iPad 9:12 PM 74%- (6%) Problem 16: There is approximately 1033 J of energy avail

Question

iPad 9:12 PM 74%- (6%) Problem 16: There is approximately 1033 J of energy available fromn the fusion of hydrogen in the world's oceans. 50% Part (a) If 0.25·1033 J of this energy were utilized, what would be the decrease in the mass of the oceans? Expr kilograms tano acosO cos) sino) cotan asin0 1 23 atan)acotan)sinh coshtanh cotanhO ODegrees O Radians Hint I give up! Hints:--deduction per hint. Hints remaining:2 Feedback: 0% deduction per feedback. 50% Part (b) How great a volume of water does this correspond to in cubic meters?

Explanation / Answer

a) Energy is given by:

E = mc^2

Total energy ET= 10^33 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 = 10^33 J/ (3.0 X 10^8 m/s)^2 = 1.11 X 10^16 Kg

Therefore m1 (or the mass of the energy from the entire ocean) is 1.11 X 10^16 Kg.

Next part, finding the mass when 0.25 X 10^33 J of the energy is used:

E1(the amount of energy if utilized)= 0.25 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 = 0.25 X 10^33 J / (3.0 X 10^8 m/s)^2 = 2.78 X 10^15 Kg

Decrease:

md= m1 - m2 = 1.11 X 10^16 Kg – 2.78 X 10^15 Kg = 8.32 X 10^15 Kg

b) To what volume of water does this correspond?

So we know that V= m/d

density of water is 1000 kg/m^3

the mass from before was 8.32 X 10^15 Kg, so:

m= 8.32 X 10^15 Kg

d= 1000 kg/m^3

V = m/d

V = (8.32 X 10^15)/1000 m^3 = 8.32 x 10^15 m^3

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