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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 18degrees​C? If some windows are left open and the time constant drops to 2 ​hr, when will the temperature inside reach 18degrees​C?

Answer :

Answer:

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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