# Thread: Inefficiencies in converting matter to energy

1. Okay I did a little research because I was wondering compared to the maximum amout of energy who much energy we are extracting, today. And I was expecting we'd be getting at leat 10% of the energy, but it turns out the number is actually .001 %. The article is HERE. I was just wondering what yall think and if somehow my numbers are off because .001 is well nothing.

2.

3. Hi Kingjacob,
... Matter is able to exist forever until it gets changed into energy at the rate of e=mc2. But in current practice, we are unable to even come close to this rate.

Currently, the efficiency of matter to energy conversion of todays fuels are beyond pitiful. According to coaleducation.org it takes 2205 lbs of coal to make 2500 kwh’s of electricity. And is from this source that I base my efficiencies. Now if we do a little math we come up with a few numbers,

-2,500 kwh the amount of electrical power gained from one tonne of coal today
- 2,500 Gwh the maximum amount of electrical power possible to derive from one tonne of coal.
Well, when I do the math I get a little different numbers (that you'll probably like!).
2205 lbs (or one metric ton) = 1000 kg
If you are just using e=mc^2, assuming all the mass (1000 kg) gets converted into energy, then I get (1000 kg)*(3.0*10^8 m/s)^2 = 9.0*10^19 kg m^2 / s^2
or 9.0*10^19 Joules.
This is 2.5*10^13 kWh,
or 2.5*10^7 GWh,
or 25 PWh. (Unless I goofed up somewhere...)

So I get a larger number for the maximum amount of energy assuming 100% efficiency yada, yada. That is, if you could convert all that mass into energy and using e=mc^2.

We may be opening up a can of worms here...
I have a bad feeling you may get some criticism on the difficulty of actually doing this....

Cheers,
william

4. Thanks, Im going to redo my math and see if/where I messed up before I edit the post.

I know very well how difficult it is to extract energy from mass, Ive spent the better part of my free time the past 8 months trying to think of a dynamo-less means of energy extraction.

5. Hey guys

One of you appears to be shovelling coal onto the fire and measuring it the other seems to be splitting atoms!

If you check the mass Loss of the sun (4 Million Tons per second) Then put that into E=MC^2 It is real fun, Now calculate the energy falling onto a square Metre of the Earth - Should be around 1000 Watts at Ground Level or 1300 Watts at the Upper atmosphere.

How much coal would you need to burn to get the same result!

6. Excellent post billco!

Here's what I get;

Using the solar luminosity (I'm allowed to do this since I'm an astrophysicist!)
L=3.826*10^33 erg/s
This luminosity is spread out over a sphere of radius 1AU at the earth's distance.
The surface area of a sphere is 4*pi*r^2.
At 1 AU (1.496*10^13 cm), the rate of energy hitting the earth is 1.36*10^6 erg/(s*cm^2)
or 1360 W/m^2 as you said.

Now, using king's coal-to-energy value of 2500 kWh per metric ton of coal,
I get 694 W/s/ton of coal.

Therefore, I get that it would take roughly 2 tons of coal just to match the solar energy hitting a mere 1 square meter on earth!!!

Astonishing.

Regards,
wm

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