First Problem

Calculate the difference between the two average crustal elevations

Second Problem

Calculate a sea floor spreading rate

Third Problem

Calculate the average sea floor subsidence rate in cm/1000 yr

Fourth Problem

Calculate the deposition rate of marine sediments in the deep-ocean basin in cm/1000 yr

Fifth Problem

Calculate the difference between the two average crustal elevations.

See the Math Problem .mov to learn how to solve the problem. The file will open in a new window. After it loads, click on the image to move to the next slide.

Calculate the difference between the average elevation of continental crust (840 m above sea level) and the average elevation of oceanic crust (3800 m below sea level). Note that elevations above sea level are positive and elevations below sea level are negative.

840 m |

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Calculate a sea floor spreading rate.

See the Math Problem .mov to learn how to solve the problem. The file will open in a new window. After it loads, click on the image to move to the next slide.

To calculate the rate of sea floor spreading for a particular lithospheric plate, the distance between two points and the difference in the ages of those two points is required. A rate is determined by the distance divided by the time: distance/time or, simply, d/t.

Rate = d/t

For example, the distance between stations A and B is 1,080 km. The difference between the ages of the rocks at station A and station B is 54,000,000 years. Calculate the average rate of sea floor spreading.

d/t = 1080 km/54,000,000 yr = 0.00002 km/yr

To better visualize this rate, it should be converted to cm/yr. First convert kilometers to meters, then meters to centimeters.

1080 km * 1000m/1 km = 1,080,000 m

1,080,000 m * 100 cm/1 m = 108,000,000 cm

Finally,

d/t = 108,000,000 cm/54,000,000 yr = 2 cm/yr

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Calculate the average the sea floor subsidence rate in cm/1000 yr.

The average increase in depth:

- 6 km - 2 km = 4 km

Therefore, the average rate of subsidence = 4 km/100,000,000 yr.

To convert the units to centimeters, change kilometers to meters and, then, meters to centimeters.

4 km/100,000,000 yr * 1000m/1 km *100 cm/1 m = 400,000 cm/100,000,000 yr = 4 cm/1000 yr.

Scientist like to use distances that are easy to visualize, so give the units in cm/1000 yrs:

400,000 cm/100,000,000 yr = 4 cm/1000 yr.

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Calculate the deposition rate of marine sediments in the deep-ocean basin.

To calculate the deposition rate of sediment for a region of sea floor, divide the total thickness of the sediment by the time required for the sediment to accumulate (i.e. the age of the sea floor). For example, if 842 m of sediment was deposited in 80,000,000 yr:

amt of sed/t = 842 m/80,000,000 yr * 100 cm/ 1 m = 0.001 cm/yr.

To better visualize this rate, multiple the annual rate by 1000 to yield 1 cm/1000 yr.

0.001 cm/yr * 1000 = 1 cm/1000 yr

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Calculate the sinking rate of sand in cm/s.

To calculate the sinking rate of a sedimentary particle, divide the distance that the particle sinks by the time required for the particle to sink to that depth. For example, if it takes 1.8 days for a sand particle to sink 4 km:

d/t = 4 km/1.8 days

To better visualize this rate, it should be converted to cm/s (centimeters per second). First convert kilometers to meters, then meters to centimeters:

4 km * 1000m/1 km = 4,000 m

4,000 m * 100 cm/1 m = 400,000 cm

Then, convert the days to hours, then the hours to minutes, and, lastly, the minutes to seconds:

1.8 days * 24 hr/day * 60 min/hr * 60 s/min = 155,520 s

Finally, divide the distance in centimeters by the time in seconds:

d/t = 400,000 cm/155,520 s = 2.6 cm/s

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Calculate the time required for the ocean to evaporate completely, if the evaporated water did not return to the ocean.

To calculate the time, divide the average depth of the ocean, 3800 m, by the average rate of evaporation, 1 m/yr:

rate = d/t

d/rate = t

3800 m/(1 m/yr) = 3800 yr

You will be asked the question only, so you must remember the average depth of the ocean, 3800 m, and the average evaporation rate, 1 m/yr.

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Calculate how much sea level could rise if the polar ice caps melt because of global warming

This question would be asked without providing values or formulas. To obtain a rough estimate of potential sea level rise, you must know that 97% of the water on Earth is in the ocean and 2% is in the polar ice caps. Given that the average depth of the ocean is 3800 m, you can calculate 2% of 3800 m:

3800 m x 0.02 = 76 m

This is a very rough estimate; however it shows you that sea level can rise well over 100 feet.

To receive full credit for this math problem, you must write all of the values in this problem, as well as solve the equation.

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Calculate the residence time of a seawater constituent

To calculate the residence time of a seawater constituent, divide the total amount of a constituent in the ocean by its input rate or its output rate. At steady state, the input rate equals the output rate.

Residence time = amount of a constituent in the ocean/input rate (or the output rate)

Calculate the resident time of Na in the ocean given that the total amount
of Na is 6.08 x 10^{20} mol and the input rate of Na by rivers is 1.18
x 10^{13} mol/yr.

