Book edition 9th

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

Pages 616 pages

ISBN 9780321977069

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Short Answer

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

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step1: important formula.

Newton’s Law of cooling is, $T (t) = M + C e^{- k t}$

Step 2: Analyzing the given statement.

The temperature inside the building is $24 ° C$ . 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 $27 ° C$ .

Newton’s Law of cooling is,

$T (t) = M + C e^{- k t}$ …… (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., $\frac{1}{k} = 4$ .

Step 2: To find the value of C in the formula of Newton’s Law of cooling to find the temperature inside the building at time, t

Using the given values in equation (1), to find the value of ,

So, at t=0,

$T (0) = 35 + C e^{0} 24 = 35 + C C = - 11$

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

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

Step 3: To find the temperature inside the building at 2:00 p.m.

Substitute t=2 and $M = 35 ° C$ in equation (2),

$T (2) = 35 - 11 e^{- \frac{2}{4}} T (2) = 28 . 3^{o} C$

Hence, the temperature inside the building at 2:00 p.m. will be 28.3°C.

Step 4: To find the temperature inside the building at 6:00 p.m.

Substitute $t = 6 and M = 35^{o}$ in equation (2),

$T (6) = 35 - 11 e^{- \frac{6}{4}} T (2) = 32 . 5^{o} C$

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

Step 5: To find the time at which the temperature inside the building will reach 27°C

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

$27 = 35 - 11 e^{- \frac{t}{4}} 8 = (11) e^{- \frac{t}{4}} e^{- \frac{t}{4}} = \frac{8}{11} - \frac{t}{4} = \ln (0.727)$

Therefore, the temperature inside the building will reach 27°C after 1.16 p.m.

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Step by Step Solution

Step1: important formula.

Step 2: Analyzing the given statement.

Step 2: To find the value of C in the formula of Newton’s Law of cooling to find the temperature inside the building at time, t

Step 3: To find the temperature inside the building at 2:00 p.m.

Step 4: To find the temperature inside the building at 6:00 p.m.

Step 5: To find the time at which the temperature inside the building will reach 27°C

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Step by Step Solution

Step1: important formula.

Step 2: Analyzing the given statement.

Step 2: To find the value of C in the formula of Newton’s Law of cooling to find the temperature inside the building at time, t

Step 3: To find the temperature inside the building at 2:00 p.m.

Step 4: To find the temperature inside the building at 6:00 p.m.

Step 5: To find the time at which the temperature inside the building will reach 27°C

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