A cold beer initially at 35°F warms up to 40°F in 3 min while sitting in a room of temperature 70°F. How warm will the beer be if left out for 20 min?

The initial temperature of a cold beer is35°F and its temperature after sitting in a room of temperature 70°F for 3 min, becomes 40°F. By using Newton’s law of cooling, we have to determine the temperature of beer after 20 minutes.

Newton’s Law of Cooling is,

$\mathit{T}\left(t\right)\mathbf{=}{\mathit{M}}_{\mathbf{0}}\mathbf{+}\left(T_0-M_0\right){\mathit{e}}^{\mathbf{-}\mathbf{k}\mathbf{t}}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\left(1\right)$

Here, we will take the values as,

The initial temperature of the beer,${\mathit{T}}_{\mathbf{0}}\mathbf{=}{\mathbf{35}}^{\mathbf{o}}\mathit{F}$

The temperature of the room,${\mathit{M}}_{\mathbf{0}}\mathbf{=}{\mathbf{70}}^{\mathbf{o}}\mathit{F}$

Temperature after 3 min,$\mathit{T}\left(3\right)\mathbf{=}{\mathbf{40}}^{\mathbf{o}}\mathit{F}$

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

$$\begin{gathered} T\left(3\right)=70+\left(35-70\right)e^{-3k} \\ 40=70+\left(-35\right)e^{-3k} \\ 40-70=\left(-35\right)e^{-3k} \\ -30=\left(-35\right)e^{-3k} \\ {e}^{3k}=\frac{35}{30} \\ 3k=ln\left(1.167\right) \\ k=\frac{ln\left(1.167\right)}{3} \\ k=0.0515 \end{gathered}$$

One will use this value of k in next step to find the temperature of beer after 20 minutes.

Substituting $\mathit{t}\mathbf{=}\mathbf{20}\mathit{m}\mathit{i}\mathit{n}$in equation (1),

$$\begin{gathered} T\left(20\right)=70+\left(35-70\right)e^{-20\left(0.0515\right)} \\ =70+\left(-35\right)e^{-1.03} \\ =57.5^{0}F \end{gathered}$$