ESE 148c Problem Set #1 Solutions
1) Isotopic Constraints on the Burial of Organic Carbon.
a) (2 pts) Assume that the Earth is a closed system in steady state with respect to carbon
(and hence carbon-13). Inputs to the atmosphere and hydrosphere through volcanic
emissions (planetary degassing) must balance removals from the atmosphere/hydrosphere
through burial of organic and inorganic (carbonate) sediments. Isotopic mass balance is
given by:
i = foo + (1 – fo)a
We know that o = -25‰, a = -1‰, and that fo = 0.30. Solve for i:
i = (0.3)(-25‰) + (1 – 0.3)(-1‰) = -8.2‰
b) (2 pts) Rearranging the equation in (a):
fo = (i - a)/(o - a)
For o = -25‰, a = 2‰, and i-8.2‰, fo = (-8.2‰ - 2‰)/(-25‰ - 2‰) = 0.38.
Today,  = a - o = -1‰ – (-25‰) = 24‰. If this fractionation remains the same, then
during the late Paleocene, o = a -  = 2‰ - 24‰ = -22‰. Under this assumption,
fo = (-8.2‰ - 2‰)/(-22‰ - 2‰) = 0.43
In justifying the assumptions, you should indicate the presence of a functional brain
residing in your head – that’s all. Alex and Ashley are happy to discuss any of your ideas
about how the carbon cycle should or should not change over geologic time.
c) (1 point) One possibility is that increased temperatures increase productivity in the
world’s oceans, thereby leading to an increased production and burial of organic carbon.
Increased weathering rates on the continents could lead to an increased flux of nutrients
to the oceans stimulating this increased productivity.
Another possibility is that increased carbon burial lead to greater methanogenesis, which
is a more powerful greenhouse gas and lead to greater warming.
You could also imagine that increased temperature caused by high concentrations of CO2 led to
fertilization of marine primary productivity.
Probably you came up with even more ideas than these. Again, the goal is that you show
evidence of an ability to use the powerful thinking machine you were born with.
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2) Carbonate Equilibrium.
a) (2 points) The total store of inorganic carbon in the ocean (also known as DIC) is equal
to the sum of the concentrations of carbonic acid (H2CO3*), bicarbonate ion, and
carbonate ion. By using the equilibrium relationship, we can rewrite this sum as:
Total Inorganic Carbon = DIC = [HCO3-][H+]/K1 + [HCO3-] + K2[HCO3-]/[H+].
Substituting [HCO3-] = 2 x 10-3, [H+] = 10-8.1, K1 = 10-6.1, K2 = 10-9.39, we find that
DIC = (2 x 10-3)(10-8.1)(106.1) + (2 x 10-3) + (10-9.39)(2 x 10-3)(108.1) = 2.12 x 10-3 mol/kg.
This implies that the whole ocean contains (2.12 x 10-3 mol/kg)(1.4 x 1021 kg) =
2.97 x 1018 mol carbon = 3.56 x 1019 g DIC. If this were all present as HCO3- (a good first
order approximation), then the ocean has (3.56 x 1019 g HCO3-)*(12 g C/61 g HCO3-) =
7.00 x 1018 g C.
The atmosphere contains only (280 x 10-6)(5.3 x 1021 g) = 1.48 x 1018 g CO2 =
4.04 x 1017 g C (multiplying again by the ratio of molecular weights to get grams carbon).
As you can see from this calculation, the ocean stores much more carbon than the
atmosphere.
b) (2 points) We now have to consider an open system where DIC is set by equilibrium
with the atmosphere. The Henry’s Law constant for CO2 is 10-1.46 mol/L atm. For the preindustrial atmosphere, 280 ppm corresponds to a CO2 partial pressure of 2.8 x 10-4 atm.
Henry’s Law tells us that the concentration of H2CO3* is (10-1.42)(2.8 x 10-4) =
1.06 x 10-5 mol/kg.
