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Chapter 1: Problem 76

The surface area and average depth of the Pacific Ocean are \(1.8 \times 10^{8} \mathrm{~km}^{2}\) and \(3.9 \times 10^{3} \mathrm{~m},\) respectively. Calculate the volume of water in the ocean in liters.

Short Answer

Expert verified
The volume of water in the Pacific Ocean is approximately \(7.02 \times 10^{20}\) liters.

Step by step solution

01

Convert area from square km to square meters

The area of the Pacific Ocean is given in \(\mathrm{km}^{2}\). We should convert it to \(\mathrm{m}^{2}\) as the given depth is in meters and we are asked to find volume in liters, not cubic kilometers. There are \(1 \times 10^{6}\) \(\mathrm{m}^{2}\) in \(\mathrm{~km}^{2}\). So, \(1.8 \times 10^{8}~\mathrm{km}^{2}\) equals to \(1.8 \times 10^{8} \times 10^{6} = 1.8 \times 10^{14}\) \(\mathrm{m}^{2}\).
02

Find Volume

Volume is calculated by multiplying area and depth. Now we have area \(\= 1.8 \times 10^{14} ~\mathrm{m}^{2}\) and depth \(= 3.9 \times 10^{3}~\mathrm{m}\). Multiply these two to get \(7.02 \times 10^{17} ~\mathrm{m}^{3}\) as the volume of Pacific Ocean.
03

Convert volume to liters

Since 1 cubic meter \(\mathrm{m}^{3}\) equals 1000 liters, we should multiply our result gained in cubic meters with 1000 to convert it into liters. Thus, \(7.02 \times 10^{17}~\mathrm{m}^{3}\) will be \(7.02 \times 10^{17} \times 1000 = 7.02 \times 10^{20}\) liters.

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Most popular questions from this chapter

The following procedure was used to determine the volume of a flask. The flask was weighed dry and then filled with water. If the masses of the empty flask and filled flask were \(56.12 \mathrm{~g}\) and \(87.39 \mathrm{~g},\) respectively, and the density of water is \(0.9976 \mathrm{~g} / \mathrm{cm}^{3}\), calculate the volume of the flask in \(\mathrm{cm}^{3}\).

The men's world record for running a mile outdoors (as of 1999 ) is 3 min 43.13 s. At this rate, how long would it take to run a \(1500-\mathrm{m}\) race? \((1 \mathrm{mi}\) \(=1609 \mathrm{~m} .)\)

Suppose that a new temperature scale has been devised on which the melting point of ethanol \(\left(-117.3^{\circ} \mathrm{C}\right)\) and the boiling point of ethanol \(\left(78.3^{\circ} \mathrm{C}\right)\) are taken as \(0^{\circ} \mathrm{S}\) and \(100^{\circ} \mathrm{S},\) respectively where \(S\) is the symbol for the new temperature scale. Derive an equation relating a reading on this scale to a reading on the Celsius scale. What would this thermometer read at \(25^{\circ} \mathrm{C}\) ?

Osmium (Os) is the densest element known (density \(=22.57 \mathrm{~g} / \mathrm{cm}^{3}\) ). Calculate the mass in pounds and in kilograms of an Os sphere \(15 \mathrm{~cm}\) in diameter (about the size of a grapefruit). [The volume of a sphere of radius \(r\) is \(\left.(4 / 3) \pi r^{3} \cdot\right]\)

A student is given a crucible and asked to prove whether it is made of pure platinum. She first weighs the crucible in air and then weighs it suspended in water (density \(=0.9986 \mathrm{~g} / \mathrm{mL}\) ). The readings are \(860.2 \mathrm{~g}\) and \(820.2 \mathrm{~g},\) respectively. Based on these measurements and given that the density of platinum is \(21.45 \mathrm{~g} / \mathrm{cm}^{3},\) what should her conclusion be? (Hint: An object suspended in a fluid is buoyed up by the mass of the fluid displaced by the object. Neglect the buoyance of air.)
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