Question

(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\(1000{\rm{ }}MW\)of electric power?

(b) Do you think it would be possible to observe this mass loss if the total mass of the fuel is\({10^4}{\rm{ }}kg\)?

Short answer

1. The total mass being destroyed is obtained as:\(1\;kg\).
2. The observation of the mass loss is insignificant.

Step-by-step solution

Given Data

The value of total mass of the fuel is:

\({10^4}{\rm{ }}kg\)

The amount of electric power needed to be produced is:\(1000{\rm{ }}MW\).

The efficiency by which the energy is converted from nuclear fission is: \(35.0\% \).

Define Special Relativity

The special theory of relativity, sometimes known as special relativity, is a physical theory that describes how space and time interact. Theoretically, this is known as STR theory.

Evaluating the mass destroyed

a. Total energy that the power plant will produce in one year is obtained as:

\(\begin{aligned}{E_1} &= (1000MW)\left( {3.15 \times {{10}^7}\;s} \right)\\ & = \left( {1000 \times {{10}^6}\;W} \right)\left( {3.15 \times {{10}^7}\;s} \right)\\ & = 3.15 \times {10^{16}}\;J\end{aligned}\)

As, this is said to be the \(35.0\% \) of the total energy that is produced during mass destruction.

Then, the total energy produced with the help of the mass is obtained as:

\(\begin{aligned}{E_{tot}} & = \dfrac{{{E_1}}}{{35\% }}\\ & = \dfrac{{\left( {3.15 \times {{10}^{16}}\;J} \right)}}{{0.35}}\\ & = 9.00 \times {10^{16}}\;J\end{aligned}\)

So, the total mass that is destroyed is obtained as:

\(\begin{aligned}{m_u} & = \dfrac{{{E_{tot{\rm{ }}}}}}{{{c^2}}}\\ & = \dfrac{{\left( {9.00 \times {{10}^{16}}\;J} \right)}}{{{{\left( {3 \times {{10}^8}\;m/s} \right)}^2}}}\\& = 1\;kg\end{aligned}\)

Therefore, the mass destroyed is: \(1\;kg\).

Explanation for part b

b. As the entire mass of the fuel is\({10^4}\;kg\), and only \(1\;kg\) is destroyed. As a result, the destroyed mass is four orders of magnitude smaller than the entire mass of the fuel. As a result, it is insignificant.

Therefore, the result obtained is negligible.