Human body Physics 2020 question

[sprint84]

Homework Statement

If a person breathes 10 liters per minute or air at 68 deg(F) and 50% relative humidity, how much water per day must the internal membranes supply to saturate the air at 98.6 Deg(F) (Assume all moisture is exhaled) If each gram of water extracts 580 calories as it is vaporized, how much daily heat loss in kilo-calories (Food calories) does the represent? (Saturation vapor pressure at 20 degrees(C) is 17g/m^3 and at 37 deg(C) 44g/m^3.

Water Per Day:
Energy per day lost:

Homework Equations

)delta)T=Q/cm
Q=cm(deltaT)

The Attempt at a Solution

I am clueless where to start

 

[zhermes]

Thats a good one.
For the first part: you need to find how much water is in a liter of air at 50% humidity at 68F, figure out what that corresponds to at 98F, and see how much more water is needed to get to 100% humidity (saturated).
% humidity is the percentage of saturated vapor pressure for a given temperature.
Once you have that, the second part is easy.

Does that make sense?

 

[sprint84]

Thats a good one.
For the first part: you need to find how much water is in a liter of air at 50% humidity at 68F, figure out what that corresponds to at 98F, and see how much more water is needed to get to 100% humidity (saturated).
% humidity is the percentage of saturated vapor pressure for a given temperature.
Once you have that, the second part is easy.

Does that make sense?

Yes it does what formula am I going to use now that I have figured them out?

 

[zhermes]

I assume you mean for the second part ('how much daily heat loss...').

Well, its essentially just a really big conversion problem:
It says we breath 10L per minute, so we need to figure out how many liters that is per day. Then we have to convert liters/day to grams/day (so you need to the density of water). Finally, it says one loses 580 cals/gram of water, so we need to convert grams/day to cal/day---> and that's the answer!

 

[rwisz]

with this problem.

As a scientist, it is important to first understand the variables and units involved in the problem. In this case, we are given the rate of air intake (10 liters per minute), the temperature (68 deg(F) and 98.6 deg(F)), the relative humidity (50%), and the amount of water that can be extracted per gram (580 calories). We are also given the saturation vapor pressure at two different temperatures (20 deg(C) and 37 deg(C)).

Using the given information, we can start by converting the rate of air intake from liters per minute to liters per day. This can be done by multiplying it by the number of minutes in a day (60 x 24 = 1440). This gives us a total of 14,400 liters of air in a day.

Next, we can calculate the amount of water that needs to be supplied by the internal membranes to saturate the air at 98.6 deg(F). This can be done by first finding the difference in saturation vapor pressure between 68 deg(F) and 98.6 deg(F). Using the given values, we get a difference of 27g/m^3. This means that for every cubic meter of air, an additional 27 grams of water needs to be supplied to reach saturation at 98.6 deg(F).

Since we know that there are 14,400 liters of air per day, we can convert this to cubic meters by multiplying by 1000 (1 m^3 = 1000 L). This gives us a total of 14.4 cubic meters of air per day. Multiplying this by 27g/m^3 gives us a total of 388.8 grams of water that needs to be supplied per day to saturate the air at 98.6 deg(F).

Finally, we can calculate the energy lost per day by multiplying the amount of water (388.8 grams) by the energy per gram (580 calories). This gives us a total of 225,504 calories lost per day. To convert this to kilo-calories, we divide by 1000, giving us a total of 225.504 kilo-calories lost per day.

In conclusion, to saturate the air at 98.6 deg(F), the internal membranes must supply 388.8 grams of water per day and this results in a loss of 225.504 kilo