when will it be reached Oxygen depletion is a serious problem in many parts of t
ID: 652 • Letter: W
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when will it be reached
Oxygen depletion is a serious problem in many parts of the ocean. Measurements have shown that the oxygen saturation has decreased between 1985 and 1995 on average from 90% to 80% in a rectangular area located between 160 degree E and 140 degree W in the Pacific and extending from 40 degree N 500 km to the North and 500 km to the South. The average depth of this oxygen depletion zone is 200 m. Assume a constant oxygen solubility of 6.0 ml_/L in that region. 3 mL/L is a critical value for fish, below which they do not have enough oxygen to live. Is the oxygen depletion already a problem for fish? If yes, when was that value reached, or if not, when will it be reached (assuming a constant absolute depletion rate). Assume that all the oxygen that "disappeared" between 1985 and 1995 was filled in tank trucks that can hold 20,000 L and are 20 m long. If these tank trucks were parked in a long line along the equator, how often would this line go around the Earth? [hint: circumference of the Earth is 40,000 km. Ignore the expansion of oxygen due to pressure].Explanation / Answer
a)let us assume we are in 2015 then from the given reference every 10 yrs saturation drops by 10%
by 2005 O2% falls from 80 to 70 by 2015 it becomes 60%
oxygen solubility = 6*0.6 = 3.6 ml/L which is greater than the critical value 3ml/L
so it is not a problem at present
value will be reached at a saturation = 3/6 = 0.5
so at 50 % saturation of O2 which occurs in 2025 the critical value will be reached at it becomes a problem for fish
b)the rectangle length is given by dist b/w 160E and 140W
it constitutes 60 degrees of earth that is 1/6 of earths circumference =40000/6 km
breadth =500+500 =1000 km
depth=200m
volume = 40/6*0.2*1*10^6 km^3 =1.33*10^15 m3 = 1.33*10^18 L
volume of o2 disapppeared = (0.9-0.8)*6* 1.33*10^18 ml =8*10^14 L
no of tanks required = 8*10^14/20000 =4*10^10
length covered by tanks = 4*10^10*20 = 8*10^11
this would go around earth for 8*10^11/(4*10^7) = 20000 times
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