# Calculate the energy in kJ that is released as heat (70%) by exactly 1 gallon of gasoline.2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g) H = –10110 kJ a. Calculate the energy in kJ that is...

Calculate the energy in kJ that is released as heat (70%) by exactly 1 gallon of gasoline.

2 C8H18 (*l*) + 25 O2 (*g*) → 16 CO2 (*g*) + 18 H2O (*g*) H = –10110 kJ

a. Calculate the energy in kJ that is released as heat (70%) by exactly 1 gallon of gasoline. The density of octane is 0.702 g/mL. Assume gasoline is pure octane (1 gallon = 3.7854 L). Show your work including any necessary formulas, all conversions, and all units.

b. Most of the heat calculated in part a goes right out the exhaust system and into the air. About 25% of the heat gets absorbed into the metal parts that make up your engine. Assume the mass of your engine is 79.8 lbs with a uniform specific heat of near 0.43 J/g x °C. If your engine begins at 25 °C, how hot will your engine be if it continuously burns 5.5 gallons of gasoline (octane) while driving? (1 lb = 0.4536 kg) To solve:

c. Let's say that the temperature at which your engine parts begin to melt is 3500 °C. Can your engine handle the temperature you calculated in part b.? Explain your answer.

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Gasoline is a hydrocarbon fuel that is used to power most automobiles and trucks worldwide. It has octane, which is an eight-carbon chain that, upon combustion with oxygen, delivers thirty-five (35) megajoules of energy per liter, or one hundred-thirty two (132)megajoules of energy per United States gallon. A megajoule (MJ) is equal to one million joules, and equal to one thousand kilojoules (kJ). So 132 times 1000 would equal 132,000 kilojoules of energy produced by one gallon of gasoline. If 70% of that is given off as heat energy, simply multiply 132,000 times .7, the decimal equivalent of 70 percent, and you get 92,400 kJ, which is what is given off as heat energy.

25% of the 132,000 kJ would be 33,000 kJ of energy, absorbed by the mass of the engine. Converting the mass of the engine to kilograms: 79.8 lbs x .4536 kg/lb = 36.2 kg. If you have 5.5 gallons of gasoline, that would be 33,000 kJ x 5.5 gallons, which would equal 181,500 kJ of heat energy. Given the specific heat of .43 J/g, I would convert the kJ to J and the kg to g, which would give 181,500,000 J and 36,200 g. Now divide the two to get Joules per gram: 181,500,000J/36,200g = 5013.8 J/g If you multiply that by the specific heat constant, you get this 5013.8 J/g x 1g/.43 J xC=11,660 degrees Celsius. That plus the 25 degrees starting temperature, would put it at 11,685 degrees Celsius.

If your engine starts to melt at 3500 degrees Celsius, it will pretty much be a goner, unless you have an immaculate cooling system to subtract the heat away from it.