OC210: Answers to Questions at the end of Lecture Notes for Topic 1A (Fall 2009)
QUESTIONS
1. Which of the three factors (temperature, salinity or pressure) primarily controls the density of surface
seawater? Which of these three factors primarily controls the in-situ density of deep seawater?
Temperature in surface ocean.
Pressure in deep sea.
2. Calculate the volume change that occurs when a seawater parcel is cooled from 20° to 0°C and the
density has increased from 1024 to 1027 kg/m3 . Assume the initial mass of seawater was 1000.0 kg and
the salinity was kept constant. What percent change in volume does this represent?
-since the initial density (  o) = Mo/Vo, then the initial volume (Vo) = Mo/  o
thus Vo= 1000 kg / 1024 kg/m3 = 0.9766 m3
-since the mass doesn’t change then the final volume (Vt) = Mo/  t
thus Vt = 1000kg/ 1027 kg/m3 = 0.9737 m3
-thus the change in volume is Vt- Vo = 0.9766 m3 - 0.9737m3 = 0.0029 m3
-the percent change = (Vt-Vo)/Vo * 100 = 0.0029m3 /0.9766m3 *100 = 0.30%
3. What would be the approximate temperature, at the sea surface, of a parcel of seawater that was raised
from 2000m and had an in-situ temperature of 4.00°C? What would be its approximate potential
temperature at 2000m?
Since the rate of change of temperature with increasing depth (a result of increasing pressure) is about
+0.1 ºC/1000m, then by raising the parcel of water by 2000m, the in-situ temperature should decrease by
about 0.2 ºC. It’s potential temperature at 2000m would be about 3.8 ºC.
4. Where does the range of surface and deep (>2000m) water plot in T-S coordinates? What is the
corresponding range in potential density? See Fig. 13.
-surface water has T range of –1 to 30 ºC and S range of 28 to 37
-deep water has T range of –1 to 4 ºC and S range of 34.4 to 35.
-the corresponding density range is about 1022 to 1028 kg/m3 and thus the range in σ = 22 to 28.
5. Calculate the stability of the water column at 10ºN in the Pacific between 200 and 500m, 500 and
1500m and 3500 and 4500m based on information in Fig 26.
Stability is the rate of change of density versus depth. It is expressed mathematically as Δρ/Δz. Between
200m and 500m the density change at the equator is from about σ = 26.5 to 26.9 estimated from Fig. 26.
This means that the stability is Δρ/Δz = (1026.5 – 1026.9 kg/m3)/ (200-500m) = 1.33x10-3 kg/m4.
Between 500 and 1500, Δρ/Δz = (1026.9 – 1027.6 kg/m3)/ (500-1500m) = 0.7x10-3 kg/m4. Between 3500
and 4500, Δρ/Δz = (1027.77 – 1027.80 kg/m3)/ (3500-4500m) = 0.03x10-3 kg/m4. Thus you see that
stability is decreasing with increasing depth in the ocean. (Your results may differ slightly from mine
depending on your reading of the data in Fig. 26.)
6. Why does  equal σ at the surface of the ocean?
Because there are no pressure effects on density at the surface of the ocean. At any depth > 0, however,
 will be greater than  .
7. Explain why pressure effects on density do not contribute to stability.
Since parcels of water that change depth by the same amount experience the same density change due to
pressure, this pressure induced density change has no effect on stability. Using the hypothetical beaker
experiment discussed in the notes, where the water in the beaker has the same T and S, there is no
stability in the water. There is no buoyancy resistance to mixing, which is another way of thinking about
stability. If you displace a parcel of water in this beaker to deeper depth it will have the same in-situ
density as its surrounding water parcels and thus this increase in in-situ density contributes zero stability.