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Q 3.3-11E

Expert-verifiedFound in: Page 108

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**During the summer the temperature inside a van reaches ${\mathbf{55}}{\mathbf{\xb0}}{\mathbf{C}}$****, while that outside is a constant ${\mathbf{35}}{\mathbf{\xb0}}{\mathbf{C}}$****. When the driver gets into the van, she turns on the air conditioner with the thermostat set at ${\mathbf{16}}{\mathbf{\xb0}}{\mathbf{C}}$****. If the time constant for the van is $\frac{\mathbf{1}}{\mathbf{k}}{\mathbf{=}}{\mathbf{2}}{\u200a}{\mathbf{hr}}$**** and that for the van with its air conditioning system is $\frac{\mathbf{1}}{{\mathbf{k}}_{\mathbf{1}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{3}}{\u200a}{\mathbf{hr}}$****, when will the temperature inside the van reach ** ${\mathbf{27}}{\mathbf{\xb0}}{\mathbf{C}}$**?**

The temperature inside the van will reach $27\xb0\mathrm{C}$after $38.04\u200a\u200a\mathrm{minutes}$.

The temperature inside a van is $55\xb0\mathrm{C}$and that outside is a constant $35\xb0\mathrm{C}$. When the driver gets into the van, she turns on the air conditioner with the thermostat set at $16\xb0\mathrm{C}$. When the driver gets into the van, she turns on the air conditioner with the thermostat set at $16\xb0\mathrm{C}$. Given the time constant for the van is $\frac{1}{\mathrm{k}}=2\u200a\mathrm{hr}$and that for the van with its air conditioning system is $\frac{1}{{\mathrm{k}}_{1}}=\frac{1}{3}\u200a\mathrm{hr}$. It has to find the time after which the temperature inside the van will reach $27\xb0\mathrm{C}$.

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 $\frac{1}{\mathrm{k}}=2\u200a\mathrm{hr}$.

The time constant for the van with its air conditioning system is $\frac{1}{{\mathrm{k}}_{1}}=\frac{1}{3}\u200a\mathrm{hr}$.

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,**

**$3+{K}_{u}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}{K}_{u}=\frac{1}{2}-3\phantom{\rule{0ex}{0ex}}{K}_{u}=\frac{1-6}{2}\phantom{\rule{0ex}{0ex}}{K}_{u}=\frac{-5}{2}\phantom{\rule{0ex}{0ex}}$ **

It will use this value in equation (1).

Now from equation (1),

$\frac{dT}{dt}=3\left(35-T\right)-\frac{5}{2}\left(16-T\right)\phantom{\rule{0ex}{0ex}}\frac{dT}{dt}=\frac{130-T}{2}\phantom{\rule{0ex}{0ex}}\frac{dT}{dt}=65-\frac{T}{2}$

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

** **

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

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

${e}^{\frac{1}{2}t}\xb7\frac{dT}{dt}+{e}^{\frac{1}{2}t}\xb7\frac{T}{2}=65\xb7{e}^{\frac{1}{2}t}\phantom{\rule{0ex}{0ex}}\frac{d}{dt}\left(T\xb7{e}^{\frac{1}{2}t}\right)=65\xb7{e}^{\frac{1}{2}t}\phantom{\rule{0ex}{0ex}}$

**Integrating both sides,**

** $T\xb7{e}^{\frac{1}{2}t}=130{e}^{\frac{1}{2}t}+C$**Where, C is an arbitrary constant.

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

$55=130+C\phantom{\rule{0ex}{0ex}}C=-75$

**Therefore, **

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

$27=130-75{e}^{-\frac{1}{2}t}\phantom{\rule{0ex}{0ex}}27-130=-75{e}^{-\frac{1}{2}t}\phantom{\rule{0ex}{0ex}}103=75{e}^{-\frac{1}{2}t}\phantom{\rule{0ex}{0ex}}t=2ln\left(1.373\right)\phantom{\rule{0ex}{0ex}}t=0.634hr\phantom{\rule{0ex}{0ex}}t=38.04min\phantom{\rule{0ex}{0ex}}$

** **

**Hence, the temperature inside the van will reach role="math" localid="1664180086516" ${\mathbf{27}}{\mathbf{\xb0}}{\mathbf{C}}$**** after ** role="math" localid="1664180100673" ${\mathbf{38}}{\mathbf{.}}{\mathbf{04}}{\u200a}{\u200a}{\mathbf{minutes}}$**.**

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