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Dissolved Oxygen Laboratory Report Essay

1. Based on the dissolved oxygen values for the three distilled water samples at different temperatures, it appears that there is a relationship between water temperature and DO concentration. It appears based on the averages of the class data that the lower the temperature of the water, the higher the DO concentration, although only to a certain point. There appears to be no significant difference between the DO concentration of the water at room temperature or when heated.

2. The DO concentration of the distilled water at room temperature is higher than that of the seawater, where the average for the distilled water is 4. 03 mg/L and for seawater is 3. 48 mg/L. This is likely to be due to the presence of other dissolved gases in the seawater, such as nitrogen and argon, which mean that less oxygen may be chemically combined with the water molecules . 3. At an average of 3. 22 mg/L, the dissolved oxygen concentration of the supersaturated water is also much lower than that of the distilled water at room temperature. It is not clear why this is the supersaturated water should contain more oxygen than could actually be dissolved by the water, therefore be at maximum dissolved oxygen levels.

Figure 2: The percentage saturation of the water taken from average of class data at two different temperatures 4. The dissolved oxygen percentage saturation is the amount of oxygen that is dissolved in the water expressed as a percentage of the total maximum amount that could be dissolved. The average of the class data indicates that the percentage saturation is lower in lower temperatures than at higher temperature, although there is no data available for heated water, so this judgment is based only on cooled and room temperature water.

This shows that at the lower temperature it is possible to have a much higher level of oxygen dissolved in the water than at room temperature, as it already has higher amounts, but still a larger capacity to dissolve more. 5. At the current DO level of 4. 82, this is 36. 67% saturation. Therefore 100% saturation would be: (4. 82/36. 67) x 100 = 13. 14 mg/L

References

Heguenin, J. E. & Colt, J. (2002). Design and Operating Guide for Aquaculture Seawater Systems – 2nd Edition. Danvers, MA: Elsevier.


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