1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Apr 30, 2008 #1
    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.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted



Similar Discussions: Approximately 1.0 X 10^34 J of energy is available from fusion of hydrogen(continued)
  1. Pressure (continued) (Replies: 0)

  2. Pressure (continued) (Replies: 0)

Loading...