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Math Help - Badly struggling with this conversion problem, please help???

  1. #1
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    Badly struggling with this conversion problem, please help???

    In this module we've been discussing fossil fuels and the impacts of their use. One of the impacts is an increase in acid deposition (or acid rain as many people refer to it). An acid can be neutralized by a base (also known as an alkali). Acid deposition can be neutralized by the addition of agricultural lime (an alkaline rock, not the tasty green fruit). The following exercise asks you to calculate how much agricultural lime it would take to increase the pH of a small lake. Remember that acids have lower pHs and bases have higher pHs, so that as I add more base, the pH will increase.
    Consider a small lake in the Adirondack region of New York state; Lake Whatchamacallit (surface area = 3.2 mi2, average depth = 21 ft). The pH of Lake Whatchamacallit has been measured to be 4.0 (an unfortunate result of acid deposition), which is just a little too acidic to support aquatic life; i.e. Lake Whatchamacallit is for all intents and purposes, dead.
    Farmer Jones, whose property adjoins the lake, knows that when the soil on her farm is too acidic she adds agricultural lime (crushed limestone i.e. calcium carbonate, CaCO3) to it in order to increase the pH of the soil and make it suitable for planting. She gathers all of her neighbors for a meeting with the state's department of natural resources. She proposes that they lime the lake in order to increase its pH and then to restock the lake with new fish.
    How many tons of lime would have to be added to Lake Whatchamacallit in order to raise the pH from 4.0 to 7.0? Use the following facts to help you determine your answer.

    • 1 oz of lime will raise the pH of 5,700 liters of lake-water from 4.0 to 7.0
    • the cost of agricultural lime is about $17 per ton
    • when lime dissolves in water heat is given off, such that when 100 g dissolves in water it gives off enough heat to increase the temperature of 3 liters of water by 1 degree C
    • 1 mi = 5280 ft; 16 oz = 1 lb; 2000 lb = 1 ton; 1 ft3 lake-water = 28.3 liters

    Hint: Start by finding the volume of the lake.

    #tons =

    cost = $

    2.
    Some of the implications of my results include the following: (Check all that apply)
    A) Heat is given off locally when the lime is dissolved in the lake and could therefore affect aquatic life at the site of the addition.
    B) This is a terrible idea because the overall temperature of the lake would increase dramatically.
    C) This is a great idea. We could add the agricultural lime and be done with it.
    D) This is a terrible idea as it is only a "bandaid" approach to solving the problem of acid deposition.
    E) This is a great idea because there are no other costs associated with this solution to the problem.
    F) If farmer Jones and her neighbors want to do this, I say "Why not; it shouldn't hurt anything."
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  2. #2
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    Re: Badly struggling with this conversion problem, please help???

    The procedure for finding the correct answer is more important to me than the answer itself.
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    Re: Badly struggling with this conversion problem, please help???

    There is a technique, usually taught in chemistry but useful much more generally, called dimensional analysis for this sort of problem.

    Here is a link. Math Skills - Dimensional Analysis

    After you read it, try getting started on your problem (maybe by trying to address the hint). If you get stuck, come back, show what you have done, and ask for additional help. If you think you have the right answer, feel free to show your work and ask us to confirm it or identify where you went astray.
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    Re: Badly struggling with this conversion problem, please help???

    Jeff, thank you! This is definitely a helpful start and exactly what I was looking for. I do have a question. I realize that the volume is calculated A*D but the area is in Miles^2 and depth in ft. Could you tell me in what unit I need to convert to?
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  5. #5
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    Re: Badly struggling with this conversion problem, please help???

    Alright, well I've officially tried everything. Any additional input? This is due by midnight tonight EST. Just about ready to give up.
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    Re: Badly struggling with this conversion problem, please help???

    Quote Originally Posted by joetomlee View Post
    Jeff, thank you! This is definitely a helpful start and exactly what I was looking for. I do have a question. I realize that the volume is calculated A*D but the area is in Miles^2 and depth in ft. Could you tell me in what unit I need to convert to?
    It is unfortunate that you posted this in the Math forum. Many many more volunteers would have seen it and been ready to help you if you had posted it in the Algebra forum. This is a volunteer site, and most volunteers have limited time to devote to the site. You want as many as possible looking at your questions to get a prompt answer.

    You can measure volume in terms of cubic miles or in terms of cubic feet or in terms of liters. Which type of measure is best depends on the problem. In this case, you have some of your data in traditional English units and some in metric units. Metric units are easier to work with so I'd figure on converting to metric. It is also easier to avoid mistakes with decimal points by using scientific notation for numbers.

    How do you use the method shown in the site I cited and scientific notation?

    $?\ liters = 21\ feet \times 3.2\ miles^2 \times \left(\dfrac{5,280 feet}{1\ mile}\right)^2 \times \dfrac{28.3\ liters}{1\ foot^3} =$

    $(2.1 \times 10^1)\ feet \times (3.2 \times 10^0)\ miles^2 \times \dfrac{(5.280 \times 10^3)^2\ feet^2}{1^2\ mile^2} \times \dfrac{(2.83 \times 10^1)\ liters}{1\ foot^3} =$

    $(2.1 \times 10^1)\ feet \times (3.2 \times 10^0)\ miles^2 \times \dfrac{(27.8784 \times 10^6)\ feet^2}{1\ mile^2} \times \dfrac{(2.83 \times 10^1)\ liters}{1\ foot^3} \approx$

    $(187 \times 10^{(1 + 0 + 6)})\ feet^3 \times \dfrac{(2.83 \times 10^1)\ liters}{1 foot^3} \approx (1.87 \times 10^2 \times 10^7)\ feet^3\times \dfrac{(2.83 \times 10^1)\ liters}{1\ foot^3} \approx$

    $(1.87 \times 10^9)\ feet^3 \times \dfrac{(2.83 \times 10^1)\ liters}{1\ foot^3} \approx (5.3 \times 10^{10})\ liters.$

    Now you have a bunch more conversions to do involving ounces, tons, and grams, etc. The same methods apply. The next conversion I would do involves determining how many ounces of lime you need. Then you can figure out how many tons that represents and calculate the cost. You can also figure out how many grams that represents and calculate the increase in heat.

    I shall be gone for most of the rest of the day. If you want help on the remaining conversions, I suggest that you do a post on any conversion you have trouble with under Algebra and show your work. You will get much quicker resposnes that way.
    Last edited by JeffM; April 27th 2014 at 09:40 AM.
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