# Homework Help: Approximately 1.0 X 10^34 J of energy is available from fusion of hydrogen(continued)

1. Apr 30, 2008

### jackster18

1. The problem statement, all variables and given/known data

Approximately 1.0 X 10^34 J of energy is available from fusion of hydrogen in the world's oceans.
a) If 1.0 X 10^33 J of this energy were utilized, what would be the decrease in the mass of the oceans?
b) To what volume of water does this correspond?

2. Relevant equations

E=mc^2

Where E is the energy (in joules)
C is the speed of light (3.0 X 10^8 m/s (I think that’s it I looked it up))
m is the mass (in Kg)

V=m/d

Where V is the volume (in what units I don’t know)
m is the mass (in Kg)
d is the density (in what units I don’t know)

3. The attempt at a solution

a) ET("energy total", I’m calling it that because it’s the total energy in the world’s oceans from fusion of hydrogen)= 1.0 X 10^34 J

c= 3.0 X 10^8 m/s

m1(the mass of all the energy from the ocean)= ?

E=mc^2
ET=m1c^2
Rearranging for m1

m1= ET/c^2
= 1.0 X 10^34 J/ (3.0 X 10^8 m/s)^2
= 1.11 X 10^17 Kg

Therefore m1 (or the mass of the energy from the entire ocean) is 1.11 X 10^17 Kg.

Next part, finding the mass when 1.0 X 10^33 J of the energy is used:

E1(the amount of energy if utilized)= 1.0 X 10^33 J

c= 3.0 X 10^8 m/s

m2 (the mass of E1's energy) = ?

E=mc^2

E1=m2c^2

Rearranging for m2:

m2=E1/c^2
= 1.0 X 10^33 J / (3.0 X 10^8 m/s)^2
= 1.11 X 10^16 Kg

So m1= 1.11 X 10^17 Kg, and m2= 1.11 X 10^16 Kg

So the question said what would be the decrease in the mass of the oceans...so i guess i just subtract the two?

md(the mass decrease- i just made this up so i dont get the masses mixed up, so i put a d beside the m)= ?

md= m1 - m2
= 1.11 X 10^17 Kg - 1.11 X 10^16 Kg
= 9.99 X 10^16 Kg

I guess that’s the amount of mass decrease is 9.99 X 10^16 Kg. I’m not sure if I have done this correctly that’s why I’m asking.

Ok, now part b) To what volume of water does this correspond?

So I know that V= m/d

Somewhere I found on the internet that the density of water is 1 gram/mL...I’m not sure if that’s correct though, and if it is, can I just put mass (Kg) divided by (g/mL) to get volume? So what would the volume units be? Just mL I guess not too sure..

So the mass from before was 9.99 X 10^16 Kg, so:

m= 9.99 X 10^16 Kg
d= 1 g/mL (there is 1000 grams in a Kg...so 0.001 Kg/mL)
V=?
V = m/d
V= 9.99 X 10^16 Kg / 0.001 Kg/mL
V= 9.99 X 10^16 Kg/Kg/mL (the Kg's just cancel out i guess so its...
V= 9.99 X 10^16 mL

If you can help me I would greatly appreciate it. I’m in grade 12 physics and this is a question from an independent study unit. I just want to make sure I am correct or if I am understanding what’s going on, plus my teacher is marking it so...it would be nice to get it right.