Unit Conversion (Example)

In this problem we are given density of an
oil and we want to know the mass of three barrels of oil. We want that mass in kg and
it tells us that one barrel of oil is 42 gallons. Then it gives us the conversion of 1 gallon
is 3.8754 L. I am going to solve this problem. This is not a problem that I looked at previously.
I am going to show you the approach I would use to solving this problem. First I would
tend to write down the information that I have in notation. In this case using a symbol
for density that is common. What I want is the mass in kg so this is my unknown. I have
three barrels and 1 barrel is 42 gallons. Then 1 gallon is 3.8754 L. This is a problem
essentially of unit conversion and the easiest way to do unit conversion is to first calculate
the volume of three barrels. It is critical to carry the units through. Three barrels
and 1 barrel is 42.0 gal. I am writing it this way since I am multiplying. Notice those
units cancel and that is the reason I wrote the 1 barrel in the denominator. Likewise
1 gallon is 3.8754 L so gallons cancel. Now I should be able to do this multiplication
to determine how many liters I have. I get 488.3 L. I want the mass so the mass is going
to be the density times the volume. I get 0.85 g per mL. 488.3 L but I need to make
the units cancel to get a mass in kg because the problem asks for kg. 10^3 mL is a L and
10^3 g is a kg so these 10^3 cancel and mL, L, and g cancel. My final answer then will
be in kg. That will be 415.05 kg. I should reduce to have the correct number of significant
figures. If there are only 2 significant figures here then probably should round this off to
420 kg in those three barrels of oil.

One thought on “Unit Conversion (Example)

  1. Could someone do the conversion of 4.18 kJ K-1 Kg-1 to x BTU F-1 lb-1 at 20 degrees Celsius please. It is supposed to be 0.99 but I keep getting 7.8 or 460 – Nowhere near!

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