   Chapter 11, Problem 40P

Chapter
Section
Textbook Problem

The excess internal energy of metabolism is exhausted through a variety of channels, such as through radiation and evaporation of perspiration. Consider another pathway for energy loss: moisture in exhaled breath. Suppose you breathe out 22.0 breaths per minute, each with a volume of 0.600 L. Suppose also that you inhale dry air and exhale air at 37°C containing water vapor with a vapor pressure of 3.20 kPa. The vapor comes from the evaporation of liquid water in your body. Model the water vapor as an ideal gas. Assume its latent heat of evaporation at 37°C is the same as its heat of vaporization at 100.°C. Calculate the rate at which you lose energy by exhaling humid air.

To determine
The rate at which energy is lost by exhaling humid air.

Explanation

Section 1:

To determine: Mass of water vapor in each exhaled breadth.

Answer: Mass of water vapor in each exhaled breadth is 1.34×105kg .

Given Info: Volume of each breath is 0.600 L, pressure of water vapor is 3.20 kPa, and the temperature of the exhaled air is 37°C .

Consider the water vapor as ideal gas. The ideal gas expression is,

PV=nRT

• P is the pressure of the gas,
• V is the volume of the gas,
• n is the number of moles present in the gas,
• R is the gas constant,
• T is the temperature of the gas,

Formula to calculate the number of moles present in each exhale water vapor is,

n=PVRT

Formula to calculate the mass of water vapor present in each exhale breath is,

m=nMw

• m is the mass of water vapor present in each exhale breath,
• Mw is the molar mass of water,

Use PV/RT for n in the above equation to rewrite m.

m=(PVRT)Mw

Substitute   3.20kPa for P , 0.600 L for V , 8.31J/molK for R , 37°C for T , and 18.0g/mol for Mw to find m .

m=((3.20kPa)(0.600L)(103m31L)(8.31J/molK)(37+273)K)(18.0gmol)(1kg103g)=1.3415×105kg1.34×105kg

Therefore, the mass of water vapor present in each exhaled breath is 1

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