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Part 1
The air in a room 10 ft by 10 ft by 10 ft is 5​% carbon monoxide. Starting at tequals​0, fresh air containing no carbon monoxide is blown into the room at a rate of 50 ft cubed​/min. If air in the room flows out through a vent at the same​ rate, when will the air in the room be 0.01​% carbon​ monoxide?

To solve this problem, we need to model the change in the concentration of carbon monoxide (CO) in the room over time using differential equations.

### Step-by-Step Breakdown:

1. **Define the Variables:**
   - Let \( V = 1000 \, \text{ft}^3 \) be the volume of the room (since it is \( 10 \times 10 \times 10 \)).
   - Let \( Q(t) \) be the amount of carbon monoxide in the room at time \( t \) (in cubic feet).
   - Initially, the air contains 5% carbon monoxide, so at \( t = 0 \), the initial amount of CO is:
     \[
     Q(0) = 0.05 \times 1000 = 50 \, \text{ft}^3
     \]

2. **Rate of Change:**
   - Fresh air, which contains no carbon monoxide, is being pumped into the room at a rate of \( 50 \, \text{ft}^3/\text{min} \).
   - At the same time, the air (with its CO) is leaving the room at the same rate.
   
   The rate at which the carbon monoxide is leaving the room depends on its concentration. The concentration at time \( t \) is \( \frac{Q(t)}{V} \).

3. **Differential Equation:**
   The rate of change of the amount of carbon monoxide in the room is given by:
   \[
   \frac{dQ}{dt} = \text{rate in} - \text{rate out}
   \]
   Since fresh air has no carbon monoxide, the rate in is zero. The rate out is proportional to the concentration of CO, and air is leaving at \( 50 \, \text{ft}^3/\text{min} \). Hence:
   \[
   \frac{dQ}{dt} = -50 \times \frac{Q(t)}{1000}
   \]
   Simplifying:
   \[
   \frac{dQ}{dt} = -\frac{Q(t)}{20}
   \]

4. **Solve the Differential Equation:**
   This is a separable differential equation. We can rewrite it as:
   \[
   \frac{dQ}{Q(t)} = -\frac{1}{20} \, dt
   \]
   Integrating both sides:
   \[
   \ln |Q(t)| = -\frac{t}{20} + C
   \]
   Solving for \( Q(t) \):
   \[
   Q(t) = Q(0) e^{-t/20}
   \]
   Since \( Q(0) = 50 \), we have:
   \[
   Q(t) = 50 e^{-t/20}
   \]

5. **Determine when \( Q(t) = 0.01\% \):**
   The air will contain 0.01% carbon monoxide when the amount of carbon monoxide is:
   \[
   Q(t) = 0.0001 \times 1000 = 0.1 \, \text{ft}^3
   \]
   Setting \( Q(t) = 0.1 \) in the equation:
   \[
   0.1 = 50 e^{-t/20}
   \]
   Solving for \( t \):
   \[
   e^{-t/20} = \frac{0.1}{50} = 0.002
   \]
   Taking the natural logarithm of both sides:
   \[
   -\frac{t}{20} = \ln(0.002)
   \]
   \[
   t = -20 \ln(0.002)
   \]

6. **Calculate \( t \):**
   Now, we can compute the time \( t \).

Let me compute the value.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can manually compute \( t = -20 \ln(0.002) \) using a calculator to find the value of \( t \). If you need help with any part of the calculation or further explanation, let me know!

Would you like more details or have any questions on the solution steps so far? Here are some related questions to explore further:

1. What are separable differential equations?
2. How do we integrate exponential functions in differential equations?
3. What are the common applications of differential equations in fluid dynamics?
4. How can we interpret the decay constant in this context?
5. How does changing the rate of air inflow/outflow affect the time needed for the CO concentration to decrease?

**Tip:** Always double-check your integration steps when solving differential equations, especially when modeling real-world scenarios.

The air in a room 10 ft by 10 ft by 10 ft is 5% carbon monoxide. Starting at t=0, fresh air containing no carbon monoxide is blown into the room at a rate of 50 ft³/min. If air in the room flows out through a vent at the same rate, when will the air in the room be 0.01% carbon monoxide?

Learn how to solve a differential equation modeling the reduction of carbon monoxide in a room. This problem involves exponential decay and separable equations. Detailed solution provided for determining when the carbon monoxide concentration drops to 0.01%. Suitable for advanced high school and college-level students.

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