Calculate the mass of co2 emitted into the atmosphere upon accelerating the suv from 0.0 mph to 67.0 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.)

mass of CO₂ emitted = 25.08 g

Explanation:

A large sport utility vehicle has a mass of 2700 kg.

Calculate the mass of CO2 emitted into the atmosphere upon accelerating the SUV from 0.0 mph to 67.0 mph. Assume that the required energy comes from the combustion of octane with 30% efficiency.

C₈H₁₈(l) + 25/2O₂(g) → 8CO₂(g) + 9H₂O(l); ΔHc= -5074.1kJ

(Hint: Use KE=1/2mv^2 to calculate the kinetic energy required for the acceleration.)

The process is combustion, and such is an exothermic reaction

Ke = 1/2MΔv²

Where M = Mass of the SUV = 2700 kg, v=  Change in velocity of the SUV = 0 - 67.0 mph = 67.0 mph and Ke = Kinetic Energy.

converting 67.0 mph to m/s

1 mile = 1609 meters  and 1 hours = 3600 seconds

= 67.0 ×1609/3600 =29.95 m/s.

Ke = 1/2 × (2700) × (29.95)²

Ke = 1350 × 29.95×29.95 = 1210953.4 J

Ke = 1210.95 kJ.

The amount of energy available for the SUV at the end of the acceleration is 1210.95 kJ. from the question, the process is 30% efficient. i.e

1210.95 × 30/100 = 363.285 kJ.

from the equation of the reaction,

8 moles of CO₂ is produce when 5074.1 kJ of heat is released

(8×363.285)/5074.1 moles of CO₂ will be produced when 363.285 kJ of heat is released

= 0.57 moles 0f CO₂ will be produced when 363.285 kJ of heat is released

No. of mole = mass of CO₂/molar mass of CO₂

∴ mass of CO₂ = no. of mole × molar mass of CO₂

  Molar mass of CO₂ = 12 + (16×2) = 12 + 32 = 44 g/mol.

   Mass of CO₂ = 44×0.57 = 25.08 g

∴ mass of CO₂ emitted = 25.08 g

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explanation:

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