Date Due: Thursday, 25 June 1998
The total maximum points were 100. Point distribution for each question noted below.
(4 pts.)
| Weather - A description of the current state of the atmosphere
Weather maps, today's high or low temperature, tomorrow's weather forecast. Climate - A longer term state of the atmosphere and its interactions with the other components of the earth-atmosphere-hydrosphere system. The "normal" climate for Madison; continental versus maritime climates or polar versus tropical. |
1b. The barometric pressure associated with one standard atmosphere
at mean sea level is
[you may round to nearest whole number]:
(5 pts - 1 each)
1 atmosphere of pressure is equivalent to:
76.0 centimeters of mercury 14.7 (or 15) pounds per square inch (psi) 1013.25 (1000) millibar (mb) 34 feet of water
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1c. The lowest recorded sea level corrected pressure in
the world was ________. [Please include units!]
| Lowest pressure: 870 mb = 25.68 inches of mercury |
| Highest pressure: 1083.8 mb = 32.01 inches of mercury |
| The range: Range = (High - Low)
= (1083.8 - 870) mb = 214 mb =(32.01 - 25.68) in = 6.33 in of mercury |
1d. What is the weight exerted by the atmosphere upon the flat,
horizontal roof of a 25 foot by 50 foot building? [Assume standard sea
level conditions; English units may be used here]. Clearly show your
work for partial credit!
| Weight = Pressure x area
Pressure = 15 pounds per square inch (approx.) Area = 25 ft x 50 ft = 1250 sq. ft. Since 1 sq. ft. = 144 sq. in. (count them - since 12 inches on each side of the square), then 1250 sq. ft. = 1500 x 144 = 180,000 sq. inches. Then: Weight = 15 lb per sq. in x 180,000 sq. in. = 2,700,000 lb or 1350 ton (If 14.7 psi were used, the weight would be 2,646,000 lb or 1323 ton) Note that units check too! While this answer may sound large, the roof does not collapse from the weight exerted by the atmosphere since the air pressure is pushing on the roof in all directions. |
1e. A football fan brought an aneroid barometer to Mile High
Stadium in Denver (elevation of 1 mile) and made a reading of 835 mb. What
would be the approximate sea level corrected pressure if we assumed that
the pressure decreases at approximately 1 mb per 10 meters ascent through
the atmosphere?
| Mile High Stadium in Denver is 5280 ft above mean sea level (MSL),
or 1600 m MSL.
Since the air pressure is assumed to decrease at a rate of 1 mb per 10 m, the pressure at the stadium should be 160 mb less than at the mean sea level directly below the stadium. Because the observed station (or in this case, stadium) pressure was 835 mb, by descending to sea level, the pressure would increase, or at the stadium would be [840 + 160] mb or 995 mb. |
How does this sea level pressure that you calculated compare with the
standard sea level pressure?
| The sea level pressure below Denver on this particular day (995 mb) is slightly less to a typical value of sea level pressure (1000 mb) and it is 18 mb less than standard sea level pressure (1013 mb). |
1f. Convert the following temperature readings:
| 41° F = 5° C = 278 K
-40° C = -40° F = 233 K 258 K = -15° C = 5° F Note: Be careful of signs! If the negative sign does not
appear in your answer where appropriate,
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1g. The record highest temperature for Madison, WI was 107°F
(41.7°C) on 14 July 1936, while the record low was -37°F (-38.3°C)
on 30 Jan 1951. What is the range of Madison's extreme temperatures?
| Range = (High - Low) = 107° F - (-37)° F = 144° F. |
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| United States | 134° F or
56.7° C |
-79.8° F or
-62.1° C (includes Alaska) -69.7° F or
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213.8° F or
118.8° C (includes Alaska) 203.7° F or
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| World | 136° F or
57.8° C |
-129° F or
-89.4° C |
265° F or
147.3° C |
1i. The National Weather Service at Madison reported the following
information for individual days during this past January. The "normal"
high and low temperatures for these days are also included and represent
the 30 year averages for the 1961-1990 climatological interval.
