(a) A nuclear power plant converts energy from nuclear fission into electricity with an efficiency of \(35.0 \%\) How much mass is destroyed in one year to produce a continuous 1000 MW of electric power? (b) Do you think it would be possible to observe this mass loss if the total mass of the fuel is \(10^{4} \mathrm{kg}\) ?

The energy-mass equivalence principle, immortalized by Einstein's equation, \( E=mc^2 \), tells us that energy can be converted into mass and vice versa. In the context of a nuclear power plant, this equation is crucial for understanding how a seemingly small amount of matter can be transformed into a tremendous amount of energy.

During nuclear fission, the nucleus of an atom splits into smaller parts, releasing a large amount of energy. This energy was originally part of the mass of the nucleus, according to the energy-mass equivalence principle. To calculate the mass that corresponds to a given amount of energy \(E\), we use the speed of light squared (\(c^2\)) as a conversion factor, where the speed of light \(c\) is approximately \(3.00\times 10^8 m/s\).

Here's the catch: not all of this energy gets turned into electricity. The rest is lost to the surroundings as waste heat. That's why the efficiency of the plant is so important—it tells us what fraction of the total energy produced is actually used for power generation.

Nuclear power plant efficiency is a measure of how well a plant converts nuclear energy into electrical energy. Typically, not all the energy released during fission can be turned into electricity. A significant portion is dissipated as heat. Efficiency is expressed as a percentage and given by the ratio of useful electrical energy output to the total nuclear energy produced during fission.

For example, with an efficiency of \(35.0\text{%}\), a nuclear power plant only converts \(35.0\text{%}\) of the nuclear fission energy into electric power. The remaining \(65\text{%}\) is lost, primarily as heat. To accurately calculate the mass loss in a power plant, one must take this efficiency into account, meaning the energy considered for mass loss conversion via \(E=mc^2\) equation is only a portion of the total energy unleashed by the nuclear fuel.

Electric power generation in nuclear plants involves converting the energy released from nuclear fission into electricity that we can use in our homes and industries. This is done using turbines and generators. When the nucleus of an atom splits during fission, it releases heat. This heat is then used to produce steam, which turns the turbines. The spinning turbines drive generators, thus converting mechanical energy into electrical energy.

Let's relate this to our problem. If a plant generates \(1000 \text{MW}\) of electric power consistently for a year, we can calculate the total amount of energy produced in that period. Since the plant’s efficiency is \(35.0\text{%}\), it only uses \(35.0\text{%}\) of the potential energy from fission to produce electricity. To find the total energy from fission, we would divide the electric energy by the efficiency—in our case, a simple mathematical adjustment gives a much larger figure for the initial nuclear energy. The key takeaway here is that a plant's capacity to generate power hinges on its efficiency and the total amount of energy its nuclear fuel can provide through fission.