On a hot Saturday morning while people are working​ inside, the air conditioner keeps the temperature inside the building at 24 degreesC. At noon the air conditioner is turned​ off, and the people go home. The temperature outside is a constant 35 degreesC for the rest of the afternoon. If the time constant for the building is 5 ​hr, what will be the temperature inside the building at 4 : 00 font size decreased by 2 Upper P . font size decreased by 2 Upper M . ​

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Answer:

30.06 degree Celsius at 4:00 PM.

Step-by-step explanation:

We have been given that on a hot Saturday morning while people are working​ inside, the air conditioner keeps the temperature inside the building at 24 degrees C. At noon the air conditioner is turned​ off, and the people go home. The temperature outside is a constant 35 degrees C for the rest of the afternoon. The time constant for the building is 5 ​hr. We are asked to find the temperature inside the building at 4:00 PM.

We will use Newton's law of cooling to solve our given problem.

[tex]T(t)=M_0+(T_0-M_0)e^{-kt}[/tex], where

[tex]T(t)[/tex] = Temperature after t hours.

[tex]M_0[/tex] = Temperature of surrounding environment,

[tex]k[/tex] = Time constant.

[tex]\frac{1}{k}=5\Rightarrow k=\frac{1}{5}[/tex]

Upon substituting our given values in Newton's cooling law, we will get:

[tex]T(t)=35+(24-35)e^{-\frac{t}{5}}[/tex]

[tex]T(t)=35-11e^{-\frac{t}{5}}[/tex]

To find the temperature at 4:00 pm, we will substitute [tex]t=4[/tex] in our formula as:

[tex]T(4)=35-11e^{-\frac{4}{5}}[/tex]

[tex]T(4)=35-\frac{11}{e^{\frac{4}{5}}}[/tex]

[tex]T(4)=35-\frac{11}{2.2255409284924674}[/tex]

[tex]T(4)=35-4.9426186052894379601[/tex]

[tex]T(4)=30.05738139471[/tex]

[tex]T(4)\approx 30.06[/tex]

Therefore, the temperature inside building would be approximately 30.06 degree Celsius at 4:00 PM.