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The air in a room whose volume is 10,000 cu ft tests \(0.15 \%\) carbon dioxide. Starting at \(t=0\), outside air testing \(0.05 \%\) carbon dioxide is admitted at the rate of 5000 cu \(\mathrm{ft} / \mathrm{min}\). (a) What is the percentage of carbon dioxide in the air in the room after \(3 \mathrm{~min} ?\) (b) When does the air in the room test \(0.1 \%\) carbon dioxide.

Short Answer

Expert verified
The percentage of CO2 in the room after 3 minutes is \(0.09 \%\), and the air in the room tests \(0.1 \%\) carbon dioxide after 10 minutes.

Step by step solution

01

Calculate initial CO2 content in the room

We know the volume of the room is \(10,000 \ \text{cu ft}\), and the room initially tests \(0.15 \%\) CO2. To find the initial CO2 in the room, multiply the room volume by the CO2 percentage: Initial CO2 content (in cu ft) = Room volume × CO2 percentage = \(10,000 \times 0.0015 = 15\)
02

Calculate admitted CO2 content each minute

We know that air outside, testing \(0.05 \%\) CO2, is admitted at a rate of \(5000 \ \text{cu ft/min}\). The new CO2 added can be calculated accordingly: CO2 added each minute = Added volume × CO2 percentage = \(5000 \times 0.0005 = 2.5\)
03

Calculate CO2 content after 3 minutes

We know that each minute, \(2.5 \ \text{cu ft}\) of CO2 is added, and the initial CO2 content was \(15 \ \text{cu ft}\). After three minutes, the total CO2 content would be: Total CO2 after 3 minutes = Initial CO2 content + (CO2 added each minute × 3 minutes) = \(15 + (2.5 \times 3) = 22.5\)
04

Calculate new room volume after 3 minutes

The initial room volume was \(10,000 \ \text{cu ft}\), and outside air is admitted at a rate of \(5000 \ \text{cu ft/min}\). After three minutes, the total volume would be: Total volume after 3 minutes = Initial room volume + (Added volume × 3 minutes) = \(10,000 + (5000 \times 3) = 25,000\)
05

Calculate the percentage of CO2 after 3 minutes (Part A)

We know the total CO2 after 3 minutes is \(22.5 \ \text{cu ft}\), and the new room volume is \(25,000 \ \text{cu ft}\). To find CO2 percentage after 3 minutes: Percentage of CO2 after 3 minutes = \(\dfrac{\text{Total CO2 after 3 minutes}}{\text{Total volume after 3 minutes}} \times 100 = \dfrac{22.5}{25,000} \times 100 = 0.09 \%\) The percentage of CO2 in the room after 3 minutes is \(0.09 \%\).
06

Set up an equation to find the time for CO2 concentration to reach \(0.1 \%\) (Part B)

We want to find the time it takes for the CO2 percentage to reach \(0.1 \%\). Let t be the time in minutes. Total volume at time t = Initial room volume + (Added volume × t) = \(10,000 + 5000t\) Total CO2 at time t = Initial CO2 content + (CO2 added each minute × t) = \(15 + 2.5t\) The percentage of CO2 at time t can be expressed as \(\dfrac{\text{Total CO2 at time t}}{\text{Total volume at time t}}\): Percentage of CO2 at time t = \(\dfrac{15 + 2.5t}{10,000 + 5000t}\) Setting this equal to \(0.1 \%\) and solving for t: \(\dfrac{15 + 2.5t}{10,000 + 5000t} = 0.001\)
07

Solve the equation for t (Part B)

Now, we'll solve the equation for t: \(15 + 2.5t = 0.001(10,000 + 5000t)\) \(15 + 2.5t = 10 + 5t\) \(0.5t = 5\) \(t = 10 \ \text{minutes}\) The air in the room tests \(0.1 \%\) carbon dioxide after 10 minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Ordinary Differential Equations (ODEs)
One of the foundational tools in mathematical modeling is the ordinary differential equation (ODE). An ODE is a equation that contains functions of one independent variable and its derivatives. They are crucial in mathematically describing the change in various phenomena over time.

