[2 points] (b) The question above relates to why, as far as sea level rise becau

Question

[2 points] (b) The question above relates to why, as far as sea level rise because of global warming is concerned, we're a lot more worried about ice that is currently on land, as opposed to ice that is already floating on water (such as the ice in the Arctic). The ice sheet on Greenland has a volume of approximately 3.0 million cubic km. The surface area of Earth is about 5.0 x 1014 m, and 70% of the surface is covered by oceans. Use these numbers to estimate the average amount of sea level rise that would happen if all the ice in Greenland melted and ran off into the ocean. (You can assume that the ice, after it melts, has a volume of 90% of what it had in solid form.) [2 points] (c) Assume the initial temperature of the Greenland ice is Tice-20 °C. How much energy is required to melt all the Greenland ice? Use a specific heat of 2 kJ/(kg C) for ice and approximate the latent heat of fusion as 300 kJ/(kg). [2 points] (d) During the summer, sunlight and warm air transfers energy to the Greenland ice sheet at the rate of approximately 2 x 1014 w. (Assume this happens for three months per year.) If this energy warms and melts the surface ice, what is the sea-level rise from one summer of melting? Assume all the melted ice flows into the ocean (fortunately, that's not what happens).

Explanation / Answer

Q2. b. Volume of ice = V= 3000000 cubic km = 3000000 x 109 cubic m

Volume of ice after it melts = 90 % of solid ice volume

= 90 % x 3000000 x 109 cubic m

= 2700000 x 109 cubic m

= 27 x 1014 cubic m

Surface area of earth = 5 x 1014 m2

Surface area of the ocean = 70% of earth surface area

= 70% x 5 x 1014 m2

= 3.5 x 1014 m2

so the average amount of sea height = 27 x 1014 cubic m / 3.5 x 1014 m2 = 7.714 m

c. Energy required to melt ice = energy required from to ice at -20 C to ice at 0 C + energy required from ice at 0 C to water at 0C

= mass of ice x specific heat of ice x (0-(-20)) + mass of ice x latent heat of fusion

= 3000000 x 109 x density x 2 x 20 + 3000000 x 109 x density x 300

= 3000000 x 109 x 917 x 2 x 20 + 3000000 x 109 x 917 x 300 kJ ( density of ice 917 kg/m3)

=1.1 x 1020 + 8.253 x 1020 kJ

= 9.353 x 1020 kJ = 9353 x 1020 J

d. heat supplied / s = 2 x 1014 J

Total time of one summer = 90 x 24 x 3600 s

Total heat supplied in one summer = heat supplied / s x Total time of one summer

= 2 x 1014 x 90 x 24 x 3600

=15.552 x 1020 J

So the % of total heat supplied in one summer = (15.552 x 1020 J/ 9353 x 1020 J ) x 100 =0.1662 %

so the sea level rise from one summer of melting = 7.714 x 0.1662 % = 0.0128 m

all the best in the course work

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