To test the effectiveness of a carbon monoxide (CO) alarm, a company begins a controlled experiment that delivers carbon monoxide to a closed room at the rate of 1.5% of the current amount in the room every 2 hours. Find an exponential model for the amount of CO in the room C (in ppm) as a function of time t (in hr) from the beginning of the experiment when there are 50 parts per million CO in the room.

The problem involves modeling the amount of carbon monoxide (CO) in a room, where the rate of increase is proportional to the current amount present in the room. This setup leads us to an exponential growth model.

Let’s break it down and find the exponential model step by step.

### Step 1: Understanding the Problem

The CO in the room increases by 1.5% every 2 hours. Initially, there are 50 ppm of CO present in the room. We need to find the function \( C(t) \), which models the amount of CO as a function of time \( t \) (in hours).

### Step 2: General Exponential Model

The general form of an exponential growth function is:

\[
C(t) = C_0 e^{kt}
\]

Where:
- \( C(t) \) is the amount of CO in the room at time \( t \),
- \( C_0 \) is the initial amount of CO,
- \( k \) is the rate of growth (per hour), and
- \( t \) is the time in hours.

### Step 3: Initial Condition

We are given that the initial amount of CO is 50 ppm. Therefore:

\[
C_0 = 50
\]

Thus, the equation becomes:

\[
C(t) = 50 e^{kt}
\]

### Step 4: Growth Rate Information

We know that the amount of CO increases by 1.5% every 2 hours. This means that after 2 hours, the amount of CO is 1.015 times the previous amount.

So, at \( t = 2 \), the amount of CO is:

\[
C(2) = 50 \times 1.015
\]

But according to the exponential model:

\[
C(2) = 50 e^{2k}
\]

Setting these equal to each other gives:

\[
50 e^{2k} = 50 \times 1.015
\]

Simplifying:

\[
e^{2k} = 1.015
\]

Taking the natural logarithm on both sides:

\[
2k = \ln(1.015)
\]

Solving for \( k \):

\[
k = \frac{\ln(1.015)}{2}
\]

Now, calculate \( k \):

\[
k \approx \frac{0.014889}{2} = 0.0074445
\]

### Step 5: Final Exponential Model

Now that we know \( k \), we can substitute it back into the model:

\[
C(t) = 50 e^{0.0074445t}
\]

This is the exponential model for the amount of CO in the room as a function of time \( t \).

### Final Answer:

\[
C(t) = 50 e^{0.0074445t}
\]

This model predicts the amount of CO in the room at any time \( t \), where \( t \) is in hours.

---

Would you like more details or explanations? Here are five related questions to consider:

1. What would the CO level be after 5 hours?
2. How long will it take for the CO level to double?
3. How does the model change if the rate of CO increase changes?
4. Can you determine the half-life of CO decrease if ventilation is introduced?
5. How would the model adjust for a different initial amount of CO?

**Tip:** Exponential growth models are often used in situations where the change in quantity is proportional to the current amount, such as population growth or decay processes!

Discover how to model the amount of carbon monoxide (CO) in a room using an exponential growth equation. This problem involves calculating the CO level starting from 50 ppm with a 1.5% increase every 2 hours. Learn how to derive the function C(t) = 50 * e^(0.0074445t) using natural logarithms. Suitable for Grade 11-12 and college-level students.

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