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Weather: Answers

Temperature Measurement

In the 1700s, G. Daniel Fahrenheit developed a scale used by meteorologists for measuring surface temperature. The scale was named for the developer, and the unit of measure has become known as degree Fahrenheit (F°). Also in the eighteenth century, a second scale was developed for measuring surface temperature; it became known as the Celsius scale. The unit of measure in the Celsius scale is the degree Celsius (C°). A third scale later developed for use by scientists became known as the Kelvin scale. This scale begins at absolute zero and is sometimes more convenient to use because it does not involve negative temperatures. (The word degree is not used in Kevin measure.)

Citizens of the United States primarily use the Fahrenheit scale, the rest of the world uses the Celsius scale, and scientist use either the Celsius or Kelvin scale. Since we can use three different scales used to measure temperature, it seems reasonable to have formulas for changing or converting from scale to the another. Here are some useful conversion formulas.

C° = (F° - 32°) ÷ 1.8

F° = 1.8 x C° + 32

K = C°+273

1. If the temperature is 75° Fahrenheit, what are the equivalent readings on the Celsius and Kelvin scales?

75°F = 23.9° C and 296.9 K

2. If the temperature is 26° Celsius, what are the equivalent readings on the Fahrenheit and Kelvin scales?

26° C = 78.8° F and 299 K

3. If the temperature is 288 Kelvin, what are the equivalent readings on the Celsius and Fahrenheit scales?

288 K = 15° C and 59° F

4. Create a formula to determine the Kelvin temperature give the degrees Fahrenheit.

K = (F – 32) ÷ 1.8 = 273

= F ÷ 1.8 + 255.2

The following table shows the normal high temperatures, in degrees Fahrenheit, for each month for three selected US cities.

City

Jan.

Feb.

Mar.

Apr.

May

June

July

Aug.

Sept.

Oct.

Nov.

Dec.

Baltimore

41

44

53

65

74

83

87

86

79

68

56

45

San Francisco

57

61

62

63

65

68

69

70

73

70

63

57

St. Louis

38

43

53

67

76

85

89

87

81

69

54

43

5. Use the information in the table to determine the mean high temperature for each of these three cities.

Baltimore’s mean high temperature is 65°F.

San Francisco’s mean high temperature is 65°F.

St. Louis’ mean high temperature is 65°F.

6. Based solely on the mean high temperature, can you easily decide in which city you might like to live? Why or why not?

It would not be possible to make a decision since all the had the same mean value.

7. Us the information in the table to determine the median high temperature for each of these three cities.

Baltimore’s median high temperature is 66.5°F.

San Francisco’s median high temperature is 64°F.

St. Louis’ median high temperature is 68°F.

8. Based solely on the median high temperature, can you easily decide in which city you might like to live? Why or why not?

It would be very difficult because the medians are relatively close in value. If you wanted the warmest you may be tempted to choose St. Louis.

9. Make a box plot of the monthly high temperatures for each city and compare them. Do the plots influence your choice of the city? If so, in what way.

10. What information do you receive from using the box plots that was unavailable from the mean or the median?

Box plots provide knowledge of five values concerning the spread of the temperatures. They indicate the two extremes (high and low) the median (middle value) and the two quartiles (the middle of the lower half and the middle of the upper half). You also get a visual picture of the spread of each set of data relative to the other two.

11. Describe another method you could use to get more information from this table that might help decide which city might have the best temperature for you.

A side-by-side bar graph would provide a visual comparison the variation of the temperatures each month.

Wind Chill

One of the most prevalent aspects of current weather reporting is the inclusion of the wind chill factor each day, particularly in the northern climates during the winter.

Wind chill factors are supposed to measure the effect of the combination of temperature and wind speed on human comfort. Remember that these factors do not have the same effect on inanimate objects, or even on other animals or plants. Nor is this effect felt by humans who are sheltered from the wind.

One formula for determining the wind chill factor in degrees Fahrenheit can be found at
http://ourworld.compuserve.com/homepages/Gene_Nygaard/windchil.htm and is:

Twc = (0.3V ^0.5 + 0.474 - 0.02V)(T - 91.4) + 91.4

Twc is the wind chill, V is the wind speed in statute miles per hour, and T is the temperature in degrees Fahrenheit.

