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Chapter 1: Problem 74

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?

Short Answer

Expert verified
It would take approximately 4227.4 years for a 1mm layer of dust to accumulate from unpolluted air.

Step by step solution

01

Convert tons per square mile per month to mg per square meter per hour

Initially, the conversion of units is done. 1 tonne is equivalent to \(1 \times 10^{12}\) milligrams (mg), 1 square mile converts to \(2.59 \times 10^6\) square meters (m^2) and 1 month has approximately \(720\) hours. So the dustfall rate is calculated as: \((10 \times 10^{12}\) mg) / (\(2.59 \times 10^6\) m^2 x 720 hours) = approximately \(53.9\) mg/m^2 hr.
02

Convert a 1mm thick layer of dust into volume per area unit

Since we need to find out how long it takes for a 1mm dust layer to accumulate, let's firstly convert this layer into a volume per area unit, which matches the units used in Step 1. 1mm can be written as 0.1cm, so every square meter would have a volume of \(1m^2 \times 0.1cm = 1000cm^3\). However, since the density is given in g/cm^3 and we are dealing with milligrams, we will convert this volume to \(1 \times 10^{6}\) cm^3.
03

Calculate the mass of a 1mm thick layer of dust

Mass can be calculated using the volume of the dust layer and the given dust density. Substituting the found volume into the formula for mass (mass = density x volume) gives us \(2 g/cm^3 x 10^6 cm^3 = 2 \times 10^6\) g, or \(2 \times 10^{9}\) mg.
04

Determine the time

Time is calculated by dividing the mass of the dust layer by the dustfall rate. Therefore, \(2 \times 10^{9}\)mg / \(53.9\) mg/m^2 hr gives us \(3.7 \times 10^{7}\) hours, which converts to approximately 4227.4 years.

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Most popular questions from this chapter

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