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Surface area problem

Question:

Assuming that 68.6% of the earth's surface is covered with water at an average depth of 1.01 miles, estimate the mass of the water on earth in kg.

Given the following:

Density of water = 1000 kg/m^3
1 mile = 1.6093 km
1.01 miles = 1.6254 km

Total surface area of earth = 509,600,000 km^2. Area of water = 509,600,00 * 68.6% = 349,585,600 km^2. The volume of water is 349,585,600 km^2 * 1.6254 km = 568,213,987 km^3. Divide by 1000 to get m^3 = 568213.987m^3.

Then, 568213.987m^3 * 1000kg/m^3 = 5.68213987E8 kg. This was wrong by my guess. Can you help me figure out what went wrong?

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Question:

Assuming that 68.6% of the earth's surface is covered with water at an average depth of 1.01 miles, estimate the mass of the water on earth in ...

Solution Summary

This solution provides the equations answering a surface area problem.

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