Cold air is dry because the saturation water vapor pressure of air decreases rapidly as temperature decreases. It is har
Posted: Sat Jul 09, 2022 11:59 am
Cold air is dry because the saturation water vapor pressure ofair decreases rapidly as temperature decreases. It is harmful tointroduce dry, cold air directly into lungs. To prevent it, therespiratory system in our body warms up the inhaled air to bodytemperature and moisturizes it to 100% relative humidity before itreaches lungs. Some of the water and heat in our body are"consumed" during this process. Thus, the exposure to cold air fora prolonged period will put a lot of stress on our body.
In this problem, we will see how much water and heat are takenfrom a human body in a single inhalation of dry, cold air.
In a particular cold morning, air temperature is at 0°C (32°F),and the absolute humidity of the air is 3.4 g/kg. A person inhales0.65 L of this cold air, which weighs 0.74 g. The atmosphericpressure is 101.3 kPa, and the molar mass of water is 18 g/mol.
What is the mass of water vapor in the 0.65 L of the cold air?(2pt)
As the inhaled air goes through the airways, its temperatureincreases to 37°C of the body temperature.
Using c = 1.0 J/(g °C) for the heat capacityof the inhaled air, calculate how much heat transfers from the bodyto the inspired air when its temperature rises from 0°C to37°C. (2pt)
As the temperature of the inhaled air increases, its volumeincreases as well, while the pressure remains the same as theatmospheric pressure.
According to Charles' law, what volume does the inhaled airincrease to at 37°C? (2pt)
At the same time as the volume of the inhaled air increases,water in the airways evaporates into the inhaled air to moisturizeit to 100% relative humidity at 37°C.
Using the volume you calculated in part (c)and Psat = 6.3 kPa for the saturationwater vapor pressure at 37°C, calculate the mass of water vaporcontained in the inhaled air after moisturization (this valueshould be larger than the value you found in part (a) because watervapor is added to the inhaled air in the process ofmoisturization). (2pt)
When water in the airways evaporates, it changes phases fromliquid water to vapor water, and it requires heat transfer from thebody.
First, calculate the mass of water evaporated from the airwaysto the inhaled air, and then calculate the amount heat transferredfrom the body to vaporize the water in the airways to moisturizethe inhaled air. Use L = 2413 J/g for the latentheat of vaporization of water at 37°C. (2pt)
In this problem, we will see how much water and heat are takenfrom a human body in a single inhalation of dry, cold air.
In a particular cold morning, air temperature is at 0°C (32°F),and the absolute humidity of the air is 3.4 g/kg. A person inhales0.65 L of this cold air, which weighs 0.74 g. The atmosphericpressure is 101.3 kPa, and the molar mass of water is 18 g/mol.
What is the mass of water vapor in the 0.65 L of the cold air?(2pt)
As the inhaled air goes through the airways, its temperatureincreases to 37°C of the body temperature.
Using c = 1.0 J/(g °C) for the heat capacityof the inhaled air, calculate how much heat transfers from the bodyto the inspired air when its temperature rises from 0°C to37°C. (2pt)
As the temperature of the inhaled air increases, its volumeincreases as well, while the pressure remains the same as theatmospheric pressure.
According to Charles' law, what volume does the inhaled airincrease to at 37°C? (2pt)
At the same time as the volume of the inhaled air increases,water in the airways evaporates into the inhaled air to moisturizeit to 100% relative humidity at 37°C.
Using the volume you calculated in part (c)and Psat = 6.3 kPa for the saturationwater vapor pressure at 37°C, calculate the mass of water vaporcontained in the inhaled air after moisturization (this valueshould be larger than the value you found in part (a) because watervapor is added to the inhaled air in the process ofmoisturization). (2pt)
When water in the airways evaporates, it changes phases fromliquid water to vapor water, and it requires heat transfer from thebody.
First, calculate the mass of water evaporated from the airwaysto the inhaled air, and then calculate the amount heat transferredfrom the body to vaporize the water in the airways to moisturizethe inhaled air. Use L = 2413 J/g for the latentheat of vaporization of water at 37°C. (2pt)