Fall 2016 Due: 10:30 a.m., Tuesday, September 13

Transcription

1 Geol 330_634 Problem Set #1_Key Fall 2016 Due: 10:30 a.m., Tuesday, September Why is the surface circulation in the central North Pacific dominated by a central gyre with clockwise circulation? (5) The easterly trade winds near the equator and the westerly winds centered at 40ºN result in convergence at ~30N due to Eckman transport, which is 90 to the right of the winds in the northern hemisphere. As a result of this doming, water tends to transport down the pressure gradient but is deflected to the right due to the Coriolis force. The result is a clockwise circulation around the dome of high pressure. 2. Why is the surface boundary current at the western boundary of the North Pacific stronger than the current on the eastern boundary? (5) The gradient in sea surface elevation is much steeper at the western boundary. 3. Where in the Deep Sea would you expect to find the water that has been isolated from the atmosphere for the longest time. (5) In the deep North Pacific 4. What process caused the salinity to decrease southward in the core of the North Atlantic Deep Water (NADW)? (5) It mixes above (Antarctic Intermediate Water) and below (Antarctic Bottom Water) with seawater with lower salinity. 5. Why does deep water only form in the North Atlantic and not in the North Pacific? (5) The surface seawater in the North Pacific does not have high enough density to sink to more than intermediate depths. 6. What determines if an ion is a major ion in seawater? (5) - Major ions are those that contribute to salinity. - If salinity can be determined to parts per thousand (g/kg) or 1 mg/kg (ppm) a major element is any ion with a concentration greater than 1 ppm. 7. Why do chemical oceanographers prefer to use mole units rather than weight units (like grams)? (5) Because when thinking about the stoichiometry of reactions it is atoms or moles that react not weight or grams 8. Why do chemical oceanographers prefer to use concentration units of moles /kg rather than moles/ltr? (5) Water is compressible (the amount of material in a liter of seawater is dependent on pressure and temperature). The mass of water in a liter varies with depth but a kg is always a kg.

2 9. Some elements in seawater are considered to be conservative. a) Propose a definition that would apply to conservative elements in seawater. (5) Conservative elements in seawater are those whose distributions are controlled by physical processes only (like mixing), and are not affected by chemical and biological processes. Also acceptable: elements that follow the law of conservative ratios. b) Which of the major elements in seawater are considered conservative? (5) Na +, K +, Cl -, SO4 2-, Br-, B(OH)3 (or B), and F c) Vanadium (V) is a trace element that has been proposed to be conservative in seawater. How would you design a study to show that this is true? (5) Collect seawater samples from a variety of locations and depths across a known salinity gradient. Measure concentrations of V in each sample and plot versus salinity, potential temperature or some other conservative property. If the points fall on a straight line, V could be considered conservative. d) If Vanadium is conservative does it have a constant concentration throughout the oceans? Explain (5) No, its concentration will change by the same magnitude that salinity changes.

3 10. Methane is a potent greenhouse gas. Global Methane reservoirs and fluxes are given in the figure below (from Bill Reeburgh; a) Is the present concentration of atmospheric methane (CH 4 ) at steady state? (5) In this box model the arrows are the fluxes and 10 y -1 means 10 Tg CH 4 per year. No. The sources and sinks of methane to the atmosphere are not in balance. Sources = 500 Tg CH 4 y -1 Sinks = 460 Tg CH 4 y -1 (There is a net increase of 40 Tg methane per year in the atmosphere.) b) What are the major sources and sinks of methane to the atmosphere? (5) Major source = several terrestrial sources including wetlands, rice production Major sink = photochemical reduction c) How does the flux of CH 4 (as C) from the ocean to the atmosphere compare with that of CO 2 from Sarmiento and Gruber? You need to convert the fluxes as CH 4 to C only. (5) Total Methane flux from the ocean = 10 Tg CH4 y-1 (10 Tg CH4 y-1)(12 mass units C/16 mass units CH4) = 7.5 Tg C y-1 (0.001 Pg/1 Tg) = PgC per year Total pre-industrial flux of C in S&G is 70.6 Pg C y-1 The C from CH4 is 0.01% (100*( PgC y-1/70.6 PgC y-1)) of the total preindustrial C flux.

