MSCI 304, Marine Geelogy Name I have followed the academic honesty policy of thi

Question

MSCI 304, Marine Geelogy Name I have followed the academic honesty policy of this course. (your signature) Many factors affect the amount of sea level rise, such as ridge spreading rate, ice volume, or thermal expansion. In this assignment, we will examine the importance of these variables to contributing to sea-level rise during the assume an ocean configuration with square sides. Do not round answers before the final calculation of the it was warmer, no ice existed, and ocean basins were spreading. For the calculations, we will question (e & part "c" in question 1). SHOW ALL YOUR WORK current ocean volume-1 370 x 10, km, current average ocean depth-380 r on 1. During the Cretaceous, the water is predicted to have been up to 15°C warmer. If the density of 10 water is 1.028053e/cm, and of 16-C water is 1.02591 g/cm?, what is the predicted geneal rise in ea level? (Hint: density s mass/volume; assume the density change from 1'C to 16 C represents the 15°C temperature change) a. Calculate the volume of the Cretaceous Occan Calculate the mass of the ourrent ocean using the density = 1.37 x 10" krm, x ( 1.028053 g/cm?"( 100cm/1m)suocon/1km)-1.40843 x 10" g Cakulate the volume of the Cretaceous ocean assuming the new density 1.40843 x 1024 g / ( 1.02591 g/crn®(100cm/1m)". ( 1000m/1km)-1372861 x 10, km What is the depth of the Cretaccous ocean (assume area of oceans has not changed)? Caloulate the depth of the Cretaceous ocean (divide by modern area (volume/depthj that b. is assumed to not changel. = (1.372861 x 10° km? / (1.370.10, kml/38 km)-3.807937733 km So, how much higher was Cretaceous sea level (in m) due to thermal expansion? Cakculate the change in depth c. (3.80793 7733-3.8)" 1000 . 7.94 m Anterctica& Greenland area 1.50 x 10kmIce Sheet Thickness 1.45 km 2. If during the Cretaceous, glacial ice is absent from these areas, what is the yertical rise in sea level (in m) due to addition of water from melting of these ice sheets? (ignore other factors like the density change as was calculated in question 1).

Explanation / Answer

mass (1 0c) = mass ( 160c)

mass (10 c) = (1.370 x 109 km3) x ( 1.028053 g/ cm3) x(100cm/1m)3(1000m/1km)3 [ volume of present ocean x density at 10c) x conversion factors]

= 1.408432 x 1024g

= mass of ocean at 160c

volume at 160c= mass at 160c / density at 160c

= 1.408432 x1024 / 1.02591 (g/ g/cm3)

= 1.372861 x 109 km3 = volume of Cretacoeus ocean

b)

depth of cretaceous ocean = volume of cretaceous ocean / area of ocean basin [ assuming area of ocean basin is constant through time]

area of ocean basin = current ocean volume / current average ocean depth

hence depth of cretaceous ocean = 1.372861 x 109 km3 / ( 1.370 x 109 / 3.8) km

= 3.80793773 km

c)

change in sea level = sea level at cretaceous - present sea level

= (3.80793773 - 3.8) km

=   (3.80793773 - 3.8) x 103 m

= 7.93773 m

2 .

assuming a uniformly thick ice cover over antartica and green land and that all of it have melted.

the volume of water locked in ice cap = area of antartica and green land x depth of ice

= 1. 50 x 107 x 1.45

= 2.175 x 107 km3 { I assume the value of area is in power of 7 , the image is not clear }

this would be an extra volume in the volum eof cretaceous ocean

hence the effective volum eof cretaceous ocean = 1. 372861 x 109 km3 + 2.175 x 107 km3

= 1.394611 x 109 km3

now this would be the cretaceous ocean volume and the corresponding change in sea level has to be measure like in question 1.c

i.e depth of cretaceous ocean = 1.394611 x 109 km3 / ( 1.370 x 109 / 3.8) km

= 3.868264 km

now calculating the change in sea level

vertical rise in sea level = 68. 26408m

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