Question

A large sport utility vehicle has a mass of 2500 kg .

Part A 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=1/2mv2 to calculate the kinetic energy required for the acceleration.) Express your answer using two significant figures.

Answer 1

First, convert mph to m/s by multiplying by .44704. You get 28.1635m/s.

Then you have the equation Work=.5*2500*(28.1635)^2-.5*2500*0^2=991…

So now you have the amount of energy that is required to reach this velocity, so you have to calculate how much octane must be combusted at 30% efficiency to yield this much energy.

Start with the balanced equation-

2C8H18+2502-->18H20+16CO2

So you have the molar ration of Co2/C8h18=8/1

Solve for the combustion needed by the equation 991.478KJ=(.3)(KJ released in reaction)

You get 3304.93 kj would have to be released by the ideal combustion, which would yield the 991.478kj at 30% efficiency.

The heat of combustion of octane is -5430kj.

So to get the answer you have 3304.93kj*(1molC8H18/5430kj)*(8molCO2/1m… 214.296 g CO2