In a formula to calculate a Celsius wind chill, V would be the wind speed in kilometers per hour, and T would be the temperature in degrees Celsius.

1. Using the formula compute the Fahrenheit wind chill for a wind speed of 5 mph and a temperature of 10 °F.

Twc = (0.3 x 5^0.5 + 0.474 - 0.02 x 5)(10 - 91.4) + 91.4 = 6.35 or 6 °F

2. If the wind chill reading were –20 °F and the wind speed were 10 mph, determine the temperature in degrees Fahrenheit.

-20 = (0.3 x 10^0.5+ 0.474 - 0.02 x 10)(T - 91.4) + 91.4

-111.4 = (0.3 x 10^0.5+ 0.474 - 0.02 x 10)(T - 91.4)

-111.4 = 474.75(T - 91.4)

-111.4 / 474.75 = (T - 91.4)

-0.2346 = (T - 91.4)

-0.2346 +91.4 = T

91.2 = T

3. Explain how the formula for Fahrenheit wind chill could be changed to create a formula for Celsius wind chill. Use your explanation to create your own formula for Celsius wind chill.

Replace the V in the formula with the conversion formula for converting miles per hour to kilometers per mile, and replace the T with the conversion formula for converting Fahrenheit to Celsius.

Twc = (0.3 x (1.609V)^0.5+ 0.474 - 0.02 x 1.609V)((T-32)x1.8) -91.4) + 91.4

= (0.38V^0.5+ 0.474 - 0.032V)((T - 32)x1.8 - 91.4) + 91.4

According to the source above, a formula for Celsius wind chill is:

T(wc) = 0.045(5.27V^0.5 + 10.45 - 0.28V) (T - 33) + 33

Remember, V is the wind speed in kilometers per hour, and T the temperature in degrees Celsius.

4. Compare the results of using the formula you created to the results from the given formula for determining Celsius wind chill. What might cause any differences.

Using 60 °F with a wind speed of 5 mph, the formula we derived gives a wind chill of 3.14 °C.

Using the given formula and the equivalent measures of 15.5°C and 8.045 kph gives a wind chill of 14.79 °C.

The difference between the two values can be attributed to the rounding of many values of the constants used in our creation of the formula. If other values are used, the formulas lead to values whose difference will be larger or smaller. If we were to carry as many decimal places as our calculator or computer would allow, there would still be a difference, though it would be smaller.

Rainfall

3. Determine the average of the extreme temperatures for each site.

State	Station	Extreme High Temperature in °F	Extreme Low Temperature in °F	Average Extreme Temperature in °F
Alabama	Mobile	104	3	53.5
Alaska	Anchorage	85	-34	25.5
Alaska	Barrow	79	-56	11.5
Arizona	Phoenix	122	17	69.5
Arkansas	Little Rock	112	-5	53.5
California	Los Angeles	112	28	70
California	San Diego	111	29	70
California	San Francisco	106	20	63
Colorado	Denver	104	-30	37
Connecticut	Hartford	102	-26	37
Delaware	Wilmington	102	-14	44
District of Columbia	Washington- National	104	-5	49
Florida	Jacksonville	105	7	56
Florida	Miami	98	30	59
Georgia	Leant	105	-8	48.5
Georgia	Savannah	106	3	54
Hawaii	Honolulu	94	53	73.5
Idaho	Boise	111	-25	43
Illinois	Chicago	104	-27	38.5
Illinois	Moline	106	-27	39.5
Indiana	Indianapolis	104	-23	40.5
Iowa	Des Moines	108	-24	42
Kentucky	Lexington	103	-21	41
Kentucky	Louisville	105	-20	42.5
Louisiana	New Orleans	102	11	56.5
Maine	Caribou	96	-41	27.5
Maine	Portland	103	-39	32
Maryland	Baltimore	105	-7	49
Massachusetts	Boston	102	-12	45
Michigan	Detroit	104	-21	41.5
Michigan	Sault Ste. Marie	98	-36	31
Minnesota	Duluth	97	-39	29
Minnesota	Minneapolis	105	-34	35.5
Mississippi	Jackson	106	2	54
Missouri	Kansas City	109	-23	43
Missouri	St. Louis	107	-18	44.5
Montana	Helena	105	-42	31.5
Nebraska	Omaha	114	-23	45.5
Nebraska	Scottsbluff	109	-42	33.5
Nevada	Reno	105	-16	44.5
New Jersey	Atlantic City	106	-11	47.5
New Mexico	Albuquerque	105	-17	44
New York	Albany	100	-28	36
New York	Buffalo	99	-20	39.5
New York	New York-La Guardia	107	-3	52
North Carolina	Asheville	100	-16	41
North Carolina	Raleigh	105	-9	48
North Dakota	Bismarck	109	-44	32.5
Ohio	Cleveland	104	-19	42.5
Ohio	Columbus	102	-19	41.5
Oregon	Portland	107	-3	52
Pennsylvania	Philadelphia	104	-7	48.5
Pennsylvania	Pittsburgh	103	-18	42.5
Rhode Island	Providence	104	-13	45.5
South Carolina	Charleston	104	6	55
South Dakota	Huron	112	-39	36.5
South Dakota	Rapid City	110	-30	40
Tennessee	Memphis	108	-13	47.5
Tennessee	Nashville	107	-17	45
Texas	Galveston	101	8	54.5
Texas	Houston	107	7	57
Utah	Salt Lake City	107	-30	38.5
Vermont	Burlington	101	-30	35.5
Virginia	Norfolk	104	-3	51
Virginia	Richmond	105	-12	46.5
Washington	Seattle-Tacoma	99	0	49.5
Washington	Spokane	103	-25	41.5
Wisconsin	Milwaukee	103	-26	38.5
Wyoming	Lander	101	-37	32

