91Ó°ÊÓ

The temperature inside and outside the lecture hall is $40°\mathrm{F}.-$At $7:00\mathrm{a}.\mathrm{m}.$, the janitor turns on the furnace with the thermostat set at $70°\mathrm{F}$. The time constant for the building is $\frac{1}{\mathrm{k}}=2 \mathrm{hr}$and that for the building along with its heating system is $\frac{1}{{\mathrm{k}}_1}=\frac{1}{2} \mathrm{hr}$.

Assuming that

Here, temperature inside the lecture hall, $T_{in}={40}^{0}F$.

Temperature outside the hall, $T_{out}={40}^{0}F$.

Temperature value on furnace, $T_f={70}^{0}F$.

The time constant for the building is $\frac{1}{\mathrm{k}}=2\mathrm{hr}$.

The time constant for the building with its heating system is $\frac{1}{{\mathrm{k}}_1}=\frac{1}{2} \mathrm{hr}$.

It will use the following formula to find the solution,

$\frac{dT}{dt}=K_1\left(T_{out}-T\right)+K_u\left(T_f-T\right)$

As it knows that,

$K_1+K_u=K$

Using values from step1,

$$\begin{gathered} 2+K_u=\frac{1}{2} \\ {K}_u=\frac{1}{2}-2 \\ {K}_u=\frac{1-4}{2} \\ {K}_u=\frac{-3}{2} \end{gathered}$$

One will use this value in equation (1).

Now from equation (1),

$$\begin{gathered} \frac{dT}{dt}=2\left(40-T\right)-\frac{3}{2}\left(70-T\right) \\ \frac{dT}{dt}=\frac{-50-T}{2} \\ \frac{dT}{dt}=-25-\frac{T}{2} \end{gathered}$$

i.e., $\frac{dT}{dt}+\frac{T}{2}=-25$ …… (2)

Integrating factor = role="math" localid="1664183452880" ${\mathit{e}}^{\mathbf{∫}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{e}}^{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{t}}$

Multiplying both sides of (2) by $e^{\frac{1}{2}t}$,

$$\begin{gathered} e^{\frac{1}{2}t}·\frac{dT}{dt}+e^{\frac{1}{2}t}·\frac{T}{2}=-25·e^{\frac{1}{2}t} \\ \frac{d}{dt}\left(T·e^{\frac{1}{2}t}\right)=-25·e^{\frac{1}{2}t} \end{gathered}$$

Integrating both sides,

$T·e^{\frac{1}{2}t}=-50e^{\frac{1}{2}t}+C$

Where, C is an arbitrary constant.

When t=0,$\mathrm{T}=40°\mathrm{F}$

$$\begin{gathered} 40=-50+C \\ C=90 \end{gathered}$$

Therefore,role="math" localid="1664183691804" $\mathit{T}\mathbf{=}\mathbf{-}\mathbf{50}\mathbf{+}\mathbf{90}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{t}}$ …… (3)

Temperature in the lecture hall when t=1 hour,

$$\begin{gathered} T=-50+90e^{-\frac{1}{2}} \\ T={98}^{0}F \end{gathered}$$

Hence, the temperature inside the lecture will reach role="math" localid="1664183714947" $\mathbf{98}\mathbf{°}\mathbf{C}$atrole="math" localid="1664183727475" $\mathbf{8}\mathbf{:}\mathbf{00}\mathbf{a}\mathbf{.}\mathbf{m}\mathbf{.}$

Now from equation (3),

Substituting $\mathrm{T}=65°\mathrm{F}$

$$\begin{gathered} T=-50+90e^{-\frac{1}{2}t} \\ 65=-50+90e^{-\frac{1}{2}t} \\ t=2ln\left(1.28\right) \\ t=0.494hour \\ t=29.64min \end{gathered}$$

Hence, the temperature inside will reach role="math" localid="1664183975018" $\mathbf{65}\mathbf{°}\mathbf{F}$afterrole="math" localid="1664183986875" $\mathbf{29}\mathbf{.}\mathbf{64} \mathbf{minutes}$.