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 \(1000 \mathrm{MW}\) (megawatts)? Recall that a watt is \(1 \mathrm{~J} / \mathrm{s}\).

Short answer

The mass of uranium-235 required to produce 1000 MW of power for a year with an efficiency of 40% can be calculated using the steps provided: 1. Total energy produced in a year: \(E_{year} = (1000 \times 10^6\ \mathrm{W}) \times (365 \times 24 \times 60 \times 60\ \mathrm{s})\) 2. Total energy required from uranium fission: \(E_{uranium} = \frac{E_{year}}{0.40}\) 3. Number of fission events required: \(N = \frac{E_{uranium}}{3 \times 10^{-11}\ \mathrm{J}}\) 4. Mass of one uranium-235 atom: \(m_{U-235} \approx 235 \times (1.66 \times 10^{-27}\ \mathrm{kg})\) 5. Total mass of uranium-235 required: \(m_{required} = N \times m_{U-235}\) Using these steps and calculations, we can determine the mass of uranium-235 needed for the given power plant.

Step-by-step solution

Convert the power output to energy produced in a year

To find the total energy produced in a year, we need to multiply the power output by the number of seconds in a year. The power output is given as 1000 MW, which is equivalent to \(1000 \times 10^6\ \mathrm{W}\). There are 365 days in a year, 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute, so there are \(365 \times 24 \times 60 \times 60\ \mathrm{s}\) in a year. Total energy produced in a year: \[E_{year} = (1000 \times 10^6\ \mathrm{W}) \times (365 \times 24 \times 60 \times 60\ \mathrm{s})\]

Find the total energy required from uranium fission events

Given that the efficiency of the power plant is 40%, the energy produced by the plant is 40% of the total energy obtained from uranium fission events. Therefore, we need to find the total energy required from uranium fission events, which can be calculated by dividing the total energy produced in a year by the power plant's efficiency. Total energy required from uranium fission: \[E_{uranium} = \frac{E_{year}}{0.40}\]

Calculate the required number of fission events

We know that the average energy released per fission event of uranium-235 is \(3 \times 10^{-11}\ \mathrm{J}\). So, we can find out how many fission events are needed by dividing the total energy required from uranium fission events by the average energy per fission event. Number of fission events required: \[N = \frac{E_{uranium}}{3 \times 10^{-11}\ \mathrm{J}}\]

Calculate the mass of uranium-235 required

Each fission of a uranium-235 nucleus results in one uranium-235 atom undergoing fission. To calculate the mass of uranium-235 required, we need to multiply the number of fission events by the mass of one atom of uranium-235. First, we need to find the mass of one uranium-235 atom. The atomic mass of uranium-235 is approximately \(235\ \mathrm{u}\), where \(1\ \mathrm{u} \approx 1.66 \times 10^{-27}\ \mathrm{kg}\). So, the mass of one uranium-235 atom is: \[m_{U-235} \approx 235 \times (1.66 \times 10^{-27}\ \mathrm{kg})\] Then, we can calculate the total mass of uranium-235 required as follows: \[m_{required} = N \times m_{U-235}\] By calculating these values, we will find the mass of uranium-235 needed to produce 1000 MW of power for a year with an efficiency of 40%.

Key concepts

Uranium-235

Uranium-235 is a critical element used in nuclear fission. It is a specific isotope of uranium and has 92 protons and 143 neutrons.
This makes its atomic mass 235, hence the name Uranium-235. It is a heavy metal used as fuel in nuclear reactors and is capable of sustaining a chain reaction of nuclear fission, which is crucial for energy production in nuclear power plants.

The process of fission involves splitting the nucleus of an atom, which releases a significant amount of energy. When Uranium-235 undergoes fission, it splits into smaller atomic nuclei and releases energy, which is then converted to electricity in a nuclear power plant. This significant release of energy makes Uranium-235 a valuable resource for power generation.

• Uranium-235 is less common than its stable counterpart, Uranium-238, comprising about 0.7% of natural uranium.
• It is the isotope used in nuclear reactors due to its ability to sustain a chain reaction.

Energy Conversion Efficiency

Energy conversion efficiency is a measure of how well energy is transformed from one form to another. In the context of a nuclear power plant, it refers to the efficiency with which nuclear energy is converted into electrical energy.
A typical nuclear power plant's efficiency may be around 40%, as mentioned in our exercise. This means that 40% of the energy produced during the uranium fission is converted into electricity, while the remaining 60% might be lost as heat during the conversion process.

Understanding this concept is important because it helps in determining how much fuel, like Uranium-235, is required to achieve a desired amount of electricity.

• Efficiency impacts the amount of fuel needed: Higher efficiency requires less fuel.
• Improving efficiency can lead to more environmentally friendly power production by reducing waste and resource use.

Nuclear Power Plant

A nuclear power plant is a facility that uses nuclear reactions, primarily fission, to generate electricity. It utilizes Uranium-235 as a fuel source.
In these plants, Uranium-235 undergoes fission in the reactor core, releasing energy in the form of heat. This heat is then used to produce steam, which drives a turbine connected to a generator that produces electricity.

The design of a nuclear power plant includes several components to safely manage the fission process and convert energy efficiently.

• Reactor Core: Contains the nuclear fuel where fission occurs.
• Cooling Systems: Used to remove heat and prevent overheating.
• Containment Structure: Ensures that no radiation escapes into the environment.

The ability to produce large amounts of energy with a relatively small amount of fuel makes nuclear power plants a popular choice for regions with high energy demands.

Atomic Mass Unit

The atomic mass unit (u or amu) is a standard unit of mass used to express atomic and molecular weights. It is defined as one twelfth of the mass of a carbon-12 atom.
In nuclear physics, it is a convenient way to express the masses of atoms and other small particles, such as protons, neutrons, and electrons.

When calculating the mass of an atom, like Uranium-235, atomic mass units are necessary. For instance, the mass of one Uranium-235 atom is approximately 235 atomic mass units. To convert this to kilograms, which is a more practical unit for large-scale calculations like those needed for power plants, we use the conversion factor:\(1 \ \text{u} \approx 1.66 \times 10^{-27} \ \text{kg}\).

• This conversion is essential for calculating the mass of Uranium-235 needed for energy production.
• In calculations, it simplifies the process of working between atomic and macroscopic scales of measurement.

Understanding atomic mass units is crucial for the accurate calculation of the mass of nuclear fuels and other substances on an atomic scale.