The BOD5 of a waste water is determined to be 240 mg/l at 20℃. The k value at 20℃ is known to be 0.23 day-1 . Calculate the BOD7 if the test were run at 10℃.ans.224.05 mg/l240 mg/l196.89 mg/l258.12 mg/l

The BOD (Biochemical Oxygen Demand) is a measure of the amount of oxygen required by bacteria to decompose the organic matter in the water under aerobic conditions. The BOD5 is the BOD after 5 days, and the BOD7 is the BOD after 7 days.

The BOD at any time t can be calculated using the formula:

BODt = BOD0 * (1 - e^(-kt))

where:

• BOD0 is the initial BOD (in this case, BOD5 = 240 mg/l),
• k is the deoxygenation rate constant (in this case, k = 0.23 day^-1 at 20℃),
• t is the time (in days),
• e is the base of the natural logarithm (approximately equal to 2.71828).

However, the rate constant k is temperature-dependent. It approximately doubles for every 10℃ increase in temperature. Therefore, if the temperature decreases from 20℃ to 10℃, the rate constant k at 10℃ would be approximately half of the rate constant at 20℃, i.e., k10℃ = 0.23 day^-1 / 2 = 0.115 day^-1.

Now, we can calculate the BOD7 at 10℃ using the formula:

BOD7 = BOD5 * (1 - e^(-k7))

where k7 = 0.115 day^-1 * 7 days = 0.805.

So, BOD7 = 240 mg/l * (1 - e^(-0.805)) = 240 mg/l * (1 - 0.447) = 240 mg/l * 0.553 = 132.72 mg/l.

Therefore, none of the provided options are correct. The BOD7 at 10℃ would be approximately 132.72 mg/l.