Did I Just Create 2.5 Kilograms of Energy?

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    E=mc^2
In summary, the conversation discusses the concept of converting mass into energy and the implications of this process in relation to nuclear reactions and explosions. It also touches on the idea that the total mass of a system is not always equal to the sum of the masses of its individual components. The conversation ends with a discussion about the nature of matter and energy.
  • #1
Algr
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Okay, I know I have no business playing around with relativity, but things got heated and I grabbed this:

That is a lot of energy! For example, if we converted 1kg of mass into energy and used it all to power a 100 watt light bulb how long could we keep it lit for? In order to answer the question the first thing to do is divide the result by watts (remember that 1 watt is 1 joule per second):
9 x 1016 J / 100W = 9 x 1014 seconds
That's a lot of seconds, but how long is that in years? A year (365.25 days) is 31,557,600 seconds, so:
9 x 1014 seconds / 31,557,600 seconds = 28,519,279 years

And turned it into:

At 100% efficiency, in order to create 1 kg of matter (about 2 lbs) you would need to consume 2.8 billion watts for a year.

Give it to me straight. Did I blow up the Earth?
 
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  • #2
Algr said:
Okay, I know I have no business playing around with relativity, but things got heated and I grabbed this:

And turned it into:

Give it to me straight. Did I blow up the Earth?
Everything looks ok to me.
Released all at once, it's only equivalent to about 21 megatons of tnt.
I believe we've detonated a slightly larger bomb in the past, without blowing up the Earth.
 
  • #3
2.8 billion watts is a typical total power output of nuclear and coal power plants (a bit less than half of that as electricity, the rest as heat).

The largest explosion was the Tsar bomba, about 50 megatons TNT equivalent.
 
  • #4
To put this in perspective, to "blow up the Earth" (Death Star style) would require 2.24e32 joules or ~2.5e15 times the energy equivalent of that 1 kg, or what you'd get from converting 2.3 times the mass of the Martian moon Deimos into energy.
 
  • #5
Oh, I get it.

Uranium absorbs a neutron and becomes 236U

Then it splits into 92KR and 141BA. But 92+141=233, not 236

The missing three are the three neutrons. So Tzar Bomba is about 2.5 kilo of neutrons turned into energy?
 
  • #6
Algr said:
Oh, I get it.

Uranium absorbs a neutron and becomes 236U

Then it splits into 92KR and 141BA. But 92+141=233, not 236

The missing three are the three neutrons.
Right, but you don't see that the mass reduced if you just look at the number of nucleons (that number is conserved). The neutrons are still around.
So Tzar Bomba is about 2.5 kilo of neutrons turned into energy?
No. 2.5 kg of its original mass. It doesn't make sense to associate this to a particular decay product.
 
  • #7
So it weighs less, but the number of particles is the same? Do particles have variable weight?
 
  • #8
Algr said:
Do particles have variable weight?

Indeed, weight is position dependent.
 
  • #9
Algr said:
So it weighs less, but the number of particles is the same?
Yes.
Do particles have variable weight?
Only when mixed together[edit: sometimes[second edit: actually, always]], to make an atom.

see: mass per nucleon graph.

mass-per-nucleon.png


I'm not very good with words, so a graph is the only way I can understand this phenomena.
 
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  • #10
Algr said:
Do particles have variable weight?
No. The mass of N particles together is not necessarily the sum of masses of those N particles. Combine a proton and an electron to make a hydrogen atom, and a tiny bit of energy is released and the hydrogen atom is a bit lighter than "proton mass + electron mass".

For nuclear reactions the difference is more pronounced but the idea is the same.
 
  • #11
Algr said:
Uranium absorbs a neutron and becomes 236U

Then it splits into 92KR and 141BA. But 92+141=233, not 236

The missing three are the three neutrons. So Tzar Bomba is about 2.5 kilo of neutrons turned into energy?
The "missing" neutrons are being absorbed by other U-235 nuclei and turning them into U-236 to keep the process going - that's what makes it a chain reaction.
 
  • #12
But eventually you run out of uranium and still have neutrons running around everywhere. Then what? Do they make heavy isotopes of any atom they can find? Decay into hydrogen and a neutrino?
 
  • #13
Nearly all are captured by other atoms in microseconds, making some of them radioactive and contributing to fallout. A few will decay, but that contribution is completely irrelevant.
 
  • #14
Hmmm...

We still have 2.5 kilos of "matter" in tzar bomba that isn't actually made of anything. Wierd.

Either that or E=mc^2 doesn't actually happen in a nuclear bomb at all.
 
  • #15
Well, total mass is not the sum of masses of its constituents. That might appear weird, but that's how the universe is.
 
  • #16
I think I get it, but it seems odd to call it "matter" when it is actually the energy needed to shove particles into a difficult atom. It's more like 2.5 kilograms of energy was released.
 

1. What is E=MC^2?

E=MC^2 is a famous equation discovered by Albert Einstein in his theory of relativity. It stands for energy (E) equals mass (M) multiplied by the speed of light (C) squared.

2. Why is E=MC^2 important?

E=MC^2 is important because it revolutionized our understanding of the relationship between mass, energy, and the speed of light. It also paved the way for advancements in nuclear energy and technology.

3. How does E=MC^2 relate to recklessness?

In terms of recklessness, E=MC^2 can be used to calculate the amount of energy released in a nuclear reaction or explosion. This energy can have dangerous and destructive consequences if not handled carefully.

4. Can E=MC^2 be used for anything other than nuclear reactions?

Yes, E=MC^2 can be used to calculate the energy released in any type of reaction or transformation involving mass. It has also been applied to various fields such as cosmology, particle physics, and even medicine.

5. Is E=MC^2 still relevant today?

Absolutely. E=MC^2 is a fundamental equation in physics and is still used in various scientific and technological advancements. It has also been confirmed through numerous experiments and continues to be a cornerstone of modern physics.

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