OCEANOGRAPHY 510 Problem Set 1 (2016) Solutions 1. The heat stored per unit area in a column of air or water with heat capacity C, thickness H, density ρ, and temperature T (in °K) is approximately ρCTH. This is only approximately true because ρ and C are likely to be functions of T; unless H is very thin, ρ and C will vary through the column. Nonetheless, we can make an assessment of the statement in this problem by estimating both the atmospheric and oceanic heat content. Let the subscripts o and a refer to the ocean and atmosphere. Then the heat stored in columns of ocean and atmosphere per unit area are (approximately) Ocean heat per unit area = ρoCoToHo = (103 kg m−3)(4×103 joules kg−1 °K−1)(3×102 °K)(1 m) = 1.2×109 joules/m2 Atmospheric heat per unit area = ρaCaTaHa −
−
−
= (0.5 kg m 3)(103 joules kg 1 °K 1)(2.5×102 °K)(1×104 m) = 1.3×109 joules/m2 Thus, to within the approximation that properties are constant in the atmosphere, this choice of parameters suggests that the upper 1 meter of the ocean stores the same amount of heat as the lower 10 km of the atmosphere. 2. The evaporation of water is a function of temperature; increasing the temperature of the atmosphere should increase the evaporation rate of water from the surface ocean. Between 0°C and 30° the latent heat required for evaporation decreases by about 15%, making evaporation easier at higher temperatures. Thus, if there is a general warming of the atmosphere, the effects of added evaporation will be larger away from the poles, even if the warming is greater at higher latitudes (because the poles will still be considerably colder than the tropics and subtropics for a long time after the onset of a generalized warming). The surface salinity of the ocean is closely tied to the evaporation rate, through the relation ∂S
∂t
t
 E(t) ⇒ S(t) − S(to ) 
∫ E(τ ) d τ
. to
For this reason, it can be expected that the tropical and subtropical (i.e., equatorward of about 30° of latitude) surface salinity of the ocean would increase (note that 50% of the surface area of the earth is contained between ± 30° of latitude). This added water vapor in the atmosphere at lower latitudes will be transported by the atmospheric circulation; on a globally-­‐averaged basis, the added water vapor will be transported poleward in both hemispheres. But since the saturation of the atmosphere is already relatively high over the oceans at higher latitudes, most of this added water vapor is likely to be lost back to the ocean through added precipitation. If this indeed occurs, the surface salinity at higher latitudes is likely to decrease in response to the general warming; if the warming causes additional melting of sea ice at high latitudes, then this will cause the surface salinity at higher latitudes to decrease even further. Also, note that since water vapor itself is a greenhouse gas, increasing concentrations of water vapor in the atmosphere at lower latitudes will have the effect of a positive feedback: the more warming, the more evaporation, and the more evaporation, the more warming. 3. (a) Zeroth order explanation. By Archimedes’ principle, the weight (i.e., mass) of water displaced by the ice (an upward force) is equal to the weight (i.e., mass) of the ice (a downward force). The density of the ice is somewhat less than the density of the water (roughly 0.9 g/cm3 for ice, compared to 1.0 g/cm3 for water). Thus, when the ice melts, the water that results from the melting (of the entire ice cube, both above and below the water line) just equals the volume of ice that was submerged. The volume of water added by the melting ice thus just equals the volume of water displaced by the ice, and (in general, at this level of complexity) there is no change in the level of water in the glass. More quantitatively, if ρi is the density of the ice and ρw is the density of the water in the glass before melting starts, then Archimedes’ principle states that ρ wVw g = ρ iVi g where Vi is the initial total volume of the ice, Vw is the initial volume of ice below the water line, and g is the acceleration of gravity. This says that the buoyancy force (weight) of the ice equals the weight of the displaced water. Assuming that the volume of the ice cube is known, the volume of water required to satisfy this equation can be found to be Vw = ( ρi/ρw)Vi. Since = ρi/ρw < 1, the volume of water formed by melting the ice will be less than the total volume of ice, just equaling the volume of the submerged ice. So, the water level won’t change. (b) First modification, due to temperature differences between the ice and the water. In reality, the melting ice will have the effect of lowering the temperature of the water in the glass, since the meltwater will have a temperature of near 0 °C, while the original water in the glass is 20 °C. We’ll assume that the glass of water is relatively large and the ice cube is relatively small, so that melting of the ice causes a small but measurable change in the temperature of the water in the glass. In this case the ideas stated above still hold, but the overall temperature of the water in the glass after melting is slightly lower than when the ice existed. Thus, since ∂ρ/∂T < 0 for temperatures above about 4 °C, the density of the water in the glass will increase after the melting takes place and the temperature is slightly lowered, and the water level in the glass will fall slightly. For a glass of water that is initially at 3 °C, the temperature will also be slightly lowered when the ice cube melts. However, at this temperature we know that ∂ρ/∂T > 0, so when the ice melts and the water is cooled slightly the density will decrease, and the water level will rise slightly. (c) Second modification, due to atmospheric pressure . The analyses in both (a) and (b) ignore the air in the space above the water. In reality, a glass of water in a room will have an atmosphere (air) above. There will be an additional Archimedes force on the ice that results from the presence of the air above. Let the total volume of the ice Vi be composed of the volume below the water (V′w) and the part above the water that is in contact with the air, Va. Thus, Vi=V′w+Va. The portion of the ice above the water will displace a volume of air Va , so that the three forces now acting on the ice are (i) downward gravity, (ii) the upward Archimedes force due to the displaced water, and (iii) the upward Archimedes force due to the displaced air. Forces (i) and (ii) are identical to their representations in part (a) above, and force (iii) can be computed knowing the volume of air displaced. Thus, we can write ρ iVi g = ρ wVw′g + ρ aVa g
⎛ ρi ⎞
⎛ρ ⎞
Vi − ⎜ a ⎟ Va < Vw .