Residence time of Na = 6.08 x 10^{20} mol Na/1.18 x 10^{13} mol Na/yr = 5.15 x 10^{7} yr = 51,500,000 yr

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Determine if a chemical equation is balanced

Matter cannot be created nor destroyed, so all matter must be accounted for. For example, for an equation to be balanced, both sides (reactants and products) must have the same number of constituents.

Example A

CO_{2} + H_{2}O = CH_{2}O + O_{2}

Reactants C - 1, O - 3, H - 2

Products C - 1, O - 3, H - 2

Equation is balanced.

Example B

CO_{2} + H_{2}O = CH_{2}O

Reactants C - 1, O - 3, H - 2

Products C - 1, O - 1, H - 2

Equation is not balanced.

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Calculate the residence time of water in the atmosphere

To calculate the residence time of a constituent in a reservoir (part of a system), divide the total amount of the constituent in the reservoir by its input rate or its output rate. At steady state, the input rate equals the output rate.

The residence time of water in the atmosphere = amount of water in the atmosphere/the input rate of water to the atmosphere (evaporation)

Calculate the residence time of water in the atmosphere given that the total
amount of water in the atmosphere is 1.3 x 10^{15} kg and the input
rate of water by evaporation is 4.0x 10^{16} kg/yr.

Residence time of water in atm. = 1.3 x 10^{15} kg/4.0x 10^{16} kg/yr
= 0.0325 yr

Because most people do not know how long 0.0325 years is, you convert the value to days: 1 year = 365 days.

0.0325 yr x 365 days/yr = 12 days

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Convert a surface current flow rate from m/s to km/yr

Given a surface current flow rate of 1 m/s, convert this rate to km/yr. You must convert the meters => kilometers and the seconds => minutes => hours => days => years. The conversions can be done together or separately.

Together:

1 m/s x 1 km/1000 m x 60 s/min x 60 min/hr x 24 hr/d x 365 d/yr = 31,536 km/yr

Separately:

1 m/s x 1 km/1000 m = 0.001 km

1 s x 1 min/60 s x 1 hr/60 min x 1 d/24 hr x 1 yr/365 = 3.171 x 10^{-8} yr

0.001 km/3.171 x 10^{-8} yr = 32,000 km/yr

Note that the math seems easier when the conversions are done together.

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Calculate the time required in years for water in the deep zone to rise to the surface

Very cold, dense water sinks from the surface primarily in the subpolar regions of the North and South Atlantic, then flows into the Indian, and Pacific. Everywhere deep water rises back to the surface. The rate at which water rises is approximately 1 cm/d. Given the average depth of the ocean, 3800 m, how long would it take water rising at a rate of 1 cm/d to return to the surface?

First convert 3800 m to centimeters:

3800 m x 100 cm/m = 380,000 cm

Calculate how many days would be required for water to rise 380,000 cm:

rate = d/t

d/rate = t

380,000 cm/1 d/cm = 380,000 d

Convert 380,000 days to years, so that the time is easier to visualize:

380,000 d x 1 yr/365 d = 1,041 yr

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Calculate the percentages of the water column for the euphotic zone and the photic zone.

It is interesting to calculate these percentages for both the average depth of the ocean, 3800 m, and the depth of the water column overlying the deep abyssal plains, 6000 m. The average depth in clear water for the euphotic zone is 70 m and the photic zone is 600 m.

*Euphotic Zone*

70 m/3800 m x 100% = 1.8%

70 m/6000 m x 100% = 1.2%

*Photic Zone*

600 m/3800 m x 100% = 16%

600 m/6000 m x 100% = 10%

*To receive credit for the problem in class and on the quiz, you must show
your work.*

Calculate the biomass in a trophic level.

To determine how much biomass is maintained in a given trophic level, you need the ecological transfer efficiency and the biomass in the first trophic level.

First trophic level biomass = 10,000 kg.

Ecological transfer efficiency = 12%

*Second Trophic Level Biomass*

x = ETE x biomass in the first trophic level

x = 0.12 x 10,000 kg

x = 1,200 kg

*Third Trophic Level Biomass*

x = ETE x biomass in the second trophic level

x = 0.12 x 1,200 kg

x = 144 kg

*Fourth Trophic Level Biomass*

x = ETE x biomass in the third trophic level

x = 0.12 x 144 kg

x = 17 kg

*To receive credit for the problem in class and on the quiz, you must show
your work. *

Convert the rate of a rip current that flows at 5 mi/hr into m/s

A mile is equal to 1.6 km.

5 mi/hr x 1.6 km/mi x 1000 m/km x 1 hr/60 min x 1 min/ 60 s = 2.2 m/s

*To receive credit for the problem in class and on the quiz, you must show
your work.*

How long would it take a tsunami wave traveling at 700 km/hr to reach Hawaii from the subduction zones near California?

If California is 2400 miles from Hawaii, how long would it take a fast tsunami wave to reach Hawaii?

First convert the miles to kilometers, 1 mi = 1.6 km:

2400 mi x 1.6 km/mi = 3840 km.

Now you know the rate at which a tsunami wave can travel, 700 km/hr, and the distance in kilometers to California, 3840 km. Rearrange the distance/time = rate equation to solve for time:

d/t = rate

d/rate = t

3840 km/700 km/hr = 5.5 hr

*To receive credit for the problem in class and on the quiz, you must show
your work. *