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Q 3.3-7E

Expert-verifiedFound in: Page 108

Book edition
9th

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

Pages
616 pages

ISBN
9780321977069

**On a hot Saturday morning while people are working inside, the air conditioner keeps the temperature inside the building at ${\mathbf{24}}{\mathbf{\xb0}}{\mathbf{C}}$****. At noon the air conditioner is turned off, and the people go home. The temperature outside is a constant ${\mathbf{35}}{\mathbf{\xb0}}{\mathbf{C}}$**** 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 ${\mathbf{27}}{\mathbf{\xb0}}{\mathbf{C}}$****?**

The temperature inside the building will be $28.3\xb0\mathrm{C}$ at 2:00 p.m. and $32.5\xb0\mathrm{C}$ at 6:00 p.m. The temperature inside the building will reach $27\xb0\mathrm{C}$ after 1.16 pm.

**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 $24\xb0\mathrm{C}$. 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 $27\xb0\mathrm{C}$.

**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,

$T\left(0\right)=35+C{e}^{0}\phantom{\rule{0ex}{0ex}}24=35+C\phantom{\rule{0ex}{0ex}}C=-11\phantom{\rule{0ex}{0ex}}$

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

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

Substitute t=2 and $\mathrm{M}=35\xb0\mathrm{C}$ in equation (2),

$T\left(2\right)=35-11{e}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{4}}}\phantom{\rule{0ex}{0ex}}T\left(2\right)=28.{3}^{o}C\phantom{\rule{0ex}{0ex}}$

**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),

$\mathrm{T}\left(6\right)=35-11{\mathrm{e}}^{-\frac{6}{4}}\phantom{\rule{0ex}{0ex}}\mathrm{T}\left(2\right)=32.{5}^{\mathrm{o}}\mathrm{C}\phantom{\rule{0ex}{0ex}}$

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

Substitute $\mathrm{T}\u200a\left(\mathrm{t}\right)=2{7}^{\mathrm{o}}\mathrm{C}\u200a\u200a\mathrm{and}\u200a\u200a\mathrm{M}={35}^{\mathrm{o}}\mathrm{C}$in equation (2),

$27=35-11{\mathrm{e}}^{-\frac{\mathrm{t}}{4}}\phantom{\rule{0ex}{0ex}}8=\left(11\right){\mathrm{e}}^{-\frac{\mathrm{t}}{4}}\phantom{\rule{0ex}{0ex}}{\mathrm{e}}^{-\frac{\mathrm{t}}{4}}=\frac{8}{11}\phantom{\rule{0ex}{0ex}}-\frac{\mathrm{t}}{4}=\mathrm{ln}\left(0.727\right)\phantom{\rule{0ex}{0ex}}$

**Therefore, the temperature inside the building will reach 27°C**

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