For a bicarbonate concentration of 1.6 x 10-3 mol/kg, we use the equilibrium expression
to solve for [H+]:
10-5.95 = (1.6 x 10-3)[H+]/1.06 x 10-5 → [H+] = 7.43 x 10-9 = 10-8.13.
The pH of the pre-industrial surface ocean is 8.13. Similar calculations are performed for
the other conditions.
Modern: CO2 partial pressure = 3.8 x 10-4 atm, [H2CO3*] = 1.44 x 10-5, pH = 8.0.
Potential Future: CO2 partial pressure = 5.0 x 10-4 atm, [H2CO3*] = 1.90 x 10-5,
pH = 7.88.
c) (2 points) Rearranging the equilibrium expressions gives carbonate ion concentration
in terms of carbonic acid and proton concentrations:
[CO3-2] = K1K2[H2CO3*]/[H+]2
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For the pre-industrial surface ocean:
[CO3-2] = (10-9.1)(10-5.95)(1.06 x 10-5)(108.13)2 = 1.72 x 10-4 mol/kg.
For the potential future ocean:
[CO3-2] = (10-9.1)(10-5.95)(1.90 x 10-5)(107.88)2 = 9.74 x 10-5 mol/kg.
The change in carbonate ion concentration from an ocean in equilibrium with the
atmosphere with 280 ppm CO2 and an ocean in equilibrium with an atmosphere with 500
ppm CO2 is 1.72x10-4 mol/kg – 9.74x10-5 mol/kg = 7.45x10-5 mol/kg. In other words, the
concentration of carbonate ion will decrease by 43%!
3) (5 points) Studying the Hydrologic Cycle through Isotopes.
Consider the problem in three stages. In the first stage, water is evaporating from the
ocean and forming a cloud. This is an open system because of the enormous size of the
ocean relative to the cloud. Assume that every water molecule that evaporates forms the
cloud so there is only a fractionation associated with evaporation from the ocean. At
15ºC,  = 1.08. Using the approximation  = 103( - 1) = c - v, and given c = 0‰,
solve for v (the delta value of the cloud). [Note that I’m using the subscript c for the
condensed phase – the liquid water – and the subscript v for the vapor phase – the cloud.]
 = 103(1.08 – 1) = 80
80 = c - v = 0 - v
v = -80‰
You’ll note that I have used approximate equations here. It is also fine (though slightly
more complicated) to use the exact relationship
 = (c + 1000)/(v + 1000)
This gives you a slightly different answer of -74‰, illustrating the difference between the
approximate and exact equations. Either method of solving for the isotopic composition
of the cloud is fine for this problem set.
In the second stage, the lake is replenished by precipitation. This precipitation forms from
condensation of 25% of the cloud. Since water that condenses in the cloud is permanently
removed from the system (as rain), this is a process governed by Rayleigh distillation (in
other words, it is a closed system and the equation used above no longer holds). You
cannot treat this problem by using mass balance, because this assumes a constant
fractionation between reactant and product. In Rayleigh distillation, the fractionation
increases as the reaction proceeds. At 10ºC,  = 1.095. First, let’s focus on the cloud
which starts at -80‰ and then loses 25% of its water through the formation of rain. We
can approximate this process by the equation:
 = (r,f - r,o)/ln(1-f)
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Here,  = 103(1.095 – 1) = 95, f = 0.25, and r,o = -80‰. Solving for r,f (the delta value
of the cloud after rainfall), we find that r,f = (95‰)*(ln0.75) - 80‰ = -107‰.
Now calculate the delta value of the precipitation. You can consider this a “pooled
product” since the variable we are interested in is the delta value of water formed and
accumulated from precipitation. Use the equation:
p,f = r,o – [(1 – f)ln(1 – f)]/f
and the value  = 103(1.095 – 1) = 95 ‰.
