Question

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 25degrees​C?

Answer 1

a) 28.8°

b) 30.9°

c) 1:04pm

Step-by-step explanation:

We are given

T_o = 23

M = 34

H(t) = u(t) = 0

Time = 1/k =4

Let's use the formula:

`T(t)=M + Ce^-^t^/^4`

Therefore

T(t) = 34 + Ce°

`23 = 34 + Ce^-^t^/^4`

23 = 34 + Ce°

C = 23 -34

C = -11

To solve for

a) at 3pm we have:

`T(3) = 34 - 11e^-^3^/^4`

T(3) =28.8°

b) at 5pm =

`T(5) = 34 - 11e^-^5^/^4`

T(5) = 30.9°

c) To find Temprature at 27°

T(t) = 27

`27=34 - 11e^-^t^/^4`;

`-11e^-^t^/^4 = 27-34`;

`-11e^-^t^/^4 = -7`;

`e^-t^/^4 =7/11`;

-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