When dealing with air quality modeling, ODEs help us to describe the rate at which pollutants like CO2 change concentration within a given volume. Considering the air quality problem in our exercise, if we had wanted to calculate the instantaneous rate of change of CO2 concentration over time, we might have set up a differential equation to model this process.

However, in our scenario, we use algebraic calculations instead of ODEs to handle the fixed rates of air exchange. If we were to encounter variable exchange rates or other complicating factors, then setting up an ODE would become necessary. This would involve differential calculus and potentially integrating the ODE to find a function that predicts CO2 concentration at any time.
Modeling Air Quality Using Mathematics
Mathematical modeling of air quality is an essential tool for understanding and predicting the behavior of pollutants in enclosed spaces. The process involves the creation of mathematical abstractions that replicate the physical processes affecting air composition.

In our exercise, we modeled the air exchange in a room as a simple additive process. We tracked CO2 levels by considering the volume of air exchanged and the concentration of CO2 in both the room and incoming air. By using these calculations, we could predict CO2 concentration at a specific time. Typically, more complex air quality models might require the use of computational techniques and advanced math such as ODEs and partial differential equations (PDEs) to account for things like varying air flow rates, temperature gradients, and chemical reactions.

Relevance in Real-World Applications

In a real-world setting, such models aid in designing ventilation systems, predicting pollutant spread in events of leakages, and ensuring compliance with health standards. Ultimately, mathematical models are a key element in environmental science, engineering, and health policy.
Exponential Decay in Environmental Systems
Exponential decay describes a process where the quantity of a substance decreases at a rate proportional to its current value. This concept is frequently encountered in physics, chemistry, and environmental science.

In context with our CO2 concentration problem, if we were to consider the room's natural air exchange without any mechanical ventilation, the CO2 concentration might decrease exponentially over time. This means that the rate of decrease would be faster initially and slow down as the concentration approaches the equilibrium or outside air concentration.

In the exercise, CO2 doesn't decay exponentially because we constantly add new air at a fixed rate. But it's still useful to understand the concept of exponential decay, as it often explains how pollutants diminish in unventilated or closed systems where natural processes are the only factors influencing concentrations.

Applicability to Other Fields

This idea also extends to other areas, such as radioactive decay, population decline, and depreciation of assets, showcasing the versatility and importance of the exponential decay model in various scientific and financial fields.

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Most popular questions from this chapter

The air in a room \(50 \mathrm{ft}\) by \(20 \mathrm{ft}\) by \(8 \mathrm{ft}\) tests \(0.2 \%\) carbon dioxide. Starting at \(t=0\), outside air testing \(0.05 \%\) carbon dioxide is admitted to the room. How many cubic feet of this outside air must be admitted per minute in order that the air in the room test \(0.1 \%\) at the end of \(30 \mathrm{~min}\) ?

A given family of curves is said to be self-orthogonal if its family of orthogonal trajectories is the same as the given family. Show that the family of parabolas \(y^{2}=2 c x+c^{2}\) is self orthogonal.

Find the value of \(K\) such that the parabolas \(y=c_{1} x^{2}+K\) are the orthogonal trajectories of the family of ellipses \(x^{2}+2 y^{2}-y=c_{2}\).

A tank initially contains 100 gal of pure water. Starting at \(t=0\), a brine containing \(4 \mathrm{lb}\) of salt per gallon flows into the tank at the rate of \(5 \mathrm{gal} / \mathrm{min}\). The mixture is kept uniform by stirring and the well-stirred mixture flows out at the slower rate of \(3 \mathrm{gal} / \mathrm{min}\). (a) How much salt is in the tank at the end of \(20 \mathrm{~min} ?\) (b) When is there \(50 \mathrm{lb}\) of salt in the tank?

An object weighing \(12 \mathrm{lb}\) is placed beneath the surface of a calm lake. The buoyancy of the object is \(30 \mathrm{lb}\); because of this the object begins to rise. If the resistance of the water (in pounds) is numerically equal to the square of the velocity (in feet per second) and the object surfaces in \(5 \mathrm{sec}\), find the velocity of the object at the instant when it reaches the surface.

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