91Ó°ÊÓ

The initial temperature of a cup of hot coffee is95°C and its temperature after sitting in a room of temperature 21°C for 5 min, becomes 80°C. By using Newton’s law of cooling, we have to determine the time after which the temperature of the coffee will reach 50°C.

Newton’s Law of Cooling is,

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

Here, we will take the values as,

Initial temperature of coffee,${\mathit{T}}_{\mathbf{0}}\mathbf{=}{\mathbf{95}}^{\mathbf{o}}\mathit{C}$

Temperature of the room,${\mathit{M}}_{\mathbf{0}}\mathbf{=}{\mathbf{21}}^{\mathbf{o}}\mathit{C}$

Temperature after 5 min,$\mathit{T}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{=}{\mathbf{80}}^{\mathbf{o}}\mathit{C}$

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

$$\begin{gathered} T\left(5\right)=21+\left(95-21\right)e^{-5k} \\ 80=21+\left(74\right)e^{-5k} \\ 80-21=\left(74\right)e^{-5k} \\ 59=\left(74\right)e^{-5k} \\ {e}^{5k}=\frac{74}{59} \\ 5k=ln\left(1.254\right) \\ k=\frac{ln\left(1.254\right)}{5} \\ k=0.0453 \end{gathered}$$

One will use this value of k in next step to find the time after which the temperature of coffee will reach 50°C.

Substituting $\mathit{T}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{50}}^{\mathbf{o}}\mathit{C}$in equation (1),

$$\begin{gathered} 50=21+\left(95-21\right)e^{-(0.0453)t} \\ 50-21=\left(74\right)e^{-(0.0453)t} \\ {e}^{(0.0453)t}=\frac{74}{29} \\ \left(0.0453\right)t=ln\left(2.552\right) \\ t=\frac{ln\left(2.552\right)}{0.0453} \\ t=20.7min \end{gathered}$$

Hence, the temperature of coffee will reach 50°C after 20.7 minutes.