Sustainable Energy 1st Edition By Richard Dunlap – Test Bank

Sustainable Energy 1st Edition By Richard Dunlap – Test Bank

Chapter 4

Environmental Consequences

of Fossil Fuel Use

4.1 If one tonne of coal is burned and the CO₂ that is produced is sequestered by reaction with CaO as CaO + CO₂ → CaCO_3. What is the mass of CaO that will be required?  Assume coal is pure carbon.

Solution The combustion process is

C + O₂ → CO₂

The relevant molecular masses are

C → 12 g/mol

so burning 12 g of carbon yields 44 g of CO₂ and 44 g of CO₂ will require 56 g of CaO for sequestration. This means that burning 1 tonne of carbon will require

(1 t) × (44 g/12 g) × (56g/44g) = 4.67 t of CaO for sequestration.

4.2 Phobos, the largest moon of Mars, is at a mean distance from the sun of 1.52 times that of the Earth. It has a very low estimated albedo of ~0.1. Calculate its temperature.

Solution The total power absorbed by a body with an albedo of a and an insolation of S is

P_absorbed = (1 –- a)S πR².

The power radiated by a black body at a temperature T is

P_radiated = 4πR²σT⁴

Equating these gives the equilibrium condition for a planet without atmosphere at a temperature given by

where σ is the Stefan Boltzmann constant. The insolation on the earth (outside the atmosphere) is 1.367 kW/m² so at the distance of Phobos the 1/r² law gives the insulation as

The temperature is

T = {[(1 –- 0.1) × 592 W/m²]/[4 × 5.67 × 10^–-8 Wm^–-2K^–-4}^{1/4 =}  220 K

Note: the actual measured average temperature is 233 K.

4.3 A 1600 kg gasoline-powered vehicle requires 3.5 MJ of primary energy to travel 1 km (note: The properties of gasoline may be approximated as the properties of octane C₈H₁₈, which has the energy of combustion of 44.4 MJ/kg). If the vehicle travels a total of 300,000 km during its life, what is the ratio of the mass of the CO₂ emitted to the mass of the vehicle?

Solution 300,000 km will require energy. This requires a total gasoline consumption over the life of the vehicle of

(1.05 × 10⁶ MJ)/(44.4 MJ/kg) = 2.36 × 10⁴ kg

Octane is C₈H₁₈ and the oxidation process is

in terms of the molecular mass

so the combustion of 2 × 114g = 228g of octane yields 16 × 44 = 704g of CO₂ so the combustion of 2.36 × 10⁴ kg of gasoline will produce

(2.36 × 10⁴ kg) × (704 g/228 g) = 7.29 × 10⁴ kg

That is about 46 times the mass of the vehicle itself.