1.Summarize the processes by which water changes from one state of matterto another. Indicate whether heat energy is absorbed or liberated.A. Heat is absorbed when icechanges to water (melting), when water changes to water vapor(evaporation), and when ice turns directly to water vapor withoutpassing through the liquid state (sublimation). Heat is liberatedduring condensation (the vapor-to-liquid phase change), freezing (theconversion of water to ice), and sublimation (the change from the vaporstate directly to ice).2. After studying Table 15.1, write a generalization relatingtemperature andthe capacity of air to hold water vapor. A. As temperature increases, the capacity of airto hold water vapor increases (at an increasing rate). 3. How do relative and specific humidity differ?A. Specific humidity incicates the amountof water vapor in the air and is expressed as the weight of water vaporper weight of air.4. Referring to Figure 15.6, answer the following questions andthen write a generalization relating changes in air temperature tochanges in relative humidity.a) During a typical day, when is the relative humidity highest?Lowest?A. It is highest near sunrise and lowestduring mid-afternoon.b) At what time of day would dew most likely form?A. When the temperature was lowest and therelative humidity highest, that is, near sunrise.5. If the temperature remains unchanged and the specifichumidity decreases, how will relative humidity change?A. With a constant specific humidity, anincrease in temperature causes a decline in relative humidity, and adrop in temperature causes a rise in relative humidity.6. On a cold winter day when the temperature is -10o Cand the relative humidity is 50 percent, what is the specific humidity(refer to Table 15.1)? What is the specific humidity for a daywhen the temperature is 20oC and the relative humidity is50 percent?A. 1 gram per kilogram of air; 7 grams perkilogram of air.8. Using the standard tables (Tables 15.2 and 15.3), determinethe relative humidity and dew-point temperature if the dry-bulbthermometer reads 16oC and the wet-bulb thermometer reads 12oC. How would the relative humidity and dew point change if thewet-bulb thermometer read 8oC? A. 62% relative humidity; dew point 9oC; 29% relative humidity; dew point -1oC9.
You are watching: Adiabatic processes cause cooling by ________.
On a warm summer day when the relative humidity is high, itmay seem even warmer than the thermometer indicates. Why do wefeel so uncomfortable on a “muggy” day?A. Since the relative humidity is high,there would be a minimum of evaporation of perspiration, the body”gnatural cooling system.
10. Why does aircool when it rises through the atmosphere?A. As air rises, it expands because airpressure decreases with an increase in altitude. When airexpands, it coolsadiabatically.11. Explain the difference between environmental lapse rate andadiabatic cooling.A. The environmental lapse rate refers tothe temperature drop with increasing altitude in the troposphere; thatis the temperature of the environment at different altitudes. Itimpliesno air movement. Adiabatic cooling is associated only withascendingair, which cools by expansion.12. If unsaturated air at 23oC were to rise, whatwould its temperature be at 500 meters? If the dew pointtemperature at the condensation level were 13oC, at whataltitude would clouds begin to form? A. 18oC (because of the dryadiabaticrate); 1000 meters.13. Why does the adiabatic rate of cooling change whencondensation begins? Why is the wet adiabatic rate not a constantfigure?A. When condensation occurs, latent heat isreleasedby the vapor/water droplets into the surrounding air. This heatsupthe surrounding air, countering some of the cooling defined by the dryadiabaticrate. The wet adiabatic rate depends on the quantity of vapor intheair.14. The contents of an aerosol can are under very high pressure.When you push the nozzle on such a can, the spray feels cold.Explain.A. When you push the nozzle on the aerosolcan,pressure in the can decreases as the contents begins to escape the can.Thedrop in pressure results in adiabatic cooling.15. How do orographic lifting and frontal wedging act to forceair to rise?A. In orographic lifting, a mountain sideservesas a barrier to cause air to ramp up. In frontal wedging, a coldmassof air acts as a barrier forcing warmer air to ramp up and rise.