The Energy in the World's Oceans: Calculations and Consequences

[jackster18]

Homework Statement

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?

Homework 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)

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 don't 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.

 

[anathema]

Thank you for your detailed response. Your calculations and understanding of the concepts involved are correct. However, I would like to point out a few things that may help you in your future scientific endeavors.

1. When using scientific notation, it is important to keep track of significant figures. In this case, the given values have 2 significant figures, so the final answer should also have 2 significant figures. Therefore, the mass decrease should be reported as 1.0 x 10^17 kg.

2. When calculating the volume, it is important to use the correct units. The density of water is indeed 1 g/mL, but when converting the mass from kg to g, you need to multiply by 1000, not divide by 0.001. So the correct calculation would be: V = 9.99 x 10^16 kg x (1000 g/kg) / (1 g/mL) = 9.99 x 10^19 mL. The units for volume would be mL, as you correctly stated.

I hope this helps. Keep up the good work in your studies!A fellow scientist

 

[Moetasim]

Thank you for your post and for providing your thought process and calculations. I would like to clarify a few things and provide some feedback on your solution.

Firstly, your calculations and use of equations are correct. You have correctly used the equation E=mc^2 to calculate the mass of energy available from fusion in the world's oceans. You have also correctly used the equation V=m/d to calculate the volume of water corresponding to the mass decrease in the oceans. However, there are a few things that I would like to point out and clarify for you.

1. The units for mass in the equation E=mc^2 are in kilograms (Kg) and the units for energy are in joules (J). Therefore, the units for c (the speed of light) should also be in meters per second (m/s). In your calculation, you have used c= 3.0 X 10^8 m/s, which is correct.

2. In your calculation for the mass of energy available from fusion in the world's oceans (m1), you have used the total energy (ET) value of 1.0 X 10^34 J. However, in the question, it is stated that only 1.0 X 10^33 J of this energy will be utilized. Therefore, your calculation for m1 should be:

m1 = E1/c^2
= (1.0 X 10^33 J) / (3.0 X 10^8 m/s)^2
= 1.11 X 10^16 Kg

This value for m1 is then used in your calculation for m2 (the mass of energy if 1.0 X 10^33 J is utilized), which is correct.

3. In your calculation for the mass decrease (md), you have correctly subtracted m2 from m1. However, you have used the incorrect values for m1 and m2 (you have used the values you calculated for the total energy and the utilized energy, instead of the mass values). The correct calculation for md should be:

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

This is the same value that you have calculated, so your solution for part a) is correct.

4. For