Question

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

Answer 1

Answer:

$m_{CO2}=260.7 g CO2$

Explanation:

First of all we need to calculate the energy required:

$KE= 0.5*m*v^2$

where:

$m=2700kg$

$v=67 mph=29.95 m/s$

$KE= 0.5*2700kg*(29.95)^2$

$KE= 1210953 J=1210.953 kJ$

Octane's combustion enthalpy: $\Delta H_{comb}=- 5450 kJ/mol$

The reaction:

$C_8H_{18} + 25/2 O_2 longrightarrow 8 CO_2 +9 H_O$

Mass of CO2:

$m_{CO2}=\frac{1210.953 kJ}{5450mol}*\frac{1}{0.3}*\frac{8 mol CO2}{1 mol}*\frac{44 g CO2}{mol CO2}$

$m_{CO2}=260.7 g CO2$