# PBS Teachers™

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 50.5 + 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 100.5 + 0.474 - 0.02 x 10)(T - 91.4) + 91.4

-111.4 = (0.3 x 100.5 + 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

= (0.38V0.5 + 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(5.27V0.5 + 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.