⎟
⎝ρ ⎠
⎝ρ ⎠
⇒ Vw′ = ⎜
Using this result, we can use the relation between Va, Vi, and V′w to solve for Va to find that Va =
(ρ
(ρ
w
w
)V
+ρ )
w
− ρa
w
i
i
. Thus, the volume of water displaced by the ice is less than in the case with no atmosphere, so the ice actually rides higher in the water than in part (a). This is due to the fact that there is an additional upward Archimedes force due to the displaced air that was not present in part (a). As can be seen, this force will be a minimum (no extra displacement) when ρa = 0, and it will increase as ρa → ρw, because the Archimedes force will increase as the density of the air increases. Since the ice rides higher in the water when the atmosphere is considered, then it displaces less water than in the case with no atmosphere; correspondingly, the water level rises when the ice melts. 4. We can find the pressure by assuming that the ocean is in hydrostatic balance. The hydrostatic relation states that ∂p
∂z
= −gρ
, where p is the pressure, z is the depth, g is the acceleration due to gravity, and ρ is the density. For the ocean, we can assume that the density is approximately constant and integrate directly (assuming that g is a constant) to find that the ocean pressure poc at any depth z below the surface is poc (z) = − ρ gz + p(0) where p(0) is a constant of integration and represents the pressure at the sea surface. Taking this constant to be zero, then poc(z) = −ρgz . For the atmosphere, the density is not constant, and we must relate density to pressure in order to proceed. To do this, we note that a good, approximate equation of state for the atmosphere is given by the Ideal Gas Law, p = ρ RT
ρ=
⇒
p
RT
, where T is the Kelvin temperature and R is the gas constant. Putting this relation for ρ into the hydrostatic relation, it is found that ∂ pat
p
= −g ρ = −g at
∂z
RT
⇒
pat (z) = pat0e
−
gz
RT
, where pat0 is a reference pressure (a constant of integration) at z = 0, the bottom of the atmosphere. We can evaluate each of these pressure functions at a distance of 5 km below the reference, since in both cases the pressure will increase downwards. For the ocean, at a depth of 5000 m it is found that poc = −(1.027×103 kg/m3)(9.8 m/sec2)(−5×103 m) = 5.032×107 kg m−2sec−2 = 5.032×107 newton/m2 = 5.032×107 pascals = 5.032×102 bar = 5.032×103 decibars. For the atmosphere, we can take pat to be 500 millibars at a height of 5 km, a value typical of Earth’s atmosphere; we note that 1 millibar = 10−3 bar = 10−2 decibar = 102 kg m−1sec−2, so that pat0 = pat (z)e
+
gz
RT
, and pat0 = (5×102×102 kg m−1sec−2) exp[(9.8 m sec−2)(5×103 m)/((2.87×102 m2sec−2 °K−1)(2.78×102°K))] = (5×104 kg m−1sec−2)(1.84) = 9.24×104 kg m−1sec−2 = 9.24×102 millibars = 9.2 decibars . Thus, the hydrostatic pressure at the bottom of a 5 km ocean is over 500 times greater than the pressure at the bottom of a 5 km column of Earth’s atmosphere. 5. Near the Equator there is often a change in the air temperature (at sea level) of 2°C, perhaps more, over the course of a day, with nighttime temperatures lower. This is due mostly to the solar heating cycle, although it is possible that there is also a very weak tidal component to this change. Since the near-­‐surface waters of the ocean are in contact with an atmosphere whose temperature varies on a regular cycle, it is not surprising that the upper 1-­‐2 meters of the ocean also show such a cycle. There is probably always enough wind to keep the upper few meters of the ocean well-­‐mixed in these regions. It is not obvious, however, that there should be a diurnal cycle in salinity, since there is no obvious forcing function on diurnal cycles that affects salinity. Precipitation at low latitudes, while regular, is hardly cyclic. While the evaporation rate and relative humidity are functions of temperature and could possibly affect the surface salinity through the diurnal heating cycle, these temperature dependencies are quite weak for the magnitude of temperature change involved here, and it seems unlikely that they could be linked to measurable diurnal variability in salinity. It does not seem surprising that there is a weak or nonexistent diurnal cycle of salinity at most low-­‐
