Question

Imagine that a reactor converts all given mass into energy and that it operates at a power level of ${10^9}W$. The mass of the fuel consumed per hour in the reactor will be (velocity of light is $c = 3 \times {10^8}m/s$)
A. $0.8g$
B. $0.96g$
C. $4 \times {10^{ - 2}}g$
D. $6.6 \times {10^{ - 5}}g$

Answer 1

Hint First calculate the energy released per hour and then from that value of energy, using energy-mass equivalence of relativity, calculate the mass of the fuel consumed.
Formulas used:
$P = \dfrac{E}{t}$ where $P$ is the power, $E$ is the energy or work done and $t$ is time.
$E = m{c^2}$ where $E$ is the relativistic energy, $m$is the mass and $c$is the speed of light.

Complete step by step answer
Einstein’s mass-energy equivalence equation shows that mass and energy are equivalent and are interchangeable. So, instead of two separate laws of conservation in classical mechanics, one for the mass and the other for energy, we should replace them in relativity by a single one, namely the law of conservation of mass-energy or total relativistic energy.
We are given $P = {10^9}W$and in one hour the energy released is
$P = \dfrac{E}{t} \\\ \Rightarrow E = P \times t \\\ \Rightarrow E = {10^9} \times 3600J \\\$
By Einstein’s mass-energy equivalence we know
$E = m{c^2} \\\ \Rightarrow m = \dfrac{E}{{{c^2}}} = \dfrac{{{{10}^9} \times 3600}}{{{{\left( {3 \times {{10}^8}} \right)}^2}}} \\\ \Rightarrow m = 4 \times {10^{ - 5}}kg = 4 \times {10^{ - 2}}g \\\$
Therefore, the $4 \times {10^{ - 2}}g$of fuel is consumed per hour by the reactor.

So the correct answer is option C.

Note The mass-energy equivalence of relativity is however only applicable to relativistic particles (particles whose velocity is close to $c$). At low speeds $v < < c$and the relativistic expression reduces to the classical one. The relativistic energy is the sum of kinetic energy and the rest mass energy whereas the total energy in classical mechanics is equal to the sum of kinetic energy and potential energy.