Problem 82 - 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 undergoing fission in a year in a plant producing 1000 megawatts can be determined by following these steps: 1. Calculate the total energy output in a year: \(E_{output} = (1000 \times 10^6 \mathrm{W}) \times (31,536,000 \mathrm{s})\) 2. Determine the total energy input needed for that output: \(E_{input} = E_{output} / 0.40\) 3. Find the required number of uranium-235 nuclei: \(N = E_{input} / (3 \times 10^{-11} \mathrm{J})\) 4. Determine the mass of uranium-235 needed for the required fission events: \(m_{U-235} = N \times (235 \times 1.66 \times 10^{-27} \mathrm{kg})\) After calculating these values, we can find the mass of uranium-235 undergoing fission in a year in the plant producing 1000 megawatts.

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

First, we need to find the total energy produced by the nuclear power plant in a year, considering its capacity (1000 megawatts). We'll start by converting the plant's power output to watts and then multiply by the number of seconds in a year. 1000 megawatts = 1000 × 10^6 W There are 365 days in a year, 24 hours per day, 60 minutes per hour, and 60 seconds per minute. Therefore, Total seconds in a year = 365 × 24 × 60 × 60 = 31,536,000 s Total energy produced in a year = Power output × Time E_output = (1000 × 10^6 W) × (31,536,000 s)

Step 2: Determine the total energy input (total energy from uranium-235)

The energy input needed for the given output can be calculated using the efficiency of conversion. Efficiency = (Energy output) / (Energy input) We are given an efficiency of 40%, or 0.40. Thus, we can solve for the energy input (E_input): E_input = E_output / Efficiency E_input = E_output / 0.40

Step 3: Find the number of uranium-235 nuclei needed for that energy input

We are given that a single uranium-235 nucleus releases an average energy of \(3 \times 10^{-11} J\). To find the total number of uranium-235 nuclei needed to provide the energy input, we will divide the total energy input by the energy released per uranium-235 nucleus. Number of uranium-235 nuclei = E_input / (Energy released per nucleus) N = E_input / \(3 \times 10^{-11} J\)

Step 4: Determine the mass of uranium-235 needed for required fission events

To find the mass of uranium-235 that corresponds to the required number of fission events, we need to know the mass of a single uranium-235 nucleus. The atomic mass of uranium-235 is approximately 235 u, where u (unified atomic mass unit) is \(1.66 \times 10^{-27} kg\). Mass of one uranium-235 nucleus = 235 u × \(1.66 \times 10^{-27} kg/u\) Now, we can find the mass of uranium-235 needed for the required fission events: Total mass of uranium-235 = Number of uranium-235 nuclei × Mass of one uranium-235 nucleus By following the steps above and plugging in the numbers provided, the mass of uranium-235 undergoing fission in a year in a plant producing 1000 megawatts can be determined.

Key Concepts

Uranium-235 Fission

In order to understand how a nuclear power plant works, it is essential to start with the basics of uranium-235 fission. Fission is the process by which the nucleus of an atom splits into two or more smaller nuclei, along with a few neutrons and a large amount of energy. For uranium-235, the nucleus captures a slow-moving neutron and becomes momentarily excited and unstable. It then splits into two smaller nuclei and releases more neutrons, which can initiate a chain reaction.

This reaction is the heart of a nuclear power plant, as it provides the heat necessary to produce steam, which then drives the turbines to generate electricity. The energy released in each fission event of uranium-235 is approximately \(3 \times 10^{-11} \) joules. This may not sound like much, but the huge number of atoms undergoing fission in a power plant produces a tremendous amount of energy.

Key Points of Uranium-235 Fission

• Uranium-235 fission is a chain reaction that releases energy needed for electricity generation.
• Each fission event releases a significant amount of energy, around \(3 \times 10^{-11} \) joules.
• The chain reaction is carefully controlled within a nuclear reactor to manage the energy output.

Nuclear Power Plant Efficiency

When discussing the efficiency of a nuclear power plant, we are referring to how well the plant converts the energy released from nuclear fission into electrical energy. However, not all the energy can be converted due to energy losses, primarily in the form of heat. This is a key concept in understanding energy conversion in nuclear facilities.

The efficiency of a nuclear power plant is typically around 30-40%. This means that only 30-40% of the energy released by fission is actually turned into electricity, while the rest is lost. The efficiency rate plays a crucial role in determining how much fuel is used and how much waste is produced, both of which have significant environmental and economic implications.

Factors Affecting Efficiency

• Thermodynamic limitations - Not all heat can be converted to electricity.
• Engineering design - The design of the plant can influence how efficiently heat is converted.
• Operational practices - How the plant is run can also impact efficiency.

Energy Conversion in Nuclear Power

Energy conversion in nuclear power is a multi-step process that starts with the fission of uranium-235 and ends with the generation of electricity. The conversion process embodies the physics and engineering principles that transform nuclear energy into usable power.

First, the heat produced by fission is used to create steam from water. The steam is then directed to spin turbines connected to generators. As these turbines rotate, they convert the mechanical energy of the spinning action into electrical energy through electromagnetic induction.

However, in all energy conversions, some energy is inevitably lost. In a nuclear power plant, these losses occur in various forms such as heat loss through the reactor's cooling system and frictional losses within the turbines and generators. Thus, the efficiency of energy conversion is a critical factor in the overall output and environmental footprint of a nuclear power plant.

Steps of Energy Conversion

• Nuclear fission produces heat.
• Heat produces steam.
• Steam spins turbines.
• Turbines drive generators to produce electricity.