2. Between 3.8 and 4.0 billion years ago, the Earth received a number of direct
ID: 287458 • Letter: 2
Question
2. Between 3.8 and 4.0 billion years ago, the Earth received a number of direct impacts by large asteroids having a diameter of about 100 miles and a mass (m) of about 1019 kg. Given that these large impactors would strike the Earth with a velocity (v) of about 40,000 mph or 18 km per second, we can calculate the energy (E) released during a collision: E-mv2-(10 19 kg) (18,000 m.s-1): 3.2×1027 kgm,2s-2-3.2×1027 J cal (3.2x1027 J) 14.1857) = 7.6x1026 cal This amount of energy is staggering. The most powerful thermonuclear bomb ever detonated released 5x10- cal (this bomb was tested by the USSR in 1961, and is rated about 3000 times more powerful than the atomic bomb dropped on Hiroshima during World War II). Thus, the energy released by the asteroid impact was equivalent to detonating about 15 billion of the most powerful nuclear bombs (a) It is estimated that about 25% of the impact energy would go into heating and evaporating seawater. Estimate the volume of water that would be vaporized as a result of the collision event. Assume that the ocean has a volume of 1.4×1021 L and an average temperature of 3°C. Also, assume that seawater has a heat capacity of I calgl.C-1 and a latent heat of vaporization of 540 calgi. The average density of seawater is 1.025 g/mL (b) Assume that the ocean basins can be represented by a rectangular tank with an average depth of 3800 m. Calculate the drop in sea level resulting from the vaporization of water caused by the impactExplanation / Answer
(a)
Volume of water in ocean = 1.4 x 1021 L or in milliliter = 1.4 x 1024
Mass of water = density x volume = 1.025 x 1.4 x 1024 = 1.435 x 1024 grams
Heat energy required to make water to 1000 C = 97 x 1.435 x 1024 (Water is currently at 30 C)
= 1.39195 x 1026 calorie
Heat energy required to boil water = 540 x 1.435 x 1024 = 7.749 x 1026 calorie
Total energy required to boil all the water of ocean= 9.14095 x 1026 calorie
Energy released during collision = 7.6 x 1026 calorie
25% of this is = 1.9 x 1026 calorie
9.14095 x 1026 calorie can boil 1.435 x 1024 grams of water
therefore, 1.9 x 1026 calorie will boil = [(1.435 x 1024)/(9.14095 x 1026)]x 1.9 x 1026
= 0.156985 x 10-2 x1.9 x 1026 = 0.298273155 x 1024 grams of water
Answer: collission will boil 0.298273155 x 1024 grams of ocean water
(b)
volume = area x depth = 1.4 x 1021
= 1.4 x 1018 x (1000liter) = area x 3800 m (As 1 m3 is equivalent to 1000 liters)
area = 3.684210526 x 1014 m2
collission will boil 0.298273155 x 1024 grams of ocean water or
(0.298273155 x 1024/ 1.025) = 2.91004053 x 1023 millileter or 2.91004053 x 1020 liter
This is equivalent to a height decrease of 2.91004053 x 1017 x (1000liter) = 3.684210526 x 1014 m2 x reduction in depth in meter
Therefore, reduction in depth in meter = 789.868145
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