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Fundamentals Of Physics
Found in: Page 1333

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Short Answer

Roughly 0.0150% of the mass of ordinary water is due to “heavy water,” in which one of the two hydrogens in an H2O molecule is replaced with deuterium, H2 . How much average fusion power could be obtained if we “burned” all the H2 in 1.00 litre of water in 1.00 day by somehow causing the deuterium to fuse via the reaction H2+H2H3e+n ?

The power output is 14.4 kW .

See the step by step solution

Step by Step Solution

Step 1: Describe the expression for average kinetic energy

Let m be the mass of 1 litter of water, then the mass of the heavy water in 1 litter is 0.000150m .

The expression for number of molecules is given by,


Here, M is the molar mass of the heavy water, and NA is the number of molecules in one mole.

Step 2: Find the average fusion power

A heavy water molecule contains one oxygen atom, one hydrogen atom and one deuterium atom so the molar mass of the heavy water will be as follows.

M=16.0+1.00+2.00g/mol =19 g/mol

Substitute all the known values in equation (1).

N=0.000151000g/mol6.022×1023mol-119 g/mol =4.75×1021

This is the number of heavy water molecules (or deuterium), the deuterium fuse with another deuterium atom, according to the following reaction.


So, the number of fusion events will be half of N .

Nfu=N2 =4.75×10212 =2.375×1021

The expression of power output is given by,

P=Et =NfuQt ..........2

Here, E is total energy, and t is time.

Substitute all the known values in equation (2).

P=2.375×10213.27MeV1.602×10-13J/MeV24×3600 s =1.44×104W =14.4 kW

Therefore, the power output is 14.4kW .

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