Of all the hydrogen in the oceans, 0.0300 % of the mass is deuterium. The oceans have a volume o

Of all the hydrogen in the oceans, 0.0300 % of the mass is...

Key Concepts

Isotopic Abundance

This concept involves understanding the relative proportion of an isotope within a sample. In the context of the question, the deuterium concentration provided as a percentage of the ocean’s hydrogen mass is used to determine the quantity of fuel available for fusion. Such calculations are essential in many areas of chemistry and nuclear physics, where the abundance of isotopes can significantly affect reaction outcomes.

Nuclear Fusion

Nuclear fusion refers to the process where light atomic nuclei, such as deuterium, combine to form a heavier nucleus like helium, releasing a substantial amount of energy in the process. This concept is fundamental in energy production in stars and is of interest for controlled energy generation on Earth. Understanding the fusion reaction, including how much energy is released per fusion event, is crucial for evaluating potential energy sources.

Mass-to-Volume Conversion

This concept deals with converting a given volume of a substance into mass using density. In the context of the problem, the volume of the oceans is used along with the density of water to calculate the total mass of the ocean, which then allows for the determination of the specific mass of deuterium present. Such conversions are widely used in physics and engineering when transitioning between different measurement systems.

Energy and Power Relationship

This concept involves understanding the relationship between energy, power, and time. Energy is quantified in joules, and power, measured in watts, describes the rate at which energy is used or produced. In the problem, after calculating the total energy released from fusion, the rate of energy consumption is used to estimate how long this energy would last for a given power demand. This relationship is central in analyzing and comparing energy resources and their sustainability over time.

Transcript

00:01 Question 34 tells us that the mass, the volume of ocean water is approximately 3 .317 million miles cubed, which i took the liberty of converting it into meters cube, of course, our si unit for volume.

00:17 And we're told that of all the hydrogen atoms of the ocean, 0 .03 % of them are deuterium.

00:24 In part a, he asks if all of the deuterium and the oceans refuse to create helium -4, how much end? energy would be produced.

00:34 So we're not told a lot, but this actually does give us enough information to figure out our total energy.

00:43 So think of what we need to solve this.

00:45 We can see that the total energy of our sample, we can determine that based on the energy per reaction times the number of reactions or number of decays that would occur.

01:07 None of these things, neither of these things, sorry, we have.

01:11 So the first thing we can solve for, actually, is energy per reaction because it does tell us in the question that these deuterium atoms do fuse to create helium -4.

01:22 Such be a hint, though, that the reaction that we're dealing with is deuterium plus deuterium gives helium four.

01:36 So for each one of these reactions, we can determine this energy to be 931 .5 mbd and multiplied by the change in energy masses per unit.

01:55 State so i chose the absolute value of course of these minus two times 2 .014.

02:06 0 which you find that the energy per each one of these reactions is 23 .85 mav per reaction n.

02:20 So now we have this term all done great.

02:24 Now all i mean to do is find the number of nuclei that are undergoing these reactions which is a little bit more complicated the step.

02:33 So we're told we want to find all of the deuterium atoms that would undergo this reaction.

02:37 So basically this n here needs to be the number of deuterium.

02:40 I don't know why pick for it.

02:41 I don't know why i'd say number of deuterium atoms.

02:43 So we can determine the number of deuterium based on the mass of the deuterium sample over the moral mass of deuterum times avogadro's number.

02:55 So now we need to solve for the mass of deuterium because obviously we know molar mass and analogous number.

03:04 So how do we do that? well, we're given volume of water.

03:10 So with that, we can determine the mass of water based on, obviously, multiplying by the density of water times the volume of water.

03:20 We can look up online that the volume, sorry, the density of ocean water is 1 ,029 kilograms per meter cubed, multiplied by this massive, massive amount of water, 1 .32 times 10 to the 18.

03:38 Meters cubed gives us a total mass of 1 .36 times 10 to the 21 kilograms.

03:58 So this is the total amount of water.

04:01 So what we can do now, because we have mass of water, we can solve for the number of water molecules for mass over molar mass of water times out of the gutters number.

04:14 So the mass of water, which i'm going to put in grams, so 1 .36 times 10 to the 24 grams, grams or not kilograms.

04:25 Take the molar mass of water, two hydrogen, one oxygen to be about 18 .02 grams per mole and multiplied by avogadur's number, which 6 .2 times 203, 10 to the 23, which we know, we can determine that the number of water atoms, water molecules, sorry, in all of the oceans is 4 .54 times 10 to the 46 atoms.

04:53 So no doubt a huge amount.

05:00 But it this is the number of water molecules, but we don't want the number of water molecules, but a number of deuterium atoms.

05:07 So we can do a little bit of rearranging here.

05:09 So in each one water molecule, we have two hydrogen atoms.

05:14 So if you multiply the number of water molecules by two, that gives us a number of hydrogen.

05:20 However, the number of deuterium atoms, it's given by the ratio of 0 .03%, so 0 .03 over 100 equals the number of hydrogen atoms.

05:34 So finally therefore we now have an expression for a number of bacterium atoms and now this equation is now whole so we almost write that out so the energy total that was released in the scenario is the energy per reaction times n d which is given by the line above 0 .03 over 100 times n h which is two times the number of water molecules but i actually recognize from this equation here, this reaction, we need two deterrent atoms per reaction.

06:16 So actually that 2 is consequential...

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