On a hot Saturday morning while people are working inside, the air conditioner keeps the temperature inside the building at 23degreesC. At noon the air conditioner is turned off, and the people go home. The temperature outside is a constant 34degreesC 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 3 : 00 font size decreased by 2 Upper P . font size decreased by 2 Upper M .? At 5 : 00 font size decreased by 2 Upper P . font size decreased by 2 Upper M .? When will the temperature inside the building reach 25degreesC?

a) 28.8°

b) 30.9°

c) 1:04pm

Explanation:

We are given

T_o = 23

M = 34

H(t) = u(t) = 0

Time = 1/k =4

Let's use the formula:

Therefore

T(t) = 34 + Ce°

23 = 34 + Ce°

C = 23 -34

C = -11

To solve for

a) at 3pm we have:

T(3) =28.8°

b) at 5pm =

T(5) = 30.9°

c) To find Temprature at 27°

T(t) = 27

;

;

;

;

-t/4 = In 7/11

t = -4 In 7/11

t = 1.08

1.08 hours after noon =

1.08 *60 mins = 64.8mins after noon

= 64.8 mins + 12:00

= 1:04pm

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answer:

explanation:

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