4 11. Trace Metal profiles (15) Two part answers for each. a. What would be the general distribution characteristics of a trace metal influenced by: b. How would the profiles differ from the Atlantic to the Pacific influenced by: i) nutrient like (5) a. Concentrations are depleted in the surface and enriched in the deep ocean. (Concentrations are low in the euphotic zone where primary production of organic material consumes most available nutrients. Concentrations increase at depth as nutrients are remineralized.) b. Concentrations are greater in the Pacific (deep water has been out of contact with the surface ocean for a longer amount of time, allowing more time for nutrients to accumulate). ii) hydrothermal inputs (5) a. There would be a mid-depth concentration maximum (around approximately 2500m, or the depth of a nearby ocean ridge crest) b. Differences would not arise strictly due to basin differences or water age, but rather with proximity to hydrothermal vents. iii) scavenging by particles settling through the water column (5) a. Concentrations are high in the surface (have a near surface maximum) and are depleted in the deep ocean. (Elements are adsorbed onto particles that sink through the water column.) b. Concentrations are greater in the Atlantic (deep water has had more time to be scavenged, and therefore concentrations are depleted as water ages).

5 12. A chemical oceanographer is sitting home on Saturday night and decides to entertain herself by making synthetic seawater with chemicals she has around the house (being a chemist, her available household chemicals are wide ranging). She wants to make a solution containing the six most abundant ions in close approximation to their natural seawater concentrations. Her plan is to weigh out the necessary chemicals into a laundry tub and then to add Friday Harbor tap water (which is so dilute that it can be considered like distilled water) until the final weight of the solution is 1 kilogram. Her goal is to have the following ionic concentrations: Ion mmol kg -1 (seawater solution) Na Mg Ca K + 10 Cl SO 4 30 She has the following chemicals NaCl common spice (Morton Table Salt) MgSO 4 fertilizer, foot soaks (Epsom salts) CaCl 2 drying agent, preservative (Dri-Z-Air) HCl liquid, stolen from lab KOH lye Mg(OH) 2 laxative (Milk of Magnesia) How many grams of each chemical should she add to get the correct ionic concentrations (note: H + and OH - will end up balancing, making H + + OH - = H 2 O)? (15) First you need to find the molar mass of each of the salts. You can look this up for the compounds themselves, or add up atomic masses from a periodic table of the elements (e.g. for NaCl = (22.99 g/mol Na) + (35.45 g/mol Cl) = g/mol NaCl): NaCl(s) g/mol (or mg/mmol) MgSO4(s) g/mol CaCl2(s) g/mol KOH(s) g/mol Mg(OH)2 (s) g/mol HCl g/mol Now you want to find the mass of each of these compounds needed to make a solution with the given ion concentrations (mmol/kg). Start with the ions that are only in one of the salts: Na, Ca, K, SO4 (470 mmol Na+)(1 mol Na+/1000 mmol)(1mol NaCl/1 mol Na+)(58.44 mg NaCl/mmol) = mg NaCl = g NaCl As a shortcut, we can use equivalent units of mg/mmol for each molecular weight, but the main unit conversion concepts remain: (10 mmol Ca 2+)(110.98mg CaCl2/mmol) = mg CaCl2 = 1.11 g CaCl2 (10 mmol K+)(56.10 mg KOH /mmol) = mg KOH = 0.56 g KOH (30 mmol SO42-)( mg MgSO4/mmol) = mg MgSO4= 3.61 g MgSO4 Next, you make up the difference in the remaining two ions by accounting for what has already been added with the other salts: Add: 50 mmol Mg2+ 30 mmol Mg2+ from MgSO4 = 20 mmol Mg2+ (20 mmol Mg2+)(58.32 mg Mg(OH)2/mmol) = 1166 mg = 1.17 g Mg(OH)2 Add: 540 mmol Cl- 469 mmol Cl- from NaCl 2(10 mmol Cl- from CaCl2) = 51mmol Cl- (51mmol Cl-)(36.46 mg HCl /mmol) = mg = 1.86 g HCl