The fusion reaction: releases 23.85 MeV of energy.World power consumption is about 10¹³ W and 0.0300% of ocean water contains deuterium, .If the deuterium in the ocean water were fused by controlled fusion with 100% efficiency,how much water would be needed every second to cover world's power consumption?

Question

Quizplus · Accepted Answer

First, we need to calculate how much power can be obtained from 1 kg of deuterium by the fusion reaction:
23.85 MeV = 3.8144 × 10^-12 J (1 MeV = 1.6 × 10^-13 J)
1 kg of deuterium contains 6.022 × 10^26 atoms (Avogadro's number)
So, the energy produced by one deuterium atom is:
E = 3.8144 × 10^-12 J / 6.022 × 10^26 = 6.340 × 10^-39 J

To produce world power consumption, we need to find how many atoms are needed:
10^17 W = 10^17 J/s
Number of atoms needed per second:
(10^17 J/s) / (6.340 × 10^-39 J/atom) = 1.578 × 10^55 atoms/s

Now, we need to find how many kg of water we need to extract 1.578 × 10^55 atoms of deuterium.
In 1 kg of ocean water, we can assume there are roughly (1/2) × 0.0300% × 6.022 × 10^23 deuterium atoms (since deuterium is twice as heavy as normal hydrogen).
So, we need (1.578 × 10^55 atoms) / [(1/2) × 0.0300% × 6.022 × 10^23 atoms/kg] = 5.22 × 10^30 kg of ocean water every second.

Therefore, the correct answer is (D) 522 kg.

The Fusion Reaction: Releases 23