Problem Set #4 ANSWER KEY Fall 2009 Due: 9:30, Monday, Nov 30

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1 OCN 520 Problem Set #4 ANSWER KEY Fall 2009 Due: 9:30, Monday, Nov Two-Box Ocean Model The B Flux Using a 2 box model like the one you have worked on in problem set #4 (question 1) assume the following values for total DIC ( C ) and alkalinity ( A ) in the surface and deep ocean. Use V river = kg /yr and V mix equal to 30 x V river. C surface = 1932 µmol kg -1 A surface = 2277 µmol kg -1 C deep = 2256 µmol kg -1 A deep = 2374 µmol kg -1 a) Calculate the magnitude of the total carbon flux from the surface ocean (B C ) and the fraction of B C preserved in sediments (f) for two cases: Case 1: no river input Case 1: no river input Case 2: regular river input with C r = 960 µmol kg -1 Are these answers reasonable? V river *C river + C deep *V mix = B DIC + C surface *V mix B DIC = V river *C river + V mix *(C deep - C surface ) For case 1, there is no river input (C river = 0), but mixing between the deep and surface still occurs, so: B DIC = V mix *(C deep - C surface ) B DIC = 30(3.7x10 16 kg/yr)(2256 µmol/kg 1932 µmol/kg) = B DIC = 3.6x10 20 µmol DIC/yr The fraction of B DIC preserved in sediments must equal the amount of carbon being put into the two box model system via rivers because at steady state the sources (river input) = sinks (burial in the sediment). V river *C river = fb DIC V river *C river /B DIC = f Or similarly, since the deep ocean box is also at steady state you can solve for f in the deep ocean box using the following equation that relates sources and sinks: C deep *V mix = (1-f)B DIC + C surface *V mix Since we have no river input in Case 1 (V river *C river = 0) the fraction of carbon preserved in sediments is 0. In other words, no sediments are buried. This does not seem very realistic, but matches our steady state assumption. Case 2: regular river input with C r = 960 µmol kg -1 V river *C river + C deep *V mix = B DIC + C surface *V mix

2 B DIC = V river *C river + V mix *(C deep - C surface ) B DIC = (3.7x10 16 kg/yr)(960 µmol kg -1 ) + 30(3.7x10 16 kg/yr)(2256 µmol/kg 1932 µmol/kg) B DIC = 3.95*10 20 µmol DIC/yr. f = C river V river /B DIC = (960 µmol/kg)( 3.7*10 16 kg/yr) / (3.95*10 20 µmol/yr) = f ~ 9% This answer is more reasonable. We assumed that our ocean is in steady state so it makes sense that in the case where there is a river input there is also a burial term since steady state implies that the sources to the two box model system must equal the sinks. b) Use an alkalinity balance for the surface box to calculate the relative contribution of CaCO 3 to the total carbon flux (B C ). Assume A r = C r. If the C of sinking particles is composed only of CaCO 3 and organic C, how does your Org C /CaCO 3 ratio compare to deep ocean sediment trap samples (see Power Point Lecture 10)? (Hint: for the alkalinity sink term B A, you'll want to assume that all the alkalinity lost this way goes out as calcium carbonate.) Sources = Sinks V river *Alk river + Alk deep *V mix = Alk surface *V mix + B alk B alk = V river *Alk river + (Alk deep - Alk surface )*V mix B alk = 1.43 x µmol alkalinity/yr. The problem says to assume all alkalinity lost is lost as CaCO 3. We calculated an alkalinity particle export from the surface ocean to the deep ocean of ~1.43 x µmol/yr. If all of this alkalinity is lost as CaCO 3, how much CO 3 2- is this? CaCO 3 Ca +2 + CO 3-2 For alkalinity, the CO 3 2 contributes 2 alkalinity equivalents; therefore; for every one mole of CaCO 3 that is exported then there are 2 moles of alkalinity that are also exported. B alk * 1mol CaCO 3 /2mol alkalinity = B alk /2 = [CO 3 2- ] = 7.15*10 19 µmol/yr of CaCO 3 This number represents the inorganic carbon exported from the surface ocean to the deep ocean in the hard parts. In part b we calculated the total carbon (B DIC ) exported from the surface (in both the hard and the soft parts). B DIC = 3.95*10 20 µmol of C B CaCO3 = 7.15*10 19 µmol CaCO 3 * 1mol C/1mol CaCO 3 = 7.15*10 19 µmol C The soft parts: B org = B DIC - B CaCO3 = 3.95* *10 19 = 3.23*10 20 µmol C Corg/CaCO 3 ratio: (3.23*10 20 µmol C/ 7.15*10 19 µmol C) = 4.5

