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Burning fossil fuels math problem

  • #1

Homework Statement



Humankind uses about 500 EJ of energy each year nearly all of it from fossil fuels. If the energy content of fossil fuels is 40 MJ/kg estimate the total mass of fossil fuels we burn each year.

Homework Equations



I am not sure if this is a relevant equation but in my text book there is a similar example showing the energy content of divided by the total thermal power extracted measured in MJ/kg

The Attempt at a Solution



I really don't know if I'm doing any of this right but I thought I needed to convert the 500 EJ into MJ. I got 5 x 10^14.

This problem is talking about a year while the text book shows per second. Do I need to convert to seconds in a year? If so i got 31, 536,000 seconds. Not sure if I even needed to do this...

I have no idea if I even started this problem right or where to go from here if I have?
 

Answers and Replies

  • #2

The Attempt at a Solution



I really don't know if I'm doing any of this right but I thought I needed to convert the 500 EJ into MJ. I got 5 x 10^14.

This problem is talking about a year while the text book shows per second. Do I need to convert to seconds in a year? If so i got 31, 536,000 seconds. Not sure if I even needed to do this...

I have no idea if I even started this problem right or where to go from here if I have?
The conversion is done right, And it was neccessary to do that. It doesn't matter if the answer comes out in years or seconds, it would be the same. But since the textbook has the answer in terms of seconds it is better to do it that way.
Now given that human kind consumes 5×1014MJ per 31,536,000 seconds what would be the consumption for 1 second?
To find out the amount of fossil fuel consumed:
40 MJ of the energy is got by 1kg of the fuel, how much for X MJ (the answer you have got from above)
Regards
 
  • #3
haruspex
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This problem is talking about a year while the text book shows per second. Do I need to convert to seconds in a year?
Converting to seconds is making the problem much more complicated than necessary. You know how many MJ a year are generated, and you know how many MJ are generated by 1 kg. It's a very simple operation to determine the number of kgs/year.
 
  • #4
Converting to seconds is making the problem much more complicated than necessary. You know how many MJ a year are generated, and you know how many MJ are generated by 1 kg. It's a very simple operation to determine the number of kgs/year.
If there are 5 x 10 ^14 MJ in a year and 40 MJ/kg then would I just divide?
 
  • #5
haruspex
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If there are 5 x 10 ^14 MJ in a year and 40 MJ/kg then would I just divide?
Yes.
 
  • #6
Yes.
I did this and got 1.24 x 10 ^13 kg/yr.

The answer in the back of the book is 12 billion tones/yr.

I am not sure how this answer was gotten.... or what step I would need to do next to get this answer.
 
  • #7
3,740
417
1 tone= 1000 kg
1 billion = 10^9 (in the US).
 
  • #8
1 tone= 1000 kg
1 billion = 10^9 (in the US).
I am confused as to what I am supposed to do with this?
 
  • #9
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To show that the two results are the same (pretty much).
 
  • #10
The second part of this question says assume that each kilogram of fuel has to be move 1000 km from its source to where it's burned at an average speed of 20 km per hour. Estimate the total mass of fuel that must be in transit at a given time.

I am not sure how to start this problem...or what formula to use.
 
  • #11
haruspex
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The second part of this question says assume that each kilogram of fuel has to be move 1000 km from its source to where it's burned at an average speed of 20 km per hour. Estimate the total mass of fuel that must be in transit at a given time.

I am not sure how to start this problem...or what formula to use.
How long does each bit of fuel spend moving? How many kg must reach their destination in that time?
 
  • #12
How long does each bit of fuel spend moving? How many kg must reach their destination in that time?
I really don't know what you mean by this. What number should I be starting with?
 

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