=\frac{3}{14}$$ $$\implies -kt=\ln\left(\frac{3}{14}\right)$$ Setting the value of $k$ from (1), we get time $t$ as follows $$t\frac{1}{10}\ln\left(\frac{4}{7}\right)=\ln\left(\frac{3}{14}\right)$$ $$t=\frac{10\ln\left(\frac{3}{14}\right)}{\ln\left(\frac{4}{7}\right)}\approx 27.53\ minutes$$