The following is a scatter plot of the average extreme temperatures and the normal precipitation for each site.

3. Describe the relationship between the average extreme temperature and the normal precipitation indicated in the scatter plot.

There is very little association apparent from the graph.

4. The lower right hand corner of this plot shows a cluster of five sites. Using both the table and graph, identify these five sites and determine if there is any relationship among them.

The five cities are San Diego, San Francisco, Los Angeles, Phoenix, and Honolulu. These five cities are among the most south and/or west of all the cities.

5. What observations can you make about the relationship between the temperature and the amount of rainfall in these five cities?

They have a high average temperature and low precipitation.

The following table is from Greener Pastures Relocation Guide, 1984.

City	Mean inches of Rainfall	Percent Sunshine
Los Angeles, CA	14	73
Salt Lake City, UT	15	70
Phoenix, AZ	7	86
Las Vegas, NV	9	84
San Francisco, CA	20	67
Denver, CO	16	70
Wichita, KS	31	65
Oklahoma City, OK	31	67
Albuquerque, NM	8	77
Houston, TX	48	57
Little Rock, AR	49	63
New Orleans, LA	57	59
Nashville, TN	46	57
Jackson, MS	49	60
Mobile, AL	60	67
Charlotte, SC	66	43
Raleigh, NC	60	43
Miami, FL	66	60
St. Louis, MO	58	36
Louisville, KY	57	43
Norfolk, VA	63	45

6. Which city has the maximum percentage of sunshine?

Phoenix, Arizona has the maximum percentage of sunshine.

7. Which three cities have the least amount of rainfall?

Phoenix, Arizona, Albuquerque, New Mexico, and Las Vegas, Nevada have the least amount of normal annual rainfall.

8. Create a scatter plot with the rainfall on the horizontal axis and the percent of sunshine on the vertical axis.

9. Describe the graph in your own words.

The graph looks like it is decreasing and somewhat linear.

10. You can notice a cluster of points in the lower right of the graph. Explain what you know about these cities simply by their locations on the graph.

These five cities have a low percentage of possible sunshine and high annual rainfall,

11. Draw a line on the graph that represents the relationship among the data.

12. Determine the slope of your line, and explain what it tells you about the relationship between the data.

The slope of the line is approximately -0.5. The slope tell us that for every one inch increase in rainfall, the percentage of possible sunshine will decrease one-half of a percentage point.

13. Describe the x- and y-intercepts in words.

The y-intercept is the predicted percentage of possible sunshine when the rainfall is zero. The x-intercept is the predicted inches of rainfall when the percent of possible sunshine is zero.

14. How can you tell how well this lines fits the data?

Younger and less experienced students will probably identify things like the points lying near the line and the line appearing to pass through the middle of the data.

More experienced students will probably use the least squares regression line and should refer to the sum of the squared residuals and the correlation coefficient. The correlation coefficient is -0.8078.