Answers to Exercises

Lab 1

1. Location (a) has a lower elevation, meaning there is typically more atmospheric mass above it relative to the higher elevation region (b). If there is more atmospheric mass, there will be greater atmospheric pressure.

2. (b) 1013.1 hPa. If 1 inHg is 33.86 hPa, 29.92 inHg is equal to 29.92 times 33.86. 29.92 (inHg) x 33.86 (hPa / inHg) = 1013.09 hPa. Note how the units of inHg x (hPa / inHg) results in the “inHg” canceling out and leaving final units of hPa.

3. (c) 214.5 hPa. following the equation, ( Partial pressure ) = ( Percent of volume of gas ) x ( total atmospheric pressure ), plug in your known values. ( Percent of volume of gas ) = 20.9% = 0.209. ( total atmospheric pressure ) = 1026.4 hPa. 0.209 x 1026.4 hPa = 214.5 hPa

4. Sample Answer: With a larger ozone hole, less ozone would be present to absorb UV radiation. Thus, more UV radiation would reach the surface. While commonly the largest Ozone Holes are over the South and North Poles, the increase in UV radiation can lead to (a) negative health impacts for people in the form of enhanced frequency of sunburn and potential for skin cancer, (b) reduction in agricultural productivity of crops, and (c) many other negative environmental outcomes.

5. Sample Answer: This seasonal fluctuation occurs due to the seasonal changes in vegetation across the planet, particularly in the Northern Hemisphere. In the NH winter, vast regions of vegetation die back, including trees dropping their leaves. This die-back and associated decomposition releases CO₂ into the atmosphere, resulting in winter CO₂ peaks. Then during late spring, summer vegetation leaf out occurs and begins to absorb large amounts of CO₂ from the atmosphere in association with photosynthesis. CO₂ levels reach minimums in early fall at the end of the warm season filled with CO₂ absorption.

6. (c) 7.2°C/1000m. start with your equation: ( lapse rate ) = ( temperature B – temperature A ) / ( elevation B – elevation A) and plug in values as you have them. ( lapse rate ) = ( 0°C – 30°C ) / ( 4150m – 10m ) = -30°C / 4140m = – 0.0072 °C / m = 7.2°C / 1000m

Lab 2

1. (a) absorbed / cooled. Evaporation is the process of changing from liquid water to water vapor phases. This change requires heat energy to occur, thus 600 cal/g of latent heat is absorbed from the surrounding air. This transfer of heat from the air to the water results in cooling of the air. Choice (a) is correct. This process is why sweating results in you feeling cooler. When sweat evaporates off of your skin, latent heat is transferred from your skin to the water (vapor). This cools your skin down, making you feel more comfortable.

2. E = 393.9 W/m². Your first step is to figure out what equation(s) you need to work with. Because you are calculating energy emitted, you need the Stephan Boltzmann equation:  E = (ε) x (σ) x (T⁴). We know ε is 1, σ is 5.67 x 10^-8, and we were given T as 288.7 K. That temperature in K is then raised to the 4th power: (288.7 K)⁴ = 6,946,837,428 K⁴ Finally, you solve the entire equation: E = (1) x (5.67 x 10-8 W/m²K⁴) x (6,946,837,428K⁴)

3. 472 W/m². Use of a +/- 25 W/m² buffer is recommended.

4. 53°. First, this entails knowing the sun is directly overhead at the equator on the vernal equinox at solar noon. The second part applies our equation of sun angle, A = 90 – |(location – directly overhead location)|   Plugging in, A = 90 – (37° – 0°)

Lab 3

1. (a) °C = ( °F – 32 ) / 1.8
   °C = ( 100 – 32) / 1.8
°C = ( 68 ) / 1.8
°C = 37.7778 = 38 °C

(b) °F = (°C x 1.8) + 32
°F = ( 15 x 1.8 ) + 32
°F = (27) + 32
°F = 59°F

   (c) ° F = ( K x 1.8 ) – 459.67
°F = ( 255 x 1.8 ) – 459.67
°F = ( 459 ) – 459.67
°F = ( -0.67 ) = -1°F

   (d) K = °C + 273.15
   K = (-25) + 273.15
K = 248.15 = 248 K

2. Example Answer: The maximum daily temperature for this location will typically occur in the middle of the afternoon (approximately) at the end of the energy surplus. Up until this point, there has been more energy being added than removed. After this point, more energy is removed than is being added, so temperature begins to decrease.

3. Example Answer: The isotherm will next move NNW towards Cincinnati, OH, continue NW across central Indiana (near Indianapolis) and then across IL into the 60° location in central IA.

4.

5. -16 °F

Lab 4

1.

2.

– For the second row, the wet bulb depression is 9.5 (or half way between 9 and 10 °F), while the dry bulb is 42.5 (or half way between 40 and 45°F). Thus, we can interpolate the relative humidity as halfway between each of these rows/columns. Rounding to the nearest whole number from these half-way points, you arrive an answer of 30% (average of 29%, 22%, 38%, 31%)

– For the third row, the wet bulb depression is 1.5°F (halfway between 1 and 2°F). For the 100°F dry bulb row, that puts this observation at half way between 1°F (96%) and 2°F (93%) = 94.5%, or 95% when rounded.

