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The total volume of ice in the Antarctic is about \(3.01 \times 10^{7} \mathrm{km}^{3} .\) If all the ice in the Antarctic were to melt completely, estimate the rise, \(h,\) in sea level that would result from the additional liquid water entering the oceans. The densities of ice and fresh water are \(0.92 \mathrm{g} / \mathrm{cm}^{3}\) and \(1.0 \mathrm{g} / \mathrm{cm}^{3},\) respectively. Assume that the oceans of the world cover an area, \(A,\) of about \(3.62 \times 10^{8} \mathrm{km}^{2}\) and that the increase in volume of the oceans can be calculated as \(A \times h\).

Short Answer

Expert verified
The sea level would rise approximately 67.32 m if all Antarctic ice were to melt completely.

Step by step solution

01

Calculate the Total Volume of Melted Ice

Since 1 g of ice melts to 1 g of water, and the densities of ice and water are 0.92 g/cm³ and 1.0 g/cm³ respectively, we can find the volume of water produced when all the ice melts by using the formula: \[ V_{water} = V_{ice} \times \frac{Density_{ice}}{Density_{water}} \] where \(V_{water}\) is the volume of water, \(V_{ice}\) is the volume of ice, \(Density_{ice}\) is density of ice and \(Density_{water}\) is the density of water. Heap in mind that initial volume of ice is given in km³ and we need to convert it to cm³ for consistency with density units. 1 km³ = \(10^{15}\) cm³.
02

Calculate the Rise in Sea Level

The volume of the resulting water will spread around the world's oceans and the sea level will rise accordingly. This rise in sea level, \(h\), can be calculated by dividing the volume of water, \(V_{water}\), by the area of the oceans, \(A_ocean\). Hence: \[h = \frac{V_{water}}{A_{ocean}}\] Again, bear in mind the units. The area of the oceans is given in km² , thus we should perform a conversion from km² to cm²: 1 km² = \(10^{10}\) cm² .
03

Substituting values

Now we substitute the given values into formulas outlined in steps 1 and 2 and calculate the final answer. Remembering about all necessary unit conversions.
04

Check the Answer

Finally, as common practice, it's always important to make sure the answer makes sense in the context of the problem. In this case, double-checking whether calculated value for sea level rise does seem plausible given the known data about the Earth's hydrosphere.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density of Ice
Density is a way to measure how much mass is contained in a unit volume of a material. For ice, this means understanding how tightly its molecules are packed together. Scientists often use the unit grams per cubic centimeter (g/cm³) to express density. Ice has a density of 0.92 g/cm³.

Because ice is less dense than water, it floats. This is why icebergs have only a small part visible above the water's surface. When you know the density of ice, you can predict how much water it would convert into when melted, simply because density is a comparative measure. For instance, the density of ice (0.92 g/cm³) allows us to deduce that it has 92% of the density of water. This is crucial in determining the volume of water that results when ice melts.
Density of Water
Water is denser than ice, with a density of 1.0 g/cm³. This higher density means that water molecules are packed closer together compared to ice molecules. The reason why water is denser than ice is because of the structure of water molecules that are tightly bonded in liquid form.

The density of water helps in calculating how ice turns into water. With the density information, one can determine changes in volume as the state changes from solid (ice) to liquid (water). By understanding these densities, we can solve questions related to the melting ice and how much liquid water would be produced.
Volume Conversion
Volume conversion is an essential part of solving problems involving changes in states of matter, such as solid to liquid. In our context, we are dealing with very large volumes initially given in cubic kilometers (km³).

To link the densities expressed in grams per cubic centimeter to these large volume units, we need to perform conversions. Specifically, we convert volumes of ice from km³ to cm³. There are a staggering number of cubic centimeters in a cubic kilometer: 1 km³ = \(10^{15}\) cm³. Carefully handling unit conversion helps maintain consistency in calculations.
Antarctic Ice Volume
The Antarctic contains a massive amount of ice, about \(3.01 \times 10^{7}\) km³. This volume represents a critical component of the Earth's long-term climate balance. When ice melts, it contributes to sea level rise, a significant concern for global climates.

To estimate the potential sea level rise if all this ice melted, one must calculate the additional water volume added to the oceans and understand how this additional volume would spread over the global ocean area. By applying the principles of density and volume, as well as conversion formulas, we can accurately predict sea level changes, connecting local ice volumes to global impacts.

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Most popular questions from this chapter

As mentioned on page \(13,\) the MCO was lost because of a mix-up in the units used to calculate the force needed to correct its trajectory. Ground-based computers generated the force correction file. On September \(29,1999,\) it was discovered that the forces reported by the ground-based computer for use in MCO navigation software were low by a factor of \(4.45 .\) The erroneous trajectory brought the MCO \(56 \mathrm{km}\) above the surface of Mars; the correct trajectory would have brought the MCO approximately \(250 \mathrm{km}\) above the surface. At \(250 \mathrm{km},\) the MCO would have successfully entered the desired elliptic orbit. The data contained in the force correction file were delivered in lb-sec instead of the required SI units of newton-sec for the MCO navigation software. The newton is the SI unit of force and is described in Appendix B. The British Engineering (gravitational) system uses a pound (lb) as a unit of force and \(\mathrm{ft} / \mathrm{s}^{2}\) as a unit of acceleration. In turn, the pound is defined as the pull of Earth on a unit of mass at a location where the acceleration due to gravity is \(32.174 \mathrm{ft} / \mathrm{s}^{2} .\) The unit of mass in this case is the slug, which is \(14.59 \mathrm{kg}\). Thus, BE unit of force \(=1\) pound \(=(\text { slug })\left(\mathrm{ft} / \mathrm{s}^{2}\right)\) Use this information to confirm that BE unit of force \(=4.45 \times\) SI unit of force 1 pound \(=4.45\) newton

How many significant figures are shown in each of the following? If this is indeterminate, explain why. (a) 450 ; (b) 98.6 ; (c) $0.0033 ; (d) 902.10 ; (e) 0.02173 ; (f) 7000 ; (g) 7.02 ; (h) 67,000,000

In the third century \(\mathrm{BC}\), the Greek mathematician Archimedes is said to have discovered an important principle that is useful in density determinations. The story told is that King Hiero of Syracuse (in Sicily) asked Archimedes to verify that an ornate crown made for him by a goldsmith consisted of pure gold and not a gold-silver alloy. Archimedes had to do this, of course, without damaging the crown in any way. Describe how Archimedes did this, or if you don't know the rest of the story, rediscover Archimedes's principle and explain how it can be used to settle the question.

A typical rate of deposit of dust ("dustfall") from unpolluted air was reported as 10 tons per square mile per month. (a) Express this dustfall in milligrams per square meter per hour. (b) If the dust has an average density of \(2 \mathrm{g} / \mathrm{cm}^{3}\), how long would it take to accumulate a layer of dust \(1 \mathrm{mm}\) thick?

A standard \(1.000 \mathrm{kg}\) mass is to be cut from a bar of steel having an equilateral triangular cross section with sides equal to 2.50 in. The density of the steel is \(7.70 \mathrm{g} / \mathrm{cm}^{3} .\) How many inches long must the section of bar be?

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