How Does Breathing Affect Water Vaporization and Caloric Loss?

[swiftmatt]

Homework Statement

If a person breathes 10 liters per minute of air at 68°F & 50% relative humidity, how much water per minute must the internal membranes supply to saturate this air at 98.6°F? If all of this moisture is subsequently exhaled, how much water per day is given off by the body in this process? If each gram of water extracts .58 kcals of energy as it is vaporized, how much daily heat loss in kcals (food calories) does this represent?

Homework Equations

Saturation Vapor Density at 68°F = 17.3g / m³
Saturation Vapor Density at 98.6°F = 44.0g / m³

The Attempt at a Solution

Yet another problem of subject matter our teacher has never touched on. I really don't understand how saturation and calories are even remotely associatable. We have some other values we need to plug in, so with any explanation on this problem, I should be able to work from there on the remaining problems.

 

[Chestermiller]

If the saturation water vapor density at 68F (20C) is 17.3 g/m^3, and the relative humidity of the water vapor in the air is 50%, what is the actual water vapor density in the inhaled air? How many grams of water are contained in 10 liters of this air?

If the saturation water vapor density at 98.6 F (37C) is 44.0 g/m^3 and the exhaled air is saturated (100% relative humidity), how many grams of water are contained in 10 liters of this exhaled air?

What is the change in the water vapor content between the inhaled air and exhaled air for 10 liters of air?

Chet

 

[swiftmatt]

A. If the saturation water vapor density at 20°C is 17.3g/m³, and the relative humidity of the water vapor in the air is 50%, what is the actual water vapor density in the inhaled air?

Inhaled Water Vapor Density = (17.3*.5) = 8.65g/m³?

B. How many grams of water are contained in 10 liters of this air?

10 liters = .01m³

8.65g/m³ * .01m³ = .0865 grams?

C. If the saturation water vapor density at 37°C is 44.0g/m³ and the exhaled air is saturated (100% relative humidity), how many grams of water are contained in 10 liters of this exhaled air?

44g/m³ * .01m³ = .44 grams?

D. What is the change in the water vapor content between the inhaled air and exhaled air for 10 liters of air?

Inhaled Water Vapor - Exhaled Water Vapor = -.3535 grams (Water Vapor leaving the body)?


Would anyone mind giving me confirmation that I am going in the right direction with this? Also, using this information, I should now be able to answer the questions in my original post, which are:

• How much water per minute must the internal membranes supply to saturate this air at 98.6°F?
Found in D.?

• How much water per day is given off by the body in this process?
Find number of minutes in a day and multiply it by what we found in D.?

• How much daily heat loss in kcals (food calories) does this represent?
Take the total water loss per day and cross multiply with my original given rate of grams of water to kcals vaporized to find total kcals of heat loss per day?

 

[Chestermiller]

All correct, except that +0.3535 gm water vapor leaving body per minute, not -0.3535. Otherwise, very nicely done.

Chet

 

[HalfLostAndSearching]

I would like to provide some guidance on this problem. First, let's define some terms and concepts to help understand the problem better.

Breathing is the process of taking in oxygen and expelling carbon dioxide from the body. This exchange of gases occurs in the lungs, where the oxygen is absorbed into the bloodstream and the carbon dioxide is released. The air that we breathe in is not just oxygen, but also contains water vapor.

Vaporization is the process of converting a liquid into a gas. In this case, we are specifically talking about the vaporization of water, which is the process of converting liquid water into water vapor.

Relative humidity is a measure of the amount of water vapor present in the air compared to the maximum amount of water vapor that the air can hold at a given temperature. A relative humidity of 50% means that the air is holding half of the maximum amount of water vapor it can hold at that temperature.

Now, let's look at the problem. The first part asks how much water per minute must the internal membranes supply to saturate the air. To answer this, we need to first calculate the amount of water vapor that the air can hold at 68°F and 50% relative humidity. Using the saturation vapor density at 68°F (17.3g/m3), we can calculate that the air can hold 8.65g of water vapor per minute. This means that the internal membranes must supply 8.65g of water per minute to saturate the air.

The second part asks how much water per day is given off by the body in this process. To answer this, we need to calculate the amount of water that is given off per minute (8.65g) and then multiply it by the number of minutes in a day (1440 minutes). This gives us a total of 12456g or 12.456kg of water given off per day.

The final part of the problem asks how much daily heat loss in kcals (food calories) does this represent. This part is a bit trickier, but we can use the fact that each gram of water extracts 0.58 kcals of energy as it is vaporized. So, if we multiply the total amount of water given off per day (12.456kg) by the energy extracted per gram (0.58 kcals), we get a total heat loss of 7229.28 kcals per day