Question

The air in a 46 cubic metre kitchen is initially clean, but when Taekyung burns his toast while making breakfast, smoke is mixed with the room's air at a rate of 0.04 mg per second. An air conditioning system exchanges the mixture of air and smoke with clean air at a rate of 6 cubic metres per minute. Assume that the pollutants are mixed uniformly throughout the room and that burnt toast is taken outside after 55 seconds. Let S(t) be the amount of smoke in mg in the room at time  t (in seconds) after the toast first began to burn.

1. Find a differential equation obeyed by S(t).
2. Find S(t) for 0 ≤ t ≤ 55 by solving the differential equation in (a) with an appropriate initial condition.
3. What is the level of pollution in mg per cubic meter after 55 seconds?
4. The time, in seconds, when the level of pollution falls to 0.002 mg per cubic metre is ................ seconds. Note that this check asks for the time since t=0 but the question part (d) asks for a time since the toast was taken outside.

Answer 1

Answer :

Given data :

Volume of kitchen (V) = 46 m³

Rate of spread of smoke R = 0.04 mg/s

Exchange rate of clean air with smokey air = Ex = 6 m³/min = 0.1 m³/s

Toast Burnt for time = 55 seconds

Smoke in air at time t = S(t) mg

(a)

So, Rate of air is cleaning the smoke r(t) :

(equation 1)

putting the values from above, we get

(equation 2)

So, Rate of smoke spread in room :

(equation 3)

putting equation 2 in equation 3,

putting the values of R from above, we get

(equation 4)

(b)

For solving the above differential equation, we have to integrate the above equation as

on further solving,

(equation 5)

Integrating equation 5,

We get,

Taking exponential on both sides, we get

(equation 6)

Now, putting the initial condition that

at time t =0 , S(0) = 0 means, there was no smoke initailly at time t =0,

so putting in equation 6, we get

(equation 7)

Putting equation 7 in equation 6, we get

on Solving, we get

(equation 8)

(c)

at t = 55 s

putting in equation 8,

on solving, we get

S(55) = 0.0451 mg/m³

(d)

when S(t) = 0.002 mg/m³

putting in equation 8,

So

On solving we get the time t,

t = 50.14 micro seconds

Please like this answer, if you found it useful, it helps in growth of our scores.

Thanks