3.

4. (c) radiation fog forms due to the emission of longwave radiation from the surface that cools the surrounding air to the dew point. This type of fog occurs most commonly on cloud free nights (allowing more longwave radiation to be emitted to space, compared to cloudy nights), where the winds are calm (decreasing the amount of mixing in the atmosphere and lessening what are called turbulent heat fluxes).

5. (d) altostratus. Based on the information that was provided, this is the best answer. Altocirrus is not an actual type of cloud. Cumulus clouds have a puffy, or heaped, appearance, not aligning with the ‘sheet-like’ description here. Nimbostratus clouds typically are thicker than altostratus and extend beyond just the middle cloud layer down into the low cloud region. They are also associated with precipitation while altostratus typically are not.

Lab 5

1. 14.1°C

Step 1: which lapse rates do we need?
Unsaturated parcel to start, saturated parcel later. Thus both dry and moist lapse rates are needed.
DALR: 1°C / 100m; MALR: 0.6 °C/100m

Step 2: how far does the parcel travel? When does it become saturated?
1000m when unsaturated; 1000m when saturated.

Step 3: Calculate the change in temperature for the amount of lift based on the lapse rate.
You can technically do this in one step, but they could be separated  for convenience.

Step 4: Calculate the final temperature:

2. -6.25°C

Step 1: which lapse rates do we need?
Parcel is sinking, so only the DALR is needed: 1.0°C / 100m

Step 2: how far the parcel travel?
Sinks from 2650 to 1225: 2650m – 1225m = 1425 meters

Step 3: calculate the change in temperature.

Step 4: calculate final temperature

3.

4. At what pressure value is the air temperature equal to -10°C?

2. (d) right to left (land to sea). Based on the isobars dipping near the land surface, the atmospheric pressure at the land surface is lower than at the ocean surface. This will cause the winds to move from high pressure location H towards low pressure location F, or from right to left (sea to land) at location G.

3. False. A Mountain Breeze is one type of Katabatic Wind, not the other way around.

4. (a) Cool in the east and warm in the west. Normally the trade winds blow from east to west and this drags the ocean surface in the same direction. Off shore of South America, this results in a displacement of water that the ocean attempts to re-balance by pulling (or upwelling) water from below. This water is much colder than the surface water and creates the gradient of warm water to the west and cool water to the east.

Lab 8

1. (a) continental Arctic
(b) maritime Polar
(c) maritime Equatorial

2. (a) 81°F
(b) 5 kts
(c) Few
(d) N
(e) 63°F
(f) 5 mb

3. (a) dry line
(b) cold front
(c) moving north to south

4. (d) Thermal Low. A thermal low does not move substantially, if at all. This is due to its formation mechanism being tied to surface heating. If it moves away from that heating, the system weakens quickly.

Lab 9

1. (a) 1200 UTC June 3, 2023
(b) None
(c) Cold front
(d) Rain, snow, mixed precipitation, and thunderstorm
(e) Low pressure, or cyclone
(f) Southeastern NM and western TX

2. (a) July 7-11, 2022
(b) Maine, New Hampshire, Vermont, Massachusetts, Rhode Island, Alaska.
(c) 60-70%
(d) 80-90%

3. (c) Winter Weather Advisory. In this region, the forecast within the next 24-48 hours for snowfall (or frozen accumulation) averaging at least 4 inches in 12 hours is considered the threshold for a winter storm warning. A winter storm watch is when that same accumulation of snowfall is forecasted for 2-3 days into the future. A winter weather advisory is for snowfall accumulations of 1-4 inches within a 24-36 hour period, or when a rapid accumulation of less than 1” of snowfall would be ill-timed, such as during rush hour. In this case, given the forecasted snowfall magnitude of 2-3 inches over 24 hours, (c) winter weather advisory, is the best selection.

4. Example Answer: This PoP means the likelihood of precipitation of at least 0.01 inches this afternoon (noon through 6pm) somewhere within the county (or forecast area) is 50%. It does not mean that 50% of the forecast area will receive precipitation. It also does not mean that there will be precipitation at your location 50% of the time.

Lab 10

1. (b) Multi-cell Thunderstorm. This is marked by a series of individual cells that span the 3 stages of development (cumulus, mature, dissipating), where higher levels of wind shear allow for the updrafts to tilt, aiding in new cells forming as older cells dissipate.

2. ~40 days per year.

3. 3.6 miles away.

4. (a) Kansas, Oklahoma, Texas all have some amount of ENH risk (orange) within the state.
(b) All but extreme western KY are rates as TSTM, meaning there is at least a 10% probability of non-severe thunderstorms

5. April. KY experiences 5 tornadoes, on average, during April, compared to 4 in May, 3 in February, and 2 in January, March, June, October, and November.

Lab 11

1. Tropical Depression

2. (b) Category 3, 4, and 5.

3. False

4. 10 years.