A red wine is brought up from the wine cellar, which is a cool 10°C, and left to breathe in a room of temperature 23°C. If it takes 10 min for the wine to reach 15°C, when will the temperature of the wine reach 18°C?

The initial temperature of red wine is10°C and left to breathe in a room oftemperature 23°C. It takes 10 min for the wine to reach 15°C. By using Newton’s law of cooling, we have to determine the time after which the temperature of the wine will reach 18°C.

Newton’s Law of Cooling is,

$T\left(t\right)=M_0+\left(T_0-M_0\right)e^{-kt}······\left(1\right)$

Here, we will take the values as,

Initial temperature,$T_0={10}^{o}C$,

The temperature of the room,$M_0={23}^{o}C$

Temperature after 10 min,$T\left(10\right)={15}^{o}C$

Using the given values in equation (1), to find the value of k,

$$\begin{gathered} T\left(10\right)=23+\left(10-23\right)e^{-10k} \\ 15=23+\left(-13\right)e^{-10k} \\ 15-23=\left(-13\right)e^{-10k} \\ -8=\left(-13\right)e^{-10k} \\ {e}^{10k}=\frac{13}{8} \\ 10k=ln\left(1.625\right) \\ k=\frac{ln\left(1.625\right)}{10} \\ k=0.0485 \end{gathered}$$

One will use this value of k in next step to find the time after which the temperature of the wine will reach 18°C.

Substituting $T\left(t\right)={18}^{o}C$in equation (1),

$$\begin{gathered} 18=23+\left(10-23\right)e^{-(0.0485)t} \\ 18-23=\left(-13\right)e^{-(0.0485)t} \\ {e}^{(0.0485)t}=\frac{13}{5} \\ \left(0.0485\right)t=ln\left(2.6\right) \\ t=\frac{ln\left(2.6\right)}{0.0485} \\ t=19.7min \end{gathered}$$

Hence, the temperature of the wine will reach 18°C after 19.7 minutes.