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| 9 Jan 1998 |
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i.) Actual Heating Degree Day Units:
| HDDU = [65° F - Average daily temperature]
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| 9 Jan 1998: 65° F - 20° F = 45 HDDU |
| 11 Jan 1998: 65° F - 4° F = 61 HDDU |
| 9 Jan: 65° F - 16° F = 49 HDDU |
| 11 Jan: 65° F - 16° F = 49 HDDU |
| Since 11 Jan 1998 was a day that was cold (daily average was +4° F) or "below normal" temperature-wise, 61 HDDU were accumulated as compared to the "normal" of 49 HDDU. Thus, more energy than normal would have to be consumed to heat your house. However, two days earlier on 9 Jan 1998 the daily average of 20°F was above "normal" meaning less heat than normal would be needed, since 45 HDDU were accumulated as compared with the typical 49 HDDU. |
a. Mark and label the North and South Poles.
b. Draw and label the Equator.
c. Draw and label the i.) Tropics of Cancer and Capricorn and ii.) the Arctic and Antarctic Circles.
d. Mark with the letter "V" that latitude where the sun appears to be directly overhead at local solar noon.
e. Mark with the letter "T" that latitude at the edge of the polar night where the sun appears to be just on the local horizon at local solar noon.
(20 pts)
21 MARCH 21 JUNE 23 SEPTEMBER 23 DECEMBERSee also satellite images (courtesy of DataStreme and Project Atmosphere, the education initiative of the American Meteorological Society)
21 MARCH 21 JUNE 23 SEPTEMBER 23 DECEMBER
(2 pts)
| 1.97 cal per sq. cm. per min or 1372 Watts per sq.
meter.
These values were provided and discussed in lecture. |
b. A mythical planet has an orbit with an average planet-sun distance exactly half that of the earth's. What would be the solar constant for this mythical planet? [Hint: make use of your answer from above.]
(4 pts)
| The inverse square law means that the planet at one half the distance
would have 4 times (2 squared) the amount of energy per unit area per unit
time.
7.88 cal per sq. cm. per min or 5488 Watts per sq. meter. |
c. What is the planetary albedo of the planet earth?
(2 pts)
| From lecture:
Planetary albedo = 30 - 31 percent . |
a. Which object would radiate more energy?
b. Which object would radiate more of its energy at a shorter wavelength?
(1 pt. each or 2 pts)
| a. A (A consequence of Stefan-Boltzmann law)
b. A (A consequence of Wien's Displacement law) |
a. What is the wind chill equivalent temperature if the ambient air temperature were 15°F and the wind speed were 10 mph?
b. What is the wind chill equivalent temperature if the ambient air temperature remained at 15°F, but the wind speed increased to 25 mph?
c. What has caused the difference between your answers a and b above? Why?
d. To what temperature does your automobile reach in part a? in part b?
(11 pts)
| Note that Table 3.3B (for English units) should be used:
a. -3° F is the wind chill equivalent temperature b.-22° F is the wind chill equivalent temperature c. The increased wind speed causes the difference in the wind chill equivalent temperatures. The convective heat loss from the human body increases with increased winds. A statement about the air being colder is not correct. d. The temperature of your automobile can only reach the ambient air temperature of 15° F in both cases, and go no lower. The wind-chill equivalent temperature is not relevant here. However, in the second case, the stronger winds would hasten the cooling process. |
a. How much energy is required to entirely melt 1 gram of ice at the ice point?
b. How much energy is required to evaporate 1 gram of liquid water at room temperature?
c. How would the temperature of 1 kilogram of liquid water originally at 20°C change if 5000 calories were used in the heating process (assume no phase transformations)? If a temperature change would take place, indicate the amount of change (and the direction of the temperature change).
[Please show your work and include units!]
(11 pts)
| a. 80 cal (This is the latent heat of melting.)
b. 590 cal (Note: The latent heat of evaporation is a function of temperature, being 590 cal at 20° C and decreasing to 540 cal at 100° C.) c. 5 deg C heating Since 1 calorie is the heat needed to raise 1 gram of water 1 Celsius degree, then 5000 calories would raise 1000 grams by 5 degrees (making the liquid warm from 20° C to 25° C) |
Last revision: 25 June 1998
Produced by Edward J. Hopkins, Ph.D. Department of Atmospheric and Oceanic Sciences University of Wisconsin-Madison, Madison, WI 53706 hopkins@meteor.wisc.eduURL Address: aos100/homework/98hmk01a.htm
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