On a hot Saturday morning while people are working inside, the air conditioner keeps the temperature inside the building at $\mathbf{24}\mathbf{°}\mathbf{C}$. At noon the air conditioner is turned off, and the people go home. The temperature outside is a constant$\mathbf{35}\mathbf{°}\mathbf{C}$for the rest of the afternoon. If the time constant for the building is 4 hr, what will be the temperature inside the building at 2:00 p.m.? At 6:00 p.m.? When will the temperature inside the building reach$\mathbf{27}\mathbf{°}\mathbf{C}$?

Newton’s Law of cooling is, $\mathit{T}\left(t\right)\mathbf{=}\mathit{M}\mathbf{+}\mathit{C}{\mathit{e}}^{\mathbf{-}\mathbf{k}\mathbf{t}}$

The temperature inside the building is $24°\mathrm{C}$. The temperature outside is a constant 35°C for the rest of the afternoon. If the time constant for the building is 4 hr. it has to find the temperature inside the building at 2:00 p.m. and at 6:00 p.m. Also, we have to find the time when the temperature will reach $27°\mathrm{C}$.

Newton’s Law of cooling is,

$T\left(t\right)=M+C{e}^{-kt}$ …… (1)

Here, it will take the values as,

Initial temperature,$T_0={24}^{o}C$,

Constant temperature outside the room, $M={35}^{o}C$.

Time constant for the building is 4 hr i.e., $\frac{1}{k}=4$.

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

So, at t=0,

$$\begin{gathered} T\left(0\right)=35+C{e}^0 \\ 24=35+C \\ C=-11 \end{gathered}$$

Thus, the temperature inside the building at time, t is

$T\left(t\right)=M-11e^{-\frac{t}{4}}$ .....................(2)

Substitute t=2 and$\mathrm{M}=35°\mathrm{C}$in equation (2),

$$\begin{gathered} T\left(2\right)=35-11e^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{4}}} \\ T\left(2\right)=28.3^{o}C \end{gathered}$$

Hence,the temperature inside the building at 2:00 p.m. will be 28.3°C.

Substitute $\mathrm{t}=6\mathrm{and}\mathrm{M}={35}^{\mathrm{o}}$in equation (2),

$$\begin{gathered} \mathrm{T}\left(6\right)=35-11{\mathrm{e}}^{-\frac{6}{4}} \\ \mathrm{T}\left(2\right)=32.5^{\mathrm{o}}\mathrm{C} \end{gathered}$$

So,the temperature inside the building at 6:00 p.m. will be 32.5°C.

Substitute $\mathrm{T} \left(\mathrm{t}\right)=27^{\mathrm{o}}\mathrm{C} \mathrm{and} \mathrm{M}={35}^{\mathrm{o}}\mathrm{C}$in equation (2),

$$\begin{gathered} 27=35-11{\mathrm{e}}^{-\frac{\mathrm{t}}{4}} \\ 8=\left(11\right){\mathrm{e}}^{-\frac{\mathrm{t}}{4}} \\ {\mathrm{e}}^{-\frac{\mathrm{t}}{4}}=\frac{8}{11} \\ -\frac{\mathrm{t}}{4}=\ln \left(0.727\right) \end{gathered}$$

Therefore, the temperature inside the building will reach 27°Cafter 1.16 p.m.