If The Matter Of My Body Was Converted Into Energy, How Much Would Yield?

Question

Given the equation e=mc^2 what is the value of energy in my body?
Please describe in a comparable context.

In theory...does this value vary between fission and fusion nuclear reactions?

BEKKI B: That's awesome...thanks!

Answer 1

You can use SI units in that equation - mass in kilograms, speed in meters per second, and energy in joules. Taking a typical adult human mass of 70 kg (154 lbs) gives you 6.3 x 10^18 joules. (mw - I think you slipped a couple of decimal places - c^2 is 9 x 10^16)

How much is that? 1.75 trillion (1.75 x 10^12) kwh (kilowatt-hours), or several months' electric power output from the entire US. In explosive terms, about 1500 megatons.

The conversion of matter into energy is the same regardless of how it happens. Both fission and fusion convert only a small part of the total mass to energy - the difference in mass between what you start with (deuterium or uranium) and what you end up with (helium plus stray neutrons, or a variety of fission products).

Answer 2

lucky3. from your pic, I would think you weigh about 100 lbs.

110 lbs x 1 kg / 2.2 lbs = 50 kg

so let's say you weigh 50 kg.

let's also igore your velocity. otherwise, if you were to consider your velocity, then you would have to us the relativistic form of the equation (e = (1/(1-v^2/c^2))
^.5 x mass x c^2)

so

e = m x c^2 = 50 kg x (3 x 10^8 m/s)^2
= 50 kg x 9 x 10^16 m^2/s^2
= 4.5 x 10^18 kg m^2/s^2
= 4.5 x 10^18 joules

= 4.5 x 10^18 joules x (1 cal / 4.184 J)
= approximately 1 x 10^18 calories

in perspective.

say you have lake containing water at room temperature. say 20C. to heat the water to 100C and vaporize it would take

m x cp x dt + m x hf = 1 g x (1cal/gC) x 80C + 1 g x 540 cal/g
= 620 cal / g

so .....

1 x 10^18 cal x (1 g/ 620 cal ) x (1 cm^3 / g) x (1 m^3/100^3 cm^3) = 1.6 x 10^9 m^3

take the cube root and thats about 1200 meters x 1200 meters by 1200 meters

in other words, the amount of energy in your body is about enough to heat and vaporize a 1 mile x 1 mile x 1 mile cube of water......

in theory, fission and fusion have nothing to do with this number. you asked how much energy was available.

now if you take your body and undergo fission or fusion processes on it, the amount of energy released would depend on the exact atomic nuclei undergoing the reaction.

for more info on fission/fusion see here....
http://science.howstuffworks.com/nuclear.htm

Answer 3

weigh yourself and convert to your mass in kg

c = speed of light (300 million meters / s)

So multiply your mass (in kg) by the square of 300 million to get your energy (in joules).

For comparison, a kilowatt hour is
1000 joules/s*3600 s/hour
= 3.6 million joules.

So express your energy in kilowatt hours. Go look at your electric bill (should tell you your kW-hours/month) and calculate how long your house could run on your energy.

If you want to calculate the energy released by a fission or fusion (or ANY) reaction, you have to calculate the mass defect

(mass reactants) - (mass products)

Multiply that mass defect by c^2 to obtain the energy released.