SME
Oceanography
Seawater chemistry
Use this for “Deep water is 2 °C with salinity 34.7 g/kg. What are its density and freezing point?”
Diagram
ρ
ρ
=
1,027
×
(1
−
1.7 ×
10
-4
×
(2
−
10)
+
7.6 ×
10
-4
×
(34.7
−
35))
ρ
=
1,028
kg/
m
3
Freezing point = −0.054 × 34.7 = −1.87 °C.
This is an example. Type your own values to start a new problem.
T
Temperature
example · °C
S
Salinity
example · g/kg
ρ
Density
calculated · kg/m³
Tf
Freezing point
calculated · °C
Formulas
ρ
=
1,027
×
(1
−
0.00017
×
(
T
−
10)
+
0.00076
×
(
S
−
35))
1,028.1626 = 1,027 × (1 − 0.00017 × (2 − 10) + 0.00076 × (34.7 − 35))
Tf
=
−0.054
×
S
−1.8738 = −0.054 × 34.7
Assumptions
•
A straight-line fit near 10 °C and 35 g/kg: ρ₀ = 1,027 kg/m³, α = 1.7 × 10⁻⁴ per °C, β = 7.6 × 10⁻⁴ per g/kg.
•
Pressure is ignored, as for water near the surface; cold water’s density really changes less with temperature.
•
Colder or saltier water is denser, so it sinks under warmer or fresher water.
Step-by-step
WE KNOW
T = 2 °C, S = 34.7 g/kg
FIND
density (ρ), freezing point (Tf)
Step 1 · Find density (ρ)
Use ρ = 1,027 × (1 − 0.00017 × (T − 10) + 0.00076 × (S − 35))
Warmer water is lighter and saltier water heavier, each by a fixed share per degree or per g/kg.
ρ
=
1,027
×
(1
−
0.00017
×
(2
−
10)
+
0.00076
×
(34.7
−
35))
ρ
=
1,027
×
(1
−
0.00017
×
(−8)
+
0.00076
×
(−0.3))
ρ
=
1,027
×
(1
−
(−0.00136)
+
(−0.000228))
ρ
=
1,027
×
1.0012
ρ
=
1,028.163
kg/
m
3
Step 2 · Find freezing point (Tf)
Use Tf = −0.054 × S
Salt lowers the freezing point by about 0.054 °C for each g/kg.
Tf = −0.054 × 34.7
Tf = −1.8738 °C
Check
Put the numbers back in. Both sides should match.
1,028.163 = 1,027 × (1 − 0.00017 × (2 − 10) + 0.00076 × (34.7 − 35)) ✓
−1.8738 = −0.054 × 34.7 ✓
Topic check
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