A white wine at room temperature 70°F is chilled in ice (32°F). If it takes 15 min for the wine to chill to 60°F, how long will it take for the wine to reach 56°F?

The initial temperature of white wine at room temperature is70°F and is chilled in ice (32°F). It takes 15 min for the wine to chill to 60°F. By using Newton’s law of cooling, we have to determine the time after which the temperature of the wine will reach 56°F.

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,$T_0={70}^{o}F$,

$M_0={32}^{o}F$

Temperature after 15 min,$T\left(15\right)={60}^{o}F$

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

$$\begin{gathered} T\left(15\right)=32+\left(70-32\right)e^{-15k} \\ 60=32+\left(38\right)e^{-15k} \\ 60-32=\left(38\right)e^{-15k} \\ 28=\left(38\right)e^{-15k} \\ {e}^{15k}=\frac{38}{28} \\ 15k=ln\left(1.357\right) \\ k=\frac{ln\left(1.357\right)}{15} \\ k=0.0203 \end{gathered}$$

One will use this value of k in next step to find the time after which the temperature of the wine will reach 56°F.

Substituting $T\left(t\right)={56}^{o}F$in equation (1),

$$\begin{gathered} 56=32+\left(70-32\right)e^{-(0.0203)t} \\ 56-32=\left(38\right)e^{-(0.0203)t} \\ {e}^{(0.0203)t}=\frac{38}{24} \\ \left(0.0203\right)t=ln\left(1.5834\right) \\ t=\frac{ln\left(1.5834\right)}{0.0203} \\ t=22.6min \end{gathered}$$

Hence, the temperature of wine will reach 56°F after 22.6 minutes.