On a mild Saturday morning while people are working inside, the furnace keeps the temperature inside the building at 21°C. At noon the furnace is turned off, and the people go home. The temperature outside is a constant 12°C for the rest of the afternoon. If the time constant for the building is 3 hr, when will the temperature inside the building reach 16°C? If some windows are left open and the time constant drops to 2 hr, when will the temperature inside reach 16°C?

The temperatureinside the buildingis21°C. The temperature outside is a constant 12°C for the rest of the afternoon. If the time constants for the building are 3 hours and 2 hours. We have to find the time when the temperature will reach 16°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, wewill take the values as,

Temperature inside the building,${\mathit{T}}_{\mathbf{0}}\mathbf{=}{\mathbf{21}}^{\mathbf{o}}\mathit{C}$,

Temperature outside the building,${\mathit{M}}_{\mathbf{0}}\mathbf{=}{\mathbf{12}}^{\mathbf{o}}\mathit{C}$

If the time constant for the building is 3 hours,$\frac{\mathbf{1}}{\mathbf{k}}\mathbf{=}\mathbf{3}$

If the time constant for the building is 2 hours,$\frac{\mathbf{1}}{\mathbf{k}}\mathbf{=}\mathbf{2}$

Substituting $T\left(t\right)={16}^{o}C$in equation (1),

$$\begin{gathered} T\left(t\right)=12+\left(21-12\right)e^{\mathbf{-}\frac{\mathbf{t}}{\mathbf{3}}} \\ 16=12+\left(21-12\right)e^{\mathbf{-}\frac{\mathbf{t}}{\mathbf{3}}} \\ 16-12=9e^{\mathbf{-}\frac{\mathbf{t}}{\mathbf{3}}} \\ 4=9e^{\mathbf{-}\frac{\mathbf{t}}{\mathbf{3}}} \\ {e}^{\frac{\mathbf{t}}{\mathbf{3}}}=\frac{9}{4} \\ \frac{t}{3}=ln\left(2.25\right) \\ t=2.43hours \end{gathered}$$

If the time constant for the building is 3 hours, the temperature inside the building will reach 16°C after 2.43 hours.

Substituting $T\left(t\right)={16}^{o}C$in equation (1),

$$\begin{gathered} T\left(t\right)=12+\left(21-12\right)e^{\mathbf{-}\frac{\mathbf{t}}{2}} \\ 16=12+\left(21-12\right)e^{\mathbf{-}\frac{\mathbf{t}}{2}} \\ 16-12=9e^{\mathbf{-}\frac{\mathbf{t}}{2}} \\ 4=9e^{\mathbf{-}\frac{\mathbf{t}}{2}} \\ {e}^{\frac{\mathbf{t}}{2}}=\frac{9}{4} \\ \frac{t}{2}=ln\left(2.25\right) \\ t=1.62hours \end{gathered}$$

If the time constant for the building is 2 hours, the temperature inside the building will reach 16°C after 1.62 hours.