Problem Set 2 Solutions - Marine Chemistry | OCEAN 520, Assignments of Geology

Material Type: Assignment; Class: MARINE CHEMISTRY; Subject: Oceanography; University: University of Washington - Seattle; Term: Autumn 2008;

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Ocean 520
Marine Chemistry
Autumn 2008
Problem Set 2 solutions
1.a. Find the volume of the lake:
Surface Area = 1 km2 = (1000 m/km)2 = 1 * 106 m2
Volume = SA * Depth = (1 * 106 m2) * (20 m) = 2 * 107 m3
Use the residence time equation:
τ = V/f
= (2 * 107 m3) / (50 m3 s-1) = 400000 s
400000 s * (1 m / 60 s) * (1 h / 60 m) * (1 d / 24 h) = 4.6 days
b. Use equation in Appendix 2.2:
C – Cin = (Co – Cin)exp-(f*t/V)
Where Cin = 10 µM L-1
f = 50 m3 s-1
t = 5 days (convert to seconds) = 5 * 60 * 60 * 24 = 432000 s
V = 2 * 107 m3
Solve for C:
C = (10 µM L-1) – (10 µM L-1) * (0.339596) = 6.6 µM L-1
c. Use same equation as above except now:
Co = 6.6 µM L-1
Cin = 1 µM L-1
t = still 5 days (from end of spill to 10 days after spill began) = 432000 s
Solve for C:
C = (6.6 µM L-1 – 1 µM L-1) * (0.339596)) + 1 µM L-1 = 2.9 µM L-1
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Ocean 520 Marine Chemistry Autumn 2008 Problem Set 2 solutions 1.a. Find the volume of the lake: Surface Area = 1 km^2 = (1000 m/km)^2 = 1 * 10^6 m^2 Volume = SA * Depth = (1 * 10^6 m^2 ) * (20 m) = 2 * 10^7 m^3 Use the residence time equation: τ = V/f = (2 * 10^7 m^3 ) / (50 m^3 s-1) = 400000 s 400000 s * (1 m / 60 s) * (1 h / 60 m) * (1 d / 24 h) = 4.6 days b. Use equation in Appendix 2.2: C – Cin = (Co – Cin)exp-(f*t/V) Where Cin = 10 μM L- f = 50 m^3 s- t = 5 days (convert to seconds) = 5 * 60 * 60 * 24 = 432000 s V = 2 * 10^7 m^3 Solve for C: C = (10 μM L-1) – (10 μM L-1) * (0.339596) = 6.6 μM L- c. Use same equation as above except now: Co = 6.6 μM L- Cin = 1 μM L- t = still 5 days (from end of spill to 10 days after spill began) = 432000 s Solve for C: C = (6.6 μM L-1^ – 1 μM L-1) * (0.339596)) + 1 μM L-1^ = 2.9 μM L-

2.a. Find volume of precipitation: Vp = (67.8 km^2 ) * (1.72 m yr-1^ precip) * (1000m/km)^2 = 1.17 * 10^8 m^3 yr-1^ (or 1.

  • 10^11 L yr-1) b. Seepage = 0.72 * (1.17 * 10^8 m^3 yr-1) = 8.40 * 10^7 m^3 (or 8.40 * 10^10 L yr-1) Evaporation = 0.28 * (1.17 * 10^8 m^3 yr-1) = 3.27 * 10^7 m^3 (or 3.27 * 10^10 L yr-1) c. Use equation 2.10: FH2Oconv = Ʃqconv / (∆T * Cp) where Ʃqconv = 6.4 * 10^8 J s- ∆T = 350 K Cp = 5.8 J s-1^ K- answer = 3.15 * 10^5 g s-1^ or 315 kg s- convert to years = answer * 60 * 60 * 24 * 365 = 9.9 * 10^9 kg yr-1^ or L yr- d. Because a temperature greater than 350 C is reached during hydrothermal circulation we can assume that all Mg is removed; thus hydrothermal uptake is the concentration of Mg in the lake multiplied by the yearly volume of the hydrothermal circulation: 9.9 * 10^9 kg yr-1or L yr-1^ * 0.0002 mol L-1^ = 1.99 * 10^6 moles Mg yr- e. Mg is removed from the lake in two ways: through seepage and through hydrothermal uptake. We can calculate the residence time by finding the total mass of Mg in the lake and dividing by annual Mg removal: Mg removed by seepage: 3.27 * 10^10 L yr-1^ * 0.0002 moles Mg L-1^ = 1.68 * 10^7 moles Mg yr- We add the amount of Mg removed by hydrothermal uptake: 1.68 * 10^7 moles Mg yr-1^ + 1.99 * 10^6 moles Mg yr-1^ = 1.88 * 10^7 moles Mg yr- The total mass in the lake: Lake volume * conc of Mg = (53.2 km^2 ) * (1000 m/km)^2 * (350 m) * 1000 L/m^3 0.0002 mol L-1^ = 3.72 * 10^9 moles of Mg Res time = (3.72 * 10^9 moles of Mg) / (1.88 * 10^7 moles Mg yr-1) = 198 years