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Energy content of automobile fuel

  1. Nov 16, 2007 #1
    1 litre of petrol contain about 8.79 KWHr (30023 BTU) of energy. A car wieghs about 1500Kg is able to run for about 10000 meters by burning 1 litre of petorl. For moving 1500 Kgs for 10000 meters the energy required is 150 Megajoules which is equal to 41.67 KWHr. How this is possible? Am I missing something? Please help me.
     
  2. jcsd
  3. Nov 16, 2007 #2

    Shooting Star

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    You have to be more specific. In what context were the two seemingly contradictory statements made?
     
  4. Nov 16, 2007 #3
    8.79 KWHr of energy in 1 litre of petrol is the data I collected from internet. 41.67 KWHr for hauling 1500 Kgs weight for 10 KMs is my calculation. I don't know what is wrong but Law of conservation of energy is missing.
     
  5. Nov 17, 2007 #4

    Shooting Star

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    How did you arrive at the value of 41.76 KWhr? What scenario did you use for your calc?
     
  6. Nov 17, 2007 #5
    Dear Shooting Star, thanks for patiently replying my silly basic questions. I hope I will be able to learn something useful here. 1 Kg of weight is approximately 10 Newton (near to 9.81 N) force. Moving 1 N force through 1 meter distance requires 1 Joule of energy. Hence moving 1500 Kgs (about 15000 N force) for 10 KM (10000 meters) requires 150000000 Joules that is 150 Mega Joules. I used the calculator available in the weblink "http://www.gordonengland.co.uk/conversion/energy.htm" and converted this 150 Mega Joules and the answer I got is 41.67 KWHr. I guess the value 150 Mega Joules is wrong but don't know how & where? I hope 150 Mega Joules is equal to 41.67 KWHr.
     
  7. Nov 18, 2007 #6

    Shooting Star

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    Oh, I get it now. You have lifted 1500 kg vertically upward through a dist of 10 Km taking g to be 10 m/s/s, which gives you 150 MJ. Oh my! But here still, the ans should be 1500 MJ.

    This has nothing to do with the chemical energy stored in petrol which, or part of which, is converted into mechanical energy when petrol is burnt. The two are unrelated.
     
  8. Dec 12, 2009 #7
    To SS: I think kbala's ~150MJ is correct, the mismatch in his calculations lie in the the wrong application of 9.81m/s^2 or (9.81 N). this is because cars travel horizontally, not vertically, so the resistance encountered is MUCH LESS then gravitational force. (i.e. tt the car's wind resistance n internal frictional resistance would amount to much less then 10 N on flat ground.)


    I believe your 2nd statement is incorrect as the 2 should add-up (more/less).

    Though its units used should be kcal rather then cal and and the marathon kcal/km travelled a bit overstated (should be 50 kcal/ Km) I believe that the following website would interest kbala (/ others) in appreciating the efficiency of foot transport.:approve:
    http://eliteclinics.com/calories.html
     
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