Elementary POW, October 7-11, 1996

Elementary POW, October 7-11,1996

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Today we are concerned about the cost of transportation, both financial costs and environmental impact. One of the major concerns is the cost of gasoline and the environmental impact of the exhaust, which consists mostly of carbon dioxide. Since the gasoline shortages of the 1970's there has been an interest in improving the efficiency of the automobile.
I'm sure you all understand the importance of saving some money, but some of you might ask, "Why is it important to reduce carbon dioxide (CO₂) releases? After all carbon dioxide is the same gas that the human body releases when breathing." Well, some scientists believe that the carbon dioxide that we are releasing into the atmosphere is contributing to global warming.
Current automobile efficiency is approximately 20 percent. This means that only 20 percent of the energy available in the gasoline is actually used to make the car go forward. Some of the energy is lost through the exhaust. (Be careful around the tail pipe of your car - it is HOT.) Some of the energy is lost through cooling of the engine cylinders. (Be careful around the car radiator - it is HOT.) Additional energy is used by your air conditioner, friction in the engine, etc.

Problem
If we could double the efficiency of your family automobile, how much money (in dollars) would you save in one year and how much less carbon dioxide (in grams) would be released to the environment in one year?
Assumptions

Your family drives 225 miles each week.
Your car can travel 20.0 miles on a gallon of gasoline.
Gasoline costs $1.60 per gallon of gasoline.
Assume that gasoline is octane (C8H18).

Hints

A gallon of gasoline is 3.785412 liters.
The density of octane is 0.7025 grams per milliliter.
Every gram of gasoline burned in your car releases approximately 3.1 grams of carbon dioxide.
There are 52 weeks in one year.
1.0 liter equals 1000.0 milliliters.

Thanks to Andre Nyce, a 10th grade student at Germantown Academy in Ruth Carver's Geometry Honors class, for the solution to this problem.

Solution
The first step is to find how much money and how much carbon dioxide is released now. The family drives 11,700 miles (225*52) in one year. The car uses 585 gallons per year (11700 miles/20 miles per gallon). The family spends $936 in one year (585*1.60). If the car were twice as efficient, it would get 40 miles to the gallon, thus the family would only use 292.5 gallons in one year, and spend only $468. So they save $468.
In the problem they give you the density of octane in grams per milliliter and the volume of gasoline in liters. By using the formula D=M/V, and substituting 0.7025 g/mL for D and 3.785412 L for V, I found that there are 2659.25 grams of gasoline per gallon. Every gram of gasoline releases 3.1 grams of CO₂, so every gallon of gasoline releases about 8243.68 grams of CO₂. In one year the car releases 4,822,554 grams of CO₂ (8243.68*585). If the car were twice as efficient, it would only release 2,411,276.688 grams of CO₂ (8243.68*292.5). So 2,411,276.688 less grams of CO₂ is released into the environment.

Correct Solutions were submitted by:

Chrislyn Doran, Jen cumming, Katie lynch, Emily G., Nathan strauss, Jo-Pete Nelson, Craig Cramer, Sara Wyszomierski, Geoffery Lawson, Ashley Cowart, Shelby Tuinder, Viay Aggarwal, Amy Forster.

Highlighted Solution

Diana M. had an exellent solution to this problem:
To solve this problem, first you had to find how many gallons of gas you use in a week. I divided to find out this answer (225 divided by 20). If you work out this problem, you receive the answer of 11.25 gallons per week.
Then, to find out how much money you spend on gas a week, you would multiply 11.25 gallons times $1.60 a gallon. I came up with $18.00 you pay each week for gasoline.
Then, to find out how much you spend on gas per year, you would multiply how much you spend per week ($18.00) by the number of weeks in a year (52). I came up with $936.00 per year on gasoline if you need 11.25 gallons of gasoline a week.
To find the double efficiency, you need to double your amount of miles per gallon of gasoline to 40 miles. So, 20 miles now equals 40 miles. Again, you need to find out how much you use in a week (5.625, how much you spend a week on gasoline ($9.00), and how much you spend on gasoline in a year ($468.00). To find out how much money you saved in a year by using the double efficiency, you would subtract: $936 (how much money you spent in one year with 20 miles on one gallon of gasoline) minus $468.00 (how much you spent by having 40 miles on one gallon of gasoline. The answer would be $468.00. This means that you save $468.00 a year.
To find out how much less carbon dioxide was released in a year, you would subtract: 11.25 (the answer of how much gasoline you use in a week in regular efficiency) minus 5.625 since that is how much gasoline you need per week in double efficiency. You should get 5.625 gallons of gasoline saved per week. If you save 5.625 gallons a week, then you multiply that by 52, which is how many weeks there are in a year. Then you would find out how many gallons were saved per year (292.500 gallons saved per year).
One of the hints provided was that one gram of gasoline burned in your car releases 3.1 grams of carbon dioxide. To calculate the total grams of carbon dioxide reduced, I had to convert gallons of gasoline to grams of gasoline using the other hints. First, to convert gallons to grams, I needed to convert gallons to liters, liters to milliliters, and milliliters to grams. You convert gallons to liters by multiplying by 3.785412 because a gallon of gasoline is equal to 3.785412 liters. To convert liters to milliliters, you multiply by 1000. Then, you would convert milliliters to grams by multiplying by 0.7025 because the density of octane is 0.7025 grams per milliliter. If you did this math correctly, you should calculate the answer 777,831,18 grams of gasoline saved a year. Then you multiply by 3.1 and calculate the number of grams of carbon dioxide reduced. You should calculate the answer 2,411,276.6 grams of carbon edioxide saved. This is the way to complete this problem.

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