Environmental Data Analysis EESC BC 3017

Homework 2 - Answer key

Unit Conversions

1.) ( 20p) Benjamin Franklin dropped oil on a lake's surface and noticed that a given amount of oil could not be introduced to spread out beyond a certain area. If the number of drops of oil was doubled, then so was the maximum area to which it would spread. His measurements revealed that 0.1 ml of oil spread to a maximum area of ~430ft2.
a) How thick is such an oil layer? Express the result in 'Angstrom'. The Angstrom is a convenient unit because lighter atoms such as hydrogen, carbon, and oxygen are on the order of 1 Angstrom in diameter. The answer is about 25 A. For the oil that Franklin used this is approximately equivalent to the thickness of one molecule. Please show how you get the result.
B. Franklin spread 0.1 ml or cm3  over an area of 430 ft2. A volume (V) is the product of area (A) and thickness or height (h), or
V = A*h. Consequently the thickness of the oil layer can be calculated as: h = V/A:
h = V/A = 0.1 cm3 / 430 ft2 * 1 m3 / 106 cm3 * 1 ft2 / 0.30482 m2 = 2.5*10-9 m = 25 A
b) Franklin actually showed that 1 teaspoon of oil would spread to cover about 0.5 acre. Determine how many cubic centimeters there are in a teaspoon.
We know from a) what quantity of oil spreads over an area of 430  ft2 . We now want to know how much oil is needed to cover a larger area, 0.5 acre.  Therefore:
V (teaspoon) / 0.1 ml = 0.5 acre / 430 ft2 = 0.5 acre / 430 ft2 * 4047 m2 / acre * 1 ft2 / 0.30482 m2 = 50.7
It follows then:
V (teaspoon) = 0.1 ml * 50.7 = 5.07 ml or about 5 ml.

2.) (8p) Estimate he average spacing (distance) between H2O molecules in liquid water by making use of two pieces of information: (a) liquid water has a density of 1 g/cm3, and (b) every 18g of water contain Avogadro's number (6.02x1023) of H2O molecules.
The density of water is about  1 g/cm3. Consequently 18g of water take up a volume of 18 cm3. I picture a cube that has this volume and contains the water molecules on a regular grid with equal distances. The length L of the side of the cube with a volume V is:

L3 = V => L = V1/3  = (18 cm3)1/3 = 2.62 cm
The number of molecules in the cube is = (number of molecules along one side)3
=> number of molecules on one side N  = (total number of molecules)1/3 = (6.02*1023)1/3 = 8.44*107
 The distance d of the molecules along one side of the cube (and everywhere else) would then be:
d = L/N = 2.62 cm / 8.44 *107 = 3.1 * 10-8 cm =  3.1 A
3.) (8p) Perform excercise #8 in the textbook, page 78 as discussed below (It is about baseball players). Print out the results as stated and hand them in:

An analyst has collected 2007 data including salary and batting average for major league players. Examine the data that have been collected.
Open Baseball.xlsx (in Chapter 2 folder) and

a.) Display only those players whose career batting average is 0.310 or greater. Print a list of those players.
b.) Add a new column to the worksheet, displaying the batting average divided by the player's salary and multiplied by 1,000,000.
c.) Sort the worksheet in descending order of the new column you created. What are the top ten players in terms of batting average per dollar? Print your results (just teh top 10).
d.) Examine the number of years played in your sorted list. Where do most of  the first-year players lie? What would account for that? (Hint: What are some of the other factors besides batting average that may account for a player's high salary?)


Answer key (pdf file)

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