Question

this is a thermodynamics class

4. We want to develop an intuition for what different amounts of energy look like in various forms. Calculate the energy in MegaJoules (MJ) present in the following systems. As a comment, many people will be surprised by some of the values, for example, a comparison of part e to parts a through c.

a) 1 liter (approximately 1 quart) of octane, a primary component of gasoline. In this case, the chemical energy is released by burning. Octane has a heat of combustion of 5.07 x 10³ kJ/mole, a density of 0.703 g/cm³, and a molecular weight of 114 g/mole.

b) A car with mass of 1500 kg (about 3300 pounds) traveling at a velocity of 100 km/hour (about 60 miles per hour). In this case, the energy is in the form of kinetic energy, calculated as (1/2)mv², where m is the mass and v is the velocity.

c) A person with mass of 75 kg at the top of a cliff of height 100 meters (over 300 feet). In this case the energy is potential energy, given by the formula mgh, where m is mass, g is the gravitational acceleration constant of 9.8 m/sec², and h is the height.

d) A 12 Volt lead acid car battery that has mass of 10 kg (just over 20 pounds). From data on battery electrical energy storage density, the energy storage capacity of a battery of this type is 0.17 MJ/kg.

e) A breakfast burrito weighing 0.25 kg (about 1/2 pound), and having an energy density of 5 food calories per gram (that is, 5 kcal/g for thermodynamic calories).

We want to develop an intuition for what different amounts of energy look like in various forms. Calculate the energy in MegaJoules (MJ) present in the following systems. As a comment, many people will be surprised by some of the values, for example, a comparison of part e to parts a through c.

a) 1 liter (approximately 1 quart) of octane, a primary component of gasoline. In this case, the chemical energy is released by burning. Octane has a heat of combustion of 5.07 x 10³ kJ/mole, a density of 0.703 g/cm³, and a molecular weight of 114 g/mole.

b) A car with mass of 1500 kg (about 3300 pounds) traveling at a velocity of 100 km/hour (about 60 miles per hour). In this case, the energy is in the form of kinetic energy, calculated as (1/2)mv², where m is the mass and v is the velocity.

c) A person with mass of 75 kg at the top of a cliff of height 100 meters (over 300 feet). In this case the energy is potential energy, given by the formula mgh, where m is mass, g is the gravitational acceleration constant of 9.8 m/sec², and h is the height.

d) A 12 Volt lead acid car battery that has mass of 10 kg (just over 20 pounds). From data on battery electrical energy storage density, the energy storage capacity of a battery of this type is 0.17 MJ/kg.

e) A breakfast burrito weighing 0.25 kg (about 1/2 pound), and having an energy density of 5 food calories per gram (that is, 5 kcal/g for thermodynamic calories).

5. (4 points) Use the data given in Question 4a to find the energy density of octane in units of kcal/g. Compare this value to our estimate of the average energy density of biomaterials (a mix of proteins, carbohydrates, fats) of about 5 kcal/g. Based on this comparison, explain the advantage of powering your car by burning gasoline versus burning dry wood.

Answer 1

a) density= mass/volume (mL)= 0.703g/mL

mass= 0.703g/mL x 1000mL = 703g

mol of octane= mass/MW= 703g/114.23g/mol= 6.154mol

Energy= 5.07x10³ kJ/mol x 6.154 mol= 31202.0485 kJ

1kJ= 0.001 MJ ----> 31.2 MJ

b) 1m/s = 3.6 km/h

100km/h= 27.77777 m/s

E= 1/2 mv² =1/2 1500kg (27.7777)²= 578703.7037 J -----> 0.5787 MJ

c) E= mgh= 75kg x 9.8 m/s² x 100 m= 73500 J ----> 0.0735 MJ

d) Energy= 0.17 MJ/kg x 10kg= 1.7 MJ

e) Energy= 5kcal/g x 250g= 1250 kcal

1 kcal= 4.184 kJ ----> 1250 kcal= 5230 kJ ----> 5.23 MJ

Question 5:

energy density= 5.07x10³ kJ/mol x (1/114.23 g/mol) x 1kcal/4.184 kJ= 10.608 kcal/g

If you burn 1 g of wood you will obtain 5kcal of energy, nut if you burn 1g of octane you will obtain 10.608 kcal, that means that burning octane is more efficient. Burning the same amount in mass gives different amount of energy, we will want to use the one that gives more kcal per gram.