latitude locations. 6. (a) At the present time, the trend in large-­‐scale imbalance in incoming and outgoing radiation as estimated by the Argo array is about 0.58 ± 0.38 watts/m2 (in 2012) and 0.71±0.10 watts/m2 (in 2016). These estimates are not inconsistent with results from climate models and reanalysis products. Over 90% of the imbalance in heat (uptake) is stored in the ocean, with only a small amount appearing elsewhere (atmosphere and land). (b) Because most of the excess heat appears in the ocean, it is the best place to try to estimate the size of the excess. 7. For simplicity, assume that the planet is not rotating. This will not qualitatively affect the result. The weight of the atmosphere is approximately its mass times the acceleration due to gravity (i.e., the force of attraction). However, force of gravity is a function of the distance from the center of the planet. Additionally, the mass at a given radius is also a function of the distance from the center. Let g(r) be the gravity as a function of radius. At any given radius above the bottom of the ocean the area of the shell at this radius is given by 4πr2; multiplying this by the vertical distance element dr yields the quantity of mass contained between any distance r and r+dr to be 4ρπr2dr. The weight of this shell is then 4ρg(r)πr2dr. Integrating this over the entire ocean, and dividing by the surface area of the planet under the ocean (4πR2), it is found that Weight per unit area = WUA =
ρ
4π R 2
R+ H
∫
4π r 2 g(r) dr =
R
ρ
R2
R+ H
∫
r 2 g(r) dr . R
The hydrostatic relation and pressure p(r) are related through the relation R
∂p
= − g (r ) ρ ⇒ p(r ) = − ρ ∫ g (r )dr = ρ
∂r
R+ H
R+ H
∫
g (r )dr
. R
The quantites WUA and p can be compared at depth H in the ocean as R+ H
WUA − p(H ) = ρ
∫
R
R+ H
R+ H
⎛ r2 ⎞
⎛ r2
⎞
g(r)
dr
−
ρ
g(r)
dr
=
ρ
∫R
∫R ⎜⎝ R2 − 1⎟⎠ g ( r ) dr .
⎜⎝ R 2 ⎟⎠
In order to examine this difference, we can examine the quantity 2
2
⎛ r2
⎞ ⎛ R +δr ⎞
⎛ δr ⎞
Δ = ⎜ 2 − 1⎟ = ⎜
⎟ − 1 = ⎜1 + ⎟ − 1
R⎠
⎝
⎝R
⎠ ⎝ R ⎠
, where it has been noted that r = R + δR, that is, the radius of a point inside the ocean is given by the radius of the earth plus the distance above the seafloor. Most of the contributions to the integral in the comparison above will come from radii near r = R+H . In this case, δR ~ H. The limiting cases for Δ then become 2
δR
δR
⎛ δR⎞
(i) Δ = ⎜1 +
−1 = 2
<< 1 for H << R ⎟ −1 : 1+ 2
R ⎠
R
R
⎝
2
⎛ δR⎞
(ii) Δ = ⎜ 1 +
⎟ − 1 : O (1) for H : R . R ⎠
⎝
Thus, in case (i), when the depth of the ocean is much less than the radius of the planet, the difference between WUA and p at the seafloor is negligible. In case (ii), where the depth of the ocean is comparable to the radius of the earth, WUA exceeds p. This difference can be explained by the presence of spherical geometry. In case (i), the ocean is so thin that locally it looks like a planar layer overlaying the solid earth, and the hydrostatic pressure is just the weight of the ocean. In case (ii), the ocean and earth are of comparable thickness, and spherical geometry must be considered. In this case, a “column” of ocean is not a cylinder, but is instead a more complicated shape similar to the frustum of a cone; because of this, the area of the column is a function of height above the solid earth, and there are outward components of the “weight” as well as components towards the center of the earth.