On a mild Saturday morning while people are working inside, the furnace keeps the temperature inside the building at 23degreesC. At noon the furnace is turned off, and the people go home. The temperature outside is a constant 14degreesC for the rest of the afternoon. If the time constant for the building is 3 hr, when will the temperature inside the building reach 18degreesC? If some windows are left open and the time constant drops to 2 hr, when will the temperature inside reach 18degreesC?

a) When the time constant is 3 hours, it takes the room 2.433 hours to cool to 18°C.

b) When the time constant is 2 hours, it takes the room 1.622 hours to cool to 18°C.

Step-by-step explanation:

Let T be the temperature inside the building at any time

T∞ be the temperature outside = 14°C

T₀ be the initial temperature of the inside of the building = 23°C

Time constant for the building = 3 hrs

On some integral analysis, the relation room cools according to a law of cooling and solving the ensuing differential equation, we obtain

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Note that k = (1/time constant) = (1/3)

a) When T = 18°C, how long will it take?

(18 - 14) = (23 - 14) e⁻ᵏᵗ

e⁻ᵏᵗ = (4/9) = 0.44444

In e⁻ᵏᵗ = In 0.44444 = -0.81093

-kt = -0.81093

(t/3) = 0.81093

t = 0.81093 × 3 = 2.433 hours.

b) When the windows are opened, the time constant falls to 2 hours.

Note that k = (1/time constant) = (1/2)

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

(18 - 14) = (23 - 14) e⁻ᵏᵗ

e⁻ᵏᵗ = (4/9) = 0.44444

In e⁻ᵏᵗ = In 0.44444 = -0.81093

-kt = -0.81093

(t/2) = 0.81093

t = 0.81093 × 2 = 1.622 hours.

Hope this Helps!!!

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step-by-step explanation:

hd extra 3-7 x d en is the answer