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A large sport utility vehicle has a mass of 2.5 * 103 kg. Calculate the mass of CO2 emitted into the atmosphere upon accelerating the SUV from 0.0 mph to 65.0 mph. Assume that the required energy comes from the combustion of octane with 30% efficiency. (Hint: Use KE = 12 mv2 to calculate the kinetic energy required for the acceleration.)

228.113

Chemistry 102

Chemistry 101

Chapter 9

Thermochemistry

Thermodynamics

Chemical reactions and Stoichiometry

Carleton College

University of Central Florida

Rice University

University of Kentucky

Lectures

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So in this problem we have the mass of SUV and we want to determine Ah, the mass of CO two emitted when this vehicle accelerates from zero miles per hour to sixty five miles per hour. So the first thing we'LL do is find out the kinetic energy this requires So we'LL take our sixty five miles per hour and multiply it by point for for seven o four, which is the conversion factor in order to get our speed in meters per second. And so we have twenty nine point zero five seven six meters per second. Now we confined kinetic energy with the K equals one half mass times velocity squared equation. So our mass is twenty five hundred kilograms and our velocity we just found you square that and that gives us a kinetic energy of one zero five five four three zero kilogram meter squared over seconds squared which is equal to one zero five, five four three zero Jules. And so we'll just divide that by a thousand to convert into Killa Jules So now we have our kinetic energy in Kila Jules. And now we're told this kinetic energy comes from ah thirty percent efficiency, So this is at thirty percent, and we want to find what it is at one hundred percent. And so to do that will take our kinetic energy and divided by zero point three that thirty percent. And that gives us the value of one hundred percent, which is three thousand five hundred eighteen point one Killer Jules. And so now that we have that we can determine the moles of octane by taking our number and dividing it by the heat of combustion of Octane, which is five thousand four hundred thirty killer Jules per mall. And that gives us zero point six four seven nine bowls of octane ch. Eighteen. So now we write the chemical formula for the combustion of octane to see H H eighteen plus twenty five o two forms eighteen h two o plus sixteen. So, too. And so now we take our zero point six four seven nine moles of octane and doing multiple ratio where we have two moles of octane for sixteen moles of co two, just our two here in our sixteen here, and that gives us that we have five point one eight three moles of co two. Soon our in our final step will take our five point one eight three moles of O two and multiplied by the Moler Mass. Forty four point zero one grams per mole. And that'LL give us two hundred twenty eight point one one three grams of o two. And this will be our final answer for the mass of CO two produced by this vehicle accelerating from zero to sixty five miles per hour.

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