91Ó°ÊÓ

The temperature inside a van is $55°\mathrm{C}$and that outside is a constant $35°\mathrm{C}$. When the driver gets into the van, she turns on the air conditioner with the thermostat set at $16°\mathrm{C}$. When the driver gets into the van, she turns on the air conditioner with the thermostat set at $16°\mathrm{C}$. Given the time constant for the van is $\frac{1}{\mathrm{k}}=2 \mathrm{hr}$and that for the van with its air conditioning system is $\frac{1}{{\mathrm{k}}_1}=\frac{1}{3} \mathrm{hr}$. It has to find the time after which the temperature inside the van will reach $27°\mathrm{C}$.

Here, temperature inside the van, $T_{in}={55}^{0}C$.

Temperature outside the van, $T_{out}={35}^{0}C$.

Temperature value on thermostat, $T_t={16}^{0}C$.

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

The time constant for the van with its air conditioning system is $\frac{1}{{\mathrm{k}}_1}=\frac{1}{3} \mathrm{hr}$.

It will use the following formula to find the solution,

$\frac{dT}{dt}=K_1\left(T_{out}-T\right)+K_u\left(T_t-T\right)$ …… (1)

As it knows that,

$K_1+K_u=K$

Using values from step 1,

$$\begin{gathered} 3+K_u=\frac{1}{2} \\ {K}_u=\frac{1}{2}-3 \\ {K}_u=\frac{1-6}{2} \\ {K}_u=\frac{-5}{2} \end{gathered}$$

It will use this value in equation (1).

Now from equation (1),

$$\begin{gathered} \frac{dT}{dt}=3\left(35-T\right)-\frac{5}{2}\left(16-T\right) \\ \frac{dT}{dt}=\frac{130-T}{2} \\ \frac{dT}{dt}=65-\frac{T}{2} \end{gathered}$$

i.e., $\frac{dT}{dt}+\frac{T}{2}=65$ …… (2)

Integrating factor =role="math" localid="1664179724944" ${\mathit{e}}^{\mathbf{∫}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{dt}}\mathbf{=}{\mathit{e}}^{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{t}}$

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

$$\begin{gathered} e^{\frac{1}{2}t}·\frac{dT}{dt}+e^{\frac{1}{2}t}·\frac{T}{2}=65·e^{\frac{1}{2}t} \\ \frac{d}{dt}\left(T·e^{\frac{1}{2}t}\right)=65·e^{\frac{1}{2}t} \end{gathered}$$

Integrating both sides,

$T·e^{\frac{1}{2}t}=130e^{\frac{1}{2}t}+C$Where, C is an arbitrary constant.

When $\mathrm{t}=0,\mathrm{T}={55}^{\mathrm{o}}\mathrm{C}$

$$\begin{gathered} 55=130+C \\ C=-75 \end{gathered}$$

Therefore,

When temperature is ${\mathbf{27}}^{\mathbf{∘}}\mathit{C}$

$$\begin{gathered} 27=130-75e^{-\frac{1}{2}t} \\ 27-130=-75e^{-\frac{1}{2}t} \\ 103=75e^{-\frac{1}{2}t} \\ t=2ln\left(1.373\right) \\ t=0.634hr \\ t=38.04min \end{gathered}$$

Hence, the temperature inside the van will reach role="math" localid="1664180086516" $\mathbf{27}\mathbf{°}\mathbf{C}$after role="math" localid="1664180100673" $\mathbf{38}\mathbf{.}\mathbf{04}  \mathbf{minutes}$.