# B Problem Solving in Environmental Science

## B.1 Dimensional Analysis

• Length/depth is one dimensional: 1-D (e.g., m)
• Area is two dimensional: 2-D (e.g., m2)
• Volume is three dimensional: 3-D (e.g., m3)
• The dimension of any one unit is related to the exponent of that unit

To convert from one dimension to another, follow the rules for exponents. For example:

• $$length×length=area = L^1 × L^1 =L^{1+1} =L^{2}$$
• $$area×length=volume=L^2 × L^1 =L^{2+1} =L^3$$
• $$volume ÷ area = length= L^3 ÷L^2 =L^{3-2} =L^1$$
• $$volume ÷ length = area = L^3 ÷ L^1 = L^{3-1} = L^2$$

## B.2 Unit Conversions

• Measurements are taken in many different units (i.e. metres, miles, inches, feet, pounds, etc.)

• ANY unit is converted in one more steps using “conversion factor(s)”

• The conversion factor is ALWAYS mathematically equal to ONE (i.e. 1 km ÷ 1000 m = 1) and all components of the conversion factor have an infinite number of significant figures.

• Cross multiplication is the key:

Given unit x desired unit ÷ given unit = desired unit

### B.2.1 Converting without Dimensional Change

If a woman has a mass of 115 lb, what is her mass in grams? Knowing that 1 lb = 435.6 g, we have:

115 lb x 453.6 g/1 lb = 52164 g = 5.22 x 104 g

Two units at the same time: The average speed of a nitrogen molecule in air at 25$$^\circ$$C is 515 m/s. Convert this to miles per hour. Knowing that 1 km = 1000 m, 1 mi = 1.6093 km, 60 s = 1 min, 60 min = 1 hour, we have:

515m/1s = 1km/1000m = 1mi/1.6093km x 60s/1min x 60min/1hr = 1152.05mi/hr = 1.15 x 103 mi/hr

Converting Across Dimensions:

Earth’s oceans contain approximately 1.36 × 109 km3 of water. Calculate the volume in liters. Knowing that 1L=10−3m3 and 1km=103 m:

$$(\frac{10^3m}{1km})^3 = \frac{10^9 m^3}{1 km^3}$$

Thus, converting from km3 to m3 to L, we have:

Volume in liters = $$(1.36 \times 10^9 km)(\frac{10^9 m^3}{1 km^3})(\frac{1 L}{10^{-3} m^3}) = 1.36 \times 10^{21} L$$