The graph below illustrates the amount of pollutant Y removed from water (in mg/L) by an absorbent material as a function of contact time (in minutes).
[Imagine a smooth curve graph: x-axis: Contact Time (minutes) from 0 to 60. y-axis: Pollutant Y Removed (mg/L) from 0 to 50. Data points: (0, 0), (10, 20), (20, 35), (30, 45), (40, 48), (50, 49), (60, 49.5)]
Based on the graph, what is the approximate amount of pollutant Y removed after 15 minutes of contact time?
Correct: B
To estimate the amount of pollutant removed at 15 minutes, we need to interpolate between the 10-minute and 20-minute data points.
- At 10 minutes, 20 mg/L is removed.
- At 20 minutes, 35 mg/L is removed.
15 minutes is halfway between 10 and 20 minutes. The curve is increasing rapidly between these points. Visually, if you draw a line from 15 minutes on the x-axis up to the curve and then across to the y-axis, the value falls roughly halfway or slightly less than halfway between 20 and 35, given the curve's shape (it's still accelerating somewhat before it starts to flatten). The difference between 35 and 20 is 15. Half of 15 is 7.5. So, 20 + 7.5 = 27.5 mg/L. This estimation aligns well with the visual interpolation of the curve.