SME
Climatology · Problem type
Ocean temperature from a shell’s δ¹⁸O
Use this for “A foram’s calcite has δ¹⁸O of 1.5 ‰ and the seawater it grew in 0.5 ‰. What was the water temperature?”
Chart
T(1) = 12.62
This is an example. Type your own values to start a new problem.
T (°C)
=
16.9
−
4.38(
δc
−
δw
)
+
0.10
(
−
)
2
δc is the shell’s calcite reading and δw the seawater’s; the picture’s x-axis is their difference Δδ.
Δδ
Shell minus water
worked out (not typed) · ‰
Formulas
Δδ = δc − δw
1 = 1.5 − 0.5
T
=
16.9
−
4.38
×
Δδ
+
0.10
×
Δδ²
12.62
=
16.9
−
4.38
×
1
+
0.10
×
1
2
Assumptions
•
The Shackleton (1974) calibration for calcite grown in step with the water: colder water puts more ¹⁸O into the shell.
•
Calcite is read on the VPDB scale and the water on VSMOW less 0.27 ‰, so the two compare directly.
•
The water’s δw rises as ice sheets lock up ¹⁶O, so it must be known separately (about +1 ‰ at the last glacial maximum).
Step-by-step
WE KNOW
δc = 1.5 ‰, δw = 0.5 ‰
FIND
shell minus water (Δδ), water temperature (T)
Step 1 · Find shell minus water (Δδ)
Use Δδ = δc − δw
The shell’s offset from the water it grew in is what the temperature sets.
Δδ = 1.5 − 0.5
Δδ = 1 ‰
Step 2 · Find water temperature (T)
Use T = 16.9 − 4.38 × Δδ + 0.10 × Δδ²
Each per mil of offset means about 4.4 °C colder water; the square term bends the line slightly.
T
=
16.9
−
4.38
×
1
+
0.10
×
1
2
4.38 × 1 = 4.38
0.10
×
1
2
=
0.1
16.9 − 4.38 + 0.1 = 12.62
T
=
12.62
°C
Check
Put the numbers back in. Both sides should match.
1 = 1.5 − 0.5 ✓
12.62
=
16.9
−
4.38
×
1
+
0.10
×
1
2
✓
Topic check
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