3 The ratio calculated here is slightly higher than that given in the notes (4:1). This indicates that in our ocean box model there is slightly more soft part export relative to hard part export from the surface ocean. c) If V mix was 50% faster during the last glacial maximum, as has been proposed, what would be the new values of total carbon flux from the surface (B C ) and the fraction of B C preserved in the sediments (f)? V mix = 1.5*V mix B DIC = V river *C river + 1.5V mix (C deep - C surface ) B DIC = 5.7x10 20 µmol/yr f = V river *C river /B DIC = = 6.2%

4 2. New Production from O 2 You have measured O 2 concentrations in the surface ocean over the course of a month during the summer. Average temperature and salinity were 25 C and S = 35, respectively, for the period of observation, while O 2 concentration was 240 µmoles kg -1. The concentration appeared to be at steady state. a) Draw a schematic box-model for the mass balance of O 2 in the surface layer. Indicate the primary sources and sinks. Assume that vertical mixing processes were negligible during the period of your study. Be sure to include gas exchange, bubble injection and biological effects in your model. Bubble injection [O 2 ]atm Gas exchange [O 2 ] Photosynthesis (P) Respiration (R) b) Calculate the air-sea gas exchange using the stagnant boundary layer model. Assume that D O2 = 1.7 x 10-5 cm 2 s -1 and z (the stagnant boundary layer thickness) = 30 µm. The saturation concentration of O 2 for the conditions of your study is 220 µmoles kg -1. Finally, please give your answer in µmol m -2 s -1! F = ( D/z (Cg Csw)) Here Cg is the concentration at the top of the film (i.e. the atmospheric concentration of O 2, and Csw is the concentration of O 2 in the mixed layer). D O2 = 1.7 x 10-5 cm 2 /s z = 30 um = cm F = ( 1.7 x 10-5 cm 2 /s / cm) * ( umol/kg) = cm µmol kg -1 s -1 (Assume 1 kg = 1 L) F = (0.113 cm µmol kg -1 s -1 ) * (1kg/1L) * (1000 L/1 m 3 ) * (1 m/100 cm) = F = 1.13 µmol m -2 s -1 out of the ocean into the atm

5 c) If bubble input was negligible, what is the magnitude of the net biological oxygen production signal (i.e., that due to photosynthesis minus respiration)? We just calculated how much oxygen is coming out of the water (it has to be coming out, since the measured concentration at this site is supersaturated). This is our signal! So how much is biological? Look at the box model. If bubble input is negligible, all of the flux is biological in origin: photosynthesis is the only possible source. Therefore, our biological signal is 1.13 µmol m -2 s -1. d) What is the equivalent carbon new production in mol C m -2 yr -1 based on this biological oxygen signal (assume that 1 kg ~ 1 L)? How does it compare to total primary productivity estimates in different parts of the world s oceans? All we need to do here is use the Redfield ratio to equate C to O 2 and then change units. (1.13 µmol m -2 s -1 )*(1 mol O 2 )/(1x10^6 µmol O 2 )*(106 mol C/154 mol O 2 )*(60s/1min)*(60min/1hr)*(24hr/1day)*( days/1yr) = 24.5 mol C/m 2 *yr (27.4 mol C/m 2 *yr if you use 106:138 C:O 2 ) From the notes (p. 234), average net primary productivity in the world s oceans is estimated at 11.5 mol C/m 2 *yr, but the range is very large, from a high value of 275 mol C/m 2 *yr in the Peru Current to a low of 1.5 mol C/m 2 *yr in the Sargasso Sea. Our value is nearly double the average value reported. This estimate is difficult to make given the dynamics of primary productivity and the effects of vertical mixing and advection on the oxygen and carbon budgets. We made our calculations over the course of a month during summer, when light availability was unlikely to be limiting. This might produce an overestimate of global annual primary production.

