Earth Science Quiz: Oceanographic Data

What this quiz covers

This quiz focuses on Oceanographic Data, giving you a quick way to practice the rules, question types, and explanations that matter most for Earth Science.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. -1.50 °C/degree latitude
2. +0.58 °C/degree latitude
3. -0.45 °C/degree latitude (correct answer)
4. -2.20 °C/degree latitude

Explanation: First, read the SST values from the graph at the specified latitudes. At the Equator (0°), the SST is approximately 28°C. At 40°N, the SST is approximately 10°C. The change in temperature is ΔT = 10°C - 28°C = -18°C. The change in latitude is ΔLat = 40°N - 0° = 40°. The rate of change is ΔT / ΔLat = -18°C / 40° = -0.45 °C/degree latitude. The negative sign indicates temperature decreases as one moves north from the equator.

Question 2

1. 1.5 meters
2. 2.0 meters
3. 3.0 meters
4. 3.5 meters (correct answer)

Explanation: First, identify the pattern. This is a semidiurnal tide, with two high and two low tides per day. The question assumes the graph shows a spring tide, which is characterized by the maximum tidal range. The tidal range is the vertical distance between a high tide and the subsequent low tide. On the first day (0 to 24 hours), the first high tide is at ~+2.0 m. The following low tide is at ~-1.5 m. The tidal range is the difference between these two levels: 2.0 m - (-1.5 m) = 3.5 m.

Question 3

1. 0.02 psu
2. 1.5 psu
3. 3.0 psu (correct answer)
4. 4.5 psu

Explanation: The graph shows the salinity gradient, which is the change in salinity per unit depth (psu/m). The y-axis value is constant at 0.02 psu/m throughout the halocline. To find the total change in salinity, multiply the gradient by the thickness of the halocline. The thickness (Δz) is 200 m - 50 m = 150 m. The total salinity increase (ΔS) is Gradient × Δz = (0.02 psu/m) × (150 m) = 3.0 psu.

Question 4

1. A decrease in water temperature, which increases the solubility of oxygen gas.
2. A major phytoplankton bloom, where photosynthetic activity produces large amounts of oxygen. (correct answer)
3. Increased wind and wave action, which mixes atmospheric oxygen into the water column.
4. The decomposition of organic matter from the previous season, which releases stored oxygen.

Explanation: The graph shows a strong correlation between the peak in chlorophyll-a (an indicator of phytoplankton biomass) and the peak in dissolved oxygen. Photosynthesis, carried out by phytoplankton, consumes carbon dioxide and produces oxygen. The large spike in chlorophyll-a in April-May signifies a spring bloom, and the corresponding spike in dissolved oxygen is a direct result of this intense primary production.

Question 5

1. The MLD would deepen and the thermocline would become weaker.
2. The MLD would become shallower and the thermocline would become stronger (sharper gradient). (correct answer)
3. The MLD would remain constant, but the thermocline would shift to a greater depth.
4. The MLD would become shallower, but the thermocline's temperature gradient would remain the same.

Explanation: Seasonal warming increases the temperature of the surface water. This increases the temperature difference between the surface and the deep water, making the thermocline stronger (a sharper gradient). A stronger thermocline creates greater density stratification, which inhibits vertical mixing. As a result, wind energy can only mix the water to a shallower depth, causing the mixed layer depth (MLD) to become shallower.

Question 6

1. A warm-core eddy is moving onshore, bringing warm, nutrient-poor water to the coast.
2. A period of intense coastal upwelling, where wind-driven currents bring cold, nutrient-rich deep water to the surface. (correct answer)
3. A large river discharge event, which lowers coastal salinity and introduces terrestrial nutrients, causing a phytoplankton bloom.
4. Seasonal surface warming, which is increasing the metabolic rate of phytoplankton and causing them to flourish.

Explanation: The graphs show a strong inverse correlation: as SST near the coast drops significantly (from 14°C to 10°C), the chlorophyll-a concentration increases dramatically. This is the classic signature of coastal upwelling. Winds parallel to the coast drive surface water offshore (due to Ekman transport), and it is replaced by cold water from below. This deep water is rich in nutrients (like nitrates and phosphates), which fertilize the phytoplankton in the sunlit surface waters, causing a massive bloom and thus a high chlorophyll-a concentration.

Question 7

1. 6 seconds
2. 9 seconds
3. 11 seconds (correct answer)
4. 14 seconds

Explanation: First, confirm the wave type. The depth is 1500 m. The condition for a deep-water wave is Depth > Wavelength/2. Here, 1500 m > 200 m / 2, or 1500 m > 100 m. The condition is met, so we use the 'Deep-Water Waves' curve. Second, find the wavelength of 200 m on the x-axis. Trace vertically up to the 'Deep-Water Waves' curve. Third, trace horizontally to the left to read the corresponding value on the y-axis for Wave Period. The value is approximately 11 seconds.

Question 8

1. The second sample is approximately 1.5 kg/m³ denser than the first.
2. The first sample is approximately 2.0 kg/m³ denser than the second. (correct answer)
3. The two samples have nearly identical densities.
4. The first sample is approximately 1.0 kg/m³ denser than the second.

Explanation: To solve this, find the density of each sample from the contour plot. For Sample 1: locate 5°C on the y-axis and 34.5 psu on the x-axis. The point falls on the contour line labeled 1027.5 kg/m³. For Sample 2: locate 15°C on the y-axis and 35.5 psu on the x-axis. This point falls on the contour line labeled 1025.5 kg/m³. The difference is 1027.5 kg/m³ - 1025.5 kg/m³ = 2.0 kg/m³. The first sample (colder, slightly less salty) is denser than the second sample (warmer, slightly saltier).