suppose a car operates so inefficently that only 16% of the thermal energy from burning fuel is converted to useful "wheel turning" (mechanical energy) how many kj of useful energy would be stored in a 20 gallon tank of gasoline?(assume that octane burning and gas burning produce the same results)
Write the equation for octane + O2 ==> CO2 +_ H2O and balance it.
Convert 20 gallons octane to L, then use the density of octane to convert to grams.
Look up the combustion energy for octane and convert to grams octane you have in the 20 gallons, multiply by 16% to arrive at the energy stored.
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To calculate the amount of useful energy stored in a 20-gallon tank of gasoline, we need to consider the energy conversion efficiency.
Given:
- The car operates at an efficiency of 16%, meaning only 16% of the thermal energy from burning fuel is converted to useful mechanical energy.
First, let's determine the energy content of gasoline. Gasoline has an approximate energy content of 120 megajoules (MJ) per gallon.
To find the total energy in the 20-gallon tank, multiply the energy content per gallon by the number of gallons:
Total energy = Energy content per gallon * Number of gallons
Total energy = 120 MJ/gallon * 20 gallons
Now we have the total energy stored in the tank, but we need to calculate the useful energy.
Useful energy = Total energy * Efficiency (in decimal form)
Useful energy = (120 MJ/gallon * 20 gallons) * 0.16
To convert megajoules (MJ) to kilojoules (kJ), multiply by 1000 (1 MJ = 1000 kJ):
Useful energy in kJ = (120 MJ/gallon * 20 gallons * 0.16) * 1000
Now we can calculate the answer:
Useful energy in kJ = (120 MJ/gallon * 20 gallons * 0.16) * 1000 = XXX kJ
By substituting the values, you will get the exact amount of useful energy stored in a 20-gallon tank of gasoline, considering the given efficiency.