So,
p,f = -80‰ – [95‰(1 – 0.25)*ln(1 – 0.25)]/0.25 = -80‰ - (71.25)(-0.29)/0.25 =
-80‰ – (-82.7‰) = 2.7‰.
Lastly, calculate the delta value of snow (and assume that there is no fractionation as the
snow replenishes the lake, so that snow = groundwater.
The snow can also be considered a pooled product, but now it is forming at -10ºC, so  =
1.155 and  = 155 ‰.
Using the same equation as for rain,
p,f = -107‰ – [155‰(1 – 0.95)*ln(1 – 0.95)]/0.95 = -107‰ – (7.75)(-3.00)/0.95 =
-107‰ – (-24.5‰) = -82.5‰.
Now use isotopic mass balance to determine what fraction of the lake water comes from
each source. Let f be the fraction coming from groundwater.
lake = fgroundwater + (1-f)rain
-70‰ = f(-82.5‰) + (1 – f)(2.7‰) → f = 0.85.
Eighty-five percent of the lake’s water comes from groundwater discharge. This probably
means that it’s a bad idea to allow the manufacturing facility to drill their well.
4) Oxygen in the Deep Ocean.
a) (3 points) Heterotrophy can be approximated by the reaction CH2O + O2 → CO2 +
H2O. Thus, for every mole of C that is remineralized, one mole of oxygen is consumed.
Heterotrophs consume 90% of the carbon that is exported from the surface, so the
number of moles of oxygen consumed is (0.9)(0.6 mol/m2 yr) = 0.54 mol/m2 yr.
Assume that average deep Atlantic water is 250 years old and that average deep Pacific
water is 750 years old (these numbers come from the mean ocean overturning time of
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1000 years). The area of the Atlantic Ocean is 82.44 x 1012 m2 and the area of the Pacific
Ocean is 165.25 x 1012 m2. Under these values, Atlantic water is found to have
consumed (0.54 mol/m2 yr)*(250 yr)*(82.44 x 1012 m2) = 1.11 x 1016 moles of O2 and the
Pacific water will consume
(0.54 mol/m2 yr)*(750 yr)*(165.25 x 1012 m2) = 6.69 x 1016 moles of O2.
Compare this with the oxygen present in newly-formed deep water. Deep water formed in
the Atlantic will have a concentration of 350 mol/kg. Since the Atlantic Ocean has a
mass of 3.30 x 1020 kg (I got this number by multiplying the volume of the Atlantic by an
average ocean density value of 1021 kg/m3) , deep water has
(350 mol/kg)*(3.30 x 1020 kg) = 1.16 x 1017 moles O2. Therefore, after heterotrophic
oxygen consumption, the average deep Atlantic will have
1.16 x 1017 mol O2 – 1.11 x 1016 mol O2 = 1.05 x 1017 mol O2. This corresponds to a
concentration of (1.05 x 1017 mol O2)/(3.30 x 1020 kg) = 318 mol O2/kg.
This water continues along the conveyor belt to form Pacific deep water. The mass of the
Pacific is 7.22 x 1020 kg, so the deep Pacific starts out with (7.22 x 1020 kg)*(318
mol/kg) =
2.30 x 1017 moles O2. After heterotrophy, the Pacific will have
2.30 x 1017 moles O2 – 6.69 x 1016 moles O2 = 1.63 x 1017 moles O2 corresponding to a
concentration of (1.63 x 1017 mol)/(7.22 x 1020 kg) = 226 mol/kg.
b) (1 point) The consumption rate of organic carbon to the deep Atlantic remains
0.54 mol/m2 yr. The deep water initially has 350 mol/kg or 3.57 x 105 mol/m3. For a
mean depth of 4000 m, this means that oxygen concentration in a square meter parcel of
water is 1428 mol O2/m2. Dividing by the rate of consumption, oxygen will be depleted
after 2644 years.
Note that many variations on this solution were accepted for full credit so long as your
logic and assumptions about the ocean were correct.
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