Question

I am needing help with this assignment for my Geography class. Can anyone help out?

Geography 01 Physical Geography East Los Angeles College Spring 2019 Exercise 4: Atmospheric Moisture Suggested Reading: Hess, pp 132-138 KEY TERMS: Absolute Humidity, Relative Humidity, Saturation Capacity, Dew Point, Warm air has relatively greater capacity to hold water vapor than does cooler air. As air warms, its capacity to hold moisture increases at an increasing rate; this means that if a source of moisture is available the air will absorb it by the process of evaporation. As air cools, its capacity to hold water vapor diminishes; it gives up moisture by the process of condensation (forming clouds or fog) and possibly precipitation. The graph below shows the water vapor capacity of air at different temperatures. This water vapor capacity is also known as the saturation capacity (Sc). 15 -10 -5 o° 5° 10° 15 20 25° 300 35°C 40 15 o 30 10 2 20 10 20° 40° 60° 80° 100°F *(There are exactly 7,000 grains in a pound) HUMIDITY Humidity, the water vapor content of air, can be expressed in a variety of ways. The most commonly known value is relative humidity (RH). This is a percentage that represents the actual amount of water vapor in the air compared to the maximum amount it can hold at that temperature. In this exercise we are going to use values of absolute humidity (AH) and saturation capacity (SC). Absolute humidity represents the actual water vapor content and saturation capacity represents the potential water vapor content at a given temperature. The dew point temperature (DPT) is the temperature at which the absolute humidity equals the saturation capacity, thus the relative humidity equals 100%. A.Н.-absolute humidity R.H.-relative humidity s.с.-saturation capacity T.- temperature in °F D.P.T dew point temperature Remember R.H. (A.H./s.С.) x 100% to the water vapor capacity of air and to the relative humidity as the preceding page, complete the following table temperature changes. Using the graph on the R.H A.H. 1.8grn/f 1.8grn/ft 1.8grn/ft 1.8grn/f3 Tem S.C an air mass at 10F 70°F 50°F 30°F cooled to cooled to cooled to 700F, A.Н.-3.6gm/ft3 Given: T Find: S.C. 2. and R.H. Given: Find: T-500p, А.Н.-3.6gm/ft3 S.C and R.H. 900F, A.Н.-5.0gm/ft3 Given: T Find: S.C and D.P.T Given: T 77"F, RH-4090 Find: А.Н. and D.P.T. Given: T 700F, R.H.-62% Find: А.Н. and D.P.T

Answer 1

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