6 3. Box Models: 234 Th as a Tracer for Particulate Export Flux The surface ocean almost invariably shows a radioactive disequilibrium of 234 Th from 238 U due to the export of particulate material from the surface ocean with scavenged 234 Th. This is one approach for estimating the flux of carbon from the euphotic zone to the deep sea. Consider the 2-box model: By knowing the activity of 238 U, and assuming secular equilibrium, we know the source of 234 Th into the surface box. By knowing the activity of 234 Th, we know the particulate flux out of the surface box for a steady state ocean. a) If 238 U and 234 Th are at secular equilibrium (equal activities) and both have an activity of 2.40 dpm l -1, what are their respective molar concentrations? For 238 U λ = / t1/2 = / 4.5 x 10 9 years = 2.95 x min -1 N = A/λ = 2.4 dpm / 2.95 x min -1 = 1.35 x 10-8 mol / l For 234 Th λ = / t1/2 = /24.1 days = 2.00 x 10-5 min -1 N = A/λ = 2.4 dpm/ 2.00 x 10-5 min -1 = 1.99 x mol / l b) If the total activity of 234 Th in the surface box is 2.10 dpm l -1 and 238 U is 2.40 dpm l -1, what is the magnitude of the export flux of 234 Th (in moles m -2 y -1 )? The decay constant for 234 Th is λ = d -1. d[ 234 Th]/dt = [ 238 U] λ 238 [ 234 Th] λ 234 Ψ 234 Where (Ψ 234 ) is the amount of thorium lost due to the adsorption of thorium onto particles in units of atoms per m 3 per day (atoms/m 3 d). Note that the concentrations are in units of atoms/m 3. Multiply the above equation by the decay constant for thorium. d(λ 234 *[ 234 Th])/dt = λ 234 [ 238 U] λ 238 λ 234 [ 234 Th] λ 234 λ 234 Ψ 234 da 234 /dt = λ 234 A 238 λ 234 A 234 Ψ 234 * Where Ψ 234 * = λ 234 Ψ 234 and has units of atoms/m 3 d 2. Assuming steady state in the surface ocean (i.e. that the concentration and thus the activity of thorium is not changing with time) we can set the above equation equal to zero and solve for the thorium export term Ψ 234 *.

7 0 = λ 234 A 238 λ 234 A 234 Ψ 234 * Ψ 234 * = λ 234 A 238 λ 234 A 234 = (A 238 A 234 )* λ 234 The above value (Ψ 234 * ) gives us the activity of thorium, in atoms/d, removed per m 3 per day. We want to know how much thorium is exported from the surface box of our model ocean to the deep ocean box so we need to sum up how much thorium activity is removed between 0m and 100m (i.e. integrate over the entire surface ocean). To simplify our calculation we make the assumption that our thorium deficiency profile (the blue line below) can be approximated by a box curve (red line) so that the removal rate of thorium is constant with depth down to 100m. A 234 Depth 100m A 238 Integrating over our 100m surface box we get: ΣΨ 234 * = 100m*(2.40 dpm/l 2.10 dpm/l)*(1000l/m 3 )*( d -1 ) = 864 dpmth/m 2 d c) Assume the 234 Th flux calculated for part b holds for the global ocean (area = 361 x 10 6 km 2 ). If the global average sinking material has an average value of 0.20 dpm 234 Th per µmol organic carbon, what is global new production? Calculate the total flux of thorium out of the surface ocean: (864 dpmth/m 2 d)*(361 x 10 6 km 2 )*(1000m/km) 2 = 3.12x10 17 dpm Th/d

8 Convert this flux to a carbon flux: (3.12x10 17 dpm Th/d)*(10-6 mol C/0.20 dpm Th) = 1.56x10 12 mol C/d Global average new production is: 1.56x10 12 mol C/d d) How does this value compare with global values of new production given in Sarmiento and Gruber or in tables from my Lecture Notes (Lecture 15)? Is this a reasonable value? Explain. We need to convert our new production value to gigatons of C per yr (GtC/yr). Note that there are 1,000,000 grams per metric ton. New Production = (1.56x10 12 mol C/d)*(365 d/yr)*(12g C/mol C)*(1ton/1,000,000g) = 6.83x10 9 ton C/yr New Production = 6.83 GtC/yr The value calculated here is slightly lower than the value obtained in a particle flux study conducted by Martin et al., 1987 (7.4 GtC/yr). One plausible reason for the discrepancy is that we make the assumption that the entire top 100m of the global ocean is being equally scavenged of thorium. In reality our box curve is likely a poor estimate of the real scavenging profile. Also our scavenging profile may not be representative of the entire ocean since certain regions likely have higher particle flux rates and thus more thorium scavenging than others.

Chemical Oceanography Spring 2000 Final Exam (Use the back of the pages if necessary)(more than one answer may be correct.)

Ocean 421 Your Name Chemical Oceanography Spring 2000 Final Exam (Use the back of the pages if necessary)(more than one answer may be correct.) 1. Due to the water molecule's (H 2 O) great abundance in