Problem 87 - Chemistry: The Central Science

Question

The average energy released in the fission of a single uranium-235 nucleus is about \(3 \times 10^{-11} \mathrm{~J}\). If the conversion of this energy to electricity in a nuclear power plant is \(40 \%\) efficient, what mass of uranium- 235 undergoes fission in a year in a plant that produces 1000 megawatts? Recall that a watt is \(1 \mathrm{~J} / \mathrm{s}\).

Step-by-Step Solution

Answer

The mass of uranium-235 that undergoes fission in a year in a plant that produces 1000 megawatts is found by following the steps outlined: 1. Calculate the energy produced in a year: \(1 \times 10^9 \, \mathrm{J/s} \times 31,536,000 \, \mathrm{s}\) 2. Calculate the energy from uranium fission considering efficiency: \((1 \times 10^9 \, \mathrm{J/s} \times 31,536,000 \, \mathrm{s}) / 0.4\) 3. Find the number of uranium-235 nuclei undergoing fission: \(\frac{(1 \times 10^9 \, \mathrm{J/s} \times 31,536,000 \, \mathrm{s}) / 0.4}{3 \times 10^{-11} \, \mathrm{J}}\) 4. Calculate the mass of uranium-235: \(\frac{ \frac{(1 \times 10^9 \, \mathrm{J/s} \times 31,536,000 \, \mathrm{s}) / 0.4}{3 \times 10^{-11} \, \mathrm{J}}}{6.022 \times 10^{23}} \times 235 \, \mathrm{g/mol}\) After performing the calculations, the mass of uranium-235 that undergoes fission in a year in a 1000 MW plant is approximately 357,097 g or 357.1 kg.

Step 1: Calculate the total energy produced by the plant in a year

In this power plant, 1000 megawatts of power are produced. Since 1 watt = 1 J/s, we have: Power output = 1000 MW * 1,000,000 (to convert MW to watts) = \(1 \times 10^9\) W In a year, there are 31,536,000 seconds (365 days * 24 hours * 60 minutes * 60 seconds). Therefore, the total energy produced by the plant in a year is: Energy produced in a year = Power output * Time Energy produced in a year = \(1 \times 10^9 \, \mathrm{J/s} \times 31,536,000 \, \mathrm{s}\)

Step 2: Calculate the energy produced by uranium-235 fission

Given that the efficiency of converting the fission energy to electricity is 40%, we can calculate the total energy produced by the uranium-235 fission by dividing the energy produced in a year by the efficiency: Energy from uranium fission = Energy produced in a year / Efficiency Energy from uranium fission = \((1 \times 10^9 \, \mathrm{J/s} \times 31,536,000 \, \mathrm{s}) / 0.4\)

Step 3: Find the number of uranium-235 nuclei undergoing fission

The average energy released in the fission of a single uranium-235 nucleus is \(3 \times 10^{-11}\) J. We can find the number of uranium-235 nuclei that must undergo fission to release the calculated energy from step 2: Number of uranium nuclei = Energy from uranium fission / Energy per nucleus Number of uranium nuclei = \(\frac{(1 \times 10^9 \, \mathrm{J/s} \times 31,536,000 \, \mathrm{s}) / 0.4}{3 \times 10^{-11} \, \mathrm{J}}\)

Step 4: Calculate the mass of uranium-235

The mass number of uranium-235 is 235, which means that one mole of uranium-235 has a mass of 235 grams. Since there are \(6.022 \times 10^{23}\) atoms in a mole (Avogadro's number), we can find the mass of uranium-235 required to have the calculated number of nuclei: Mass of uranium-235 = Number of uranium nuclei * Mass of one uranium atom Mass of uranium-235 = \(\frac{ \frac{(1 \times 10^9 \, \mathrm{J/s} \times 31,536,000 \, \mathrm{s}) / 0.4}{3 \times 10^{-11} \, \mathrm{J}}}{6.022 \times 10^{23}} \times 235 \, \mathrm{g/mol}\) Now, you can calculate the mass of uranium-235 that undergoes fission in a year in a plant that produces 1000 megawatts.

Key Concepts

Uranium-235 Fission

Uranium-235 (\textsuperscript{235}U) fission is a nuclear reaction in which the nucleus of a \textsuperscript{235}U atom splits into smaller parts, releasing a significant amount of energy. During the process, a neutron collides with the uranium nucleus, causing it to become unstable and split. This fission reaction releases energy in the form of heat and additional neutrons. These newly released neutrons can then initiate further fission reactions in a chain reaction.

Each fission event releases a quantifiable amount of energy, approximately 200 MeV, which is equivalent to around \(3 \times 10^{-11}\) joules. This energy is captured in a nuclear power plant and converted to electricity through various processes, including the use of steam turbines. The concept of uranium-235 fission is critical in understanding how nuclear power plants generate electricity and provides the foundation on which calculations of fuel requirements, like the mass of uranium needed for a given energy output, are based.

Nuclear Power Plant Efficiency

The efficiency of a nuclear power plant refers to its ability to convert the energy released from nuclear fission into electrical energy. The efficiency is a percentage value that indicates what portion of the fission energy is actually converted into usable electricity. The remaining energy is not wasted; it is typically lost as heat, which cannot be utilized to produce electricity.

In our exercise, the conversion efficiency is given as 40%. This means that for every 100 joules of energy released by fission, only 40 joules are turned into electricity, and 60 joules are dissipated as heat. The efficiency factor is crucial when calculating the amount of fuel needed. Higher efficiency means less fuel is required to produce the same amount of electricity. It's important to note that while 40% may not seem very high, it is within the typical range for thermal efficiency in power plants using the Rankine or Brayton cycle.

Calculating Mass of Uranium-235

To calculate the mass of uranium-235 needed for a certain energy output, we must use the information provided about the energy released per fission event and the efficiency of the power plant. After determining the total electrical energy a plant is to produce in a given period, like a year, we account for the efficiency to find the total energy that must be released by fission reactions.

Since we know the energy released per fission of a single \textsuperscript{235}U atom, we can calculate the total number of fissions required to produce that energy. With Avogadro's number, we can relate the number of uranium atoms to a mass quantity. The mass number of \textsuperscript{235}U indicates that one mole (or \(6.022 \times 10^{23}\) atoms) has a mass of 235 grams. By using these values, we can calculate the mass of \textsuperscript{235}U that would undergo fission in a given period, allowing nuclear engineers to determine the amount of nuclear fuel needed for operation.