Question

Biochemical oxygen demand i.e. BOD) is a measure of the potential damage that can be done to the dissolved oxygen content of a body of water by organic waste. A BOD is also the amount of O2required to degrade or decompose the wastes. Its name comes from the "demand" that organic carbon has for O2 through biochemical (enzyme) oxidation. BOD – demanded oxygen – can be likened to a debt that must be paid. Dissolved oxygen (DO) can be thought of as the currency used to pay the debt. If DO (assets) exceeds BOD (debt), aerobic conditions (economic solvency) will prevail. If BOD exceeds DO, bankruptcy may ensue. Bankruptcy, in this case, is the depletion of water's chief asset, O2. Over a period of time, the water's oxygen level is usually replenished by oxygen from the air. The length of this time period depends on the severity of the initial loss of oxygen. If we assume that the part of the organic waste that is being degraded is carbon, C ( a good assumption), then we can write the following chemical formula: C + O2 ? CO2 If we were to do a mass analysis (i.e. gram for gram, pound for pound, etc.) of this equation, we would find that for every 3 parts of C, we would need 8 parts of O2. If you're thinking that it looks like it should be a one for one deal, you're absolutely correct. It does in fact require one MOLECULE of carbon to react with one MOLECULE of oxygen, but if we would weigh those molecules we'd see that different molecules weigh different amounts and therefore based on mass, we'd get the 3:8 ratio. Do not overthink this problem. If you think along the lines of a is to b as x is to y, you'll be on the right track. Enter ONLY the answer in the provided answer box. QUESTION 1 Consider a small lake, 0.3 miles across and 8 ft deep, that contains a total of 40 tons of dissolved oxygen. Calculate the BOD (i.e.calculate the amount of oxygen required to degrade the amount of organic waste given) if 6 tons of carbon-based waste is dumped into the lake. Round your answer to zero places past the decimal. Question 2 Repeat the BOD calculation as done in #1, but this time let's dump in 32 tons of carbon-based waste. Round your answer to zero places past the decimal. QUESTION 3 The differences between the two cases above include all of the following: (Select all that apply) A. In the first case there is plenty of oxygen to go around but not enough in the second case. B. The BOD in the second case is larger than that in the first case. C. The BOD in the first case is exactly the same as the COD. D. In the first case there is a chance that the lake could recover. E. Chances are, in the second case the lake will not recover.

Answer 1

3 unit mass of carbon requires 8 unit mass of oxygen.

so 1 unit mass of carbon requires 8/3 unit mass of oxygen

Case 1

6 ton carbon dumped requires 6*(8/3) ton of oxygen

so the BOD is 16 ton  

Case 2

32 ton carbon dumped requires 32*(8/3) ton of oxygen

so the BOD is 85 ton

A. Lake contains a total of 40 tons of dissolved oxygen so in the first case there is plenty of oxygen to go around (16 ton < 40 ton) but not enough in the second case (85 ton> 40 ton)

B. BOD in the second case is larger than in the first case (85 ton> 16 ton)

C.   Chemical Oxygen Demand is the total measurement of all chemicals in the water that can be oxidized so is always more than BOD. Generally, we take BOD as about 60% of COD

D. In the first case, the lake is safe as it has an excess of oxygen and can always replenish the oxygen lost by dissolving from the air. It has 100% chance to recover

E. The lake can always recover given sufficient time as it can always replenish the oxygen lost by dissolving from the air.

So the correct options that apply are A, B, and D.