For every cubic meter of 30celsius saturated air that is cooled to 16celsius, how many grams of water vapor will be lost?
To determine the amount of water vapor lost in this scenario, we need to calculate the difference in specific humidity between the initial and final states.
Specific humidity is the mass of water vapor present in a given mass of air. It is usually measured in grams of water vapor per kilogram of air.
Before we proceed, we need to know the saturated specific humidity at both 30°C and 16°C, which can be obtained using available meteorological data or by using specialized formulas. For simplicity, let's assume that the saturated specific humidity at 30°C is 20 grams of water vapor per kilogram of air and at 16°C is 10 grams of water vapor per kilogram of air.
Now, let's calculate the change in specific humidity:
Change in specific humidity = Saturated specific humidity at 30°C - Saturated specific humidity at 16°C
= 20 g/kg - 10 g/kg
= 10 g/kg
The change in specific humidity tells us that for every kilogram of air cooled from 30°C to 16°C, 10 grams of water vapor will be lost.
Since we are given the initial volume of air as 1 cubic meter, we need to convert it to kilograms using the density of dry air at the given temperature. The density of dry air at 30°C is approximately 1.164 kg/m^3.
Mass of air in 1 cubic meter = Volume x Density of air
= 1 m^3 x 1.164 kg/m^3
= 1.164 kg
Finally, we can calculate the total amount of water vapor lost:
Amount of water vapor lost = Mass of air x Change in specific humidity
= 1.164 kg x 10 g/kg
= 11.64 grams
Therefore, approximately 11.64 grams of water vapor will be lost when 1 cubic meter of 30°C saturated air is cooled to 16°C.