A water sample is found to have 0.016 % deuterium content...

A water sample is found to have $0.016 \%$ deuterium content (that is, $0.016 \%$ of the hydrogen nuclei in the water are ${ }^{2} \mathrm{H}$ ). If the fusion reaction $\left({ }^{2} \mathrm{H}+{ }^{2} \mathrm{H}\right)$ yields $3.65 \mathrm{MeV}$ of energy on average, how much energy could you get from $1.00 \mathrm{~L}$ of the water? (There are two reactions with approximately equal probabilities; one yields $4.03 \mathrm{MeV}$ and the other $3.27 \mathrm{MeV}$.) Assume that you are able to extract and fuse $87.0 \%$ of the deuterium in the water. Give your answer in $\mathrm{kW} \cdot \mathrm{h}$.

Key Concepts

Efficiency of Extraction/Fusion

Efficiency here refers to the practical yield after accounting for losses in extracting and fusing the available deuterium. An efficiency factor is applied to the total number of deuterium nuclei to determine the fraction that actually undergoes fusion, directly impacting the final energy calculation.

Percentage Composition

This concept involves understanding the fraction of a certain isotope or component present in a substance. Here, it is applied to water, where a specified percentage of the hydrogen nuclei is deuterium. It is crucial in determining the number of deuterium atoms available for reaction from the overall water volume.

Nuclear Fusion Energy Yield

Nuclear fusion energy yield refers to the amount of energy released when light atomic nuclei, such as deuterium, fuse together. The reaction specifics include the energy given in units like MeV, which must be carefully converted into commonly used energy units. This concept is pivotal for calculating the total available energy from fusion reactions.

Molar Quantity and Avogadro's Number

This key concept deals with converting between macroscopic quantities and the number of individual particles or atoms. In the context of water, it involves using the density and molecular weight of water to determine the number of water molecules, and subsequently the number of deuterium atoms, via Avogadro's number. It is vital for scaling reaction energy yield to a liter of water.

Unit Conversion

Unit conversion is essential to translate the energy from one measurement unit to another—specifically from the microscopic energy unit (MeV) to macroscopic energy units such as joules and ultimately kilowatt-hours (kW·h). This process ensures that the calculated energy is expressed in a format suitable for practical use and comparison.

Transcript

00:01 So the way that we're going to tackle this question is to first find out how many deuterium nuclei there are in the water.

00:12 Right? we call this nd over here.

00:15 This is the number of deuterium molecules.

00:18 And then we'll look at how much energy is being produced by 0 .87 of this nd of this deuterium molecules because we're given that only 80%, 87 % of this deuterium will fuse.

00:38 Alright, and we know that for each of these reactions, each reaction we get approximately 3 .65 m .ev, 6 .3 .3 .65 mv.

00:58 Right, and then we want the answer to be in kilowatt hours.

01:03 So to convert from mev to kilowatt hours, we first convert it into juice.

01:12 Right? to convert into juice, multiple pattern of power 6.

01:18 This is mev, sorry, this is ev per mev.

01:27 Then we multiply it by 6, 0 times 10 to power minus 19 joules per ev.

01:40 And then to convert from joules to kilowatt hours, we know that each kilowatt hours is 3 .6 times 10 to power 6 joules.

01:53 So we divide 1 kilowatt hour, 1 .3 .6 times 10 to power 6 joules.

02:04 So our final answer will be in kilowatt hours.

02:09 Now going back to finding how many deuterium atoms.

02:13 Or deuterium nuclei there are, we are given the mass of water to be 1 ,000 grams or 1 kilogram because we're dealing with 1 liter of water, 1 liter, and water has a density of 1 gram per milliliter, right? so we have 1 ,000 grams.

02:45 Now for each water, note that we have 2 hydrogens...

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