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Make a plot of age vs. the three variables Essay

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2) Make a plot of age vs. the three variables. (?18Ow, SST, ?18Oc) Examine the phasing of ?18Ow, SST, and ?18Oc in the plot. For the major transitions in the record, what order do these variables change in? (Like which one is first the red or the green or blue?—just explain why you chose that order.)What does this imply about how the climate system is changing at these times?

According to the graphical representation, a fluctuation in temperature, as designated by the red line, appears to occur slightly before changes in both the 18O in the water and 18Oc.   The changes in oxygen concentrations appear to be inversely proportional to changes in the temperature of the water.  The higher the temperature of the water, the lower amount of oxygen is dissolved into the water, and vice versa.

Based on the information provided, it appears that at least three warming then cooling events have been recorded.  The warming periods appear to be gradual in nature, spanning on average 80 – 100 ky.  The cooling spans, however, are markedly more rapid, spanning only approximately 25 -45 ky.   This indicates that the oceans experienced a very gradual warming, and a decrease in the amount of oxygen in the water, which in turn indicates that the oxygen was being utilized either by animal life, or was actively oxidizing different elements or minerals within the ocean environment, such as iron or other metalics.

3) Using the change in ?18Ow from the last glacial maximum to today, determine the change in sea level that corresponds to this ?18Ow change. Assume that the ice sheet mean ?18O was -30 per mil (SMOW). You will need some other information to do this, but it should be readily available

Vc (d18Oc) = Vg(d18Og) + (Vi)(d18Og)

Vc = Volume of current ocean

Vg = Volume of glacial ocean ~ 20 kya

Vi =  Volume of water removed to form ice

d18Oc = d18O in current ocean

d18Og = d18O in glacial ocean ~ 20 kya

 0.152446379 = d18Oc

1.057403921 = d18Og

Pacific Ocean

Average Depth = 4280m = 4.280 km

Surface Area = 155,557,000 km2

Vc = (4.280)(155,557,000)

Vc = 6.67 x 108 km3

Vg = (Average Depth – Depth Change)(Surface Area)

Vg = (4.280 – 0.00079)(155,557,000)

Vg = 6.66 x 108 km3

Solve for Vi

Vc (d18Oc) = Vg(d18Og) + (Vi)(d18Og)

[Vc(d18Oc) – Vg(d18Og)]/(d180g) = Vi

[6.67 x 108 (0.152446379) – 6.66 x 108(1.057403921)]/ 1.057403921 = Vi

[1.01 x 108 -7.04 x 108]/1.057403921 = Vi

-5.703 x 108 km3 = Vi

Volume of Water Removed for Ice (Vi)  is 5.703 x 108 km3

Change in Sea Level = Current Depth – [Vi/Surface Area]

Change in Sea Level = 4.280  – [5.703 x 108 / 155,557,000]

Change in Sea Level = 4.280 – 3.661

Change in Sea Level =  0.619 km