Q63PE
Question
(a) Using data from Table \(7.1\), find the mass destroyed when the energy in a barrel of crude oil is released.
(b) Given these barrels contain \(200\) litres and assuming the density of crude oil is\(750{\rm{ }}kg/{m^3}\), what is the ratio of mass destroyed to original mass, \(\Delta m/m\)?
Step-by-Step Solution
Verifieda. The mass destroyed is obtained as: \(6.5{\rm{ }}\times{\rm{ }}{10^{ - 8}}{\rm{ }}kg\).
b. The ratio of the mass destroyed to original mass is obtained as:\(4.4{\rm{ }}\times{\rm{ }}{10^{ - 10}}\).
Quantity of crude oil is:
\(200litres\)
Density of crude oil is:\(750kg/{m^3}\).
Regarding the nature and behaviour of light, space, time, and gravity, relativity is the idea that different physical events rely on the relative motion of the observer and the observed objects.
a. The energy of one barrel of oil is:
\(E{\rm{ }} = {\rm{ }}5.9{\rm{ }} \times {\rm{ }}{10^9}{\rm{ }}J\).
The speed of light is:\(c{\rm{ }} = {\rm{ }}3{\rm{ }}\times{\rm{ }}{10^8}{\rm{ }}m/s\).
So, the mass destroyed is obtained as:
\(\begin{aligned}\Delta m{\rm{ }} = {\rm{ }}\dfrac{E}{{{c^2}}}\\ = \dfrac{{5.9{\rm{ }} \times {\rm{ }}{{10}^9}{\rm{ }}J}}{{{{(3{\rm{ }}\times{\rm{ }}{{10}^8}{\rm{ }}m/s)}^2}}}\\ = 6.5{\rm{ }}\times{\rm{ }}{10^{ - 8}}{\rm{ }}kg\end{aligned}\)
Therefore, the mass destroyed is:\(6.5{\rm{ }}\times{\rm{ }}{10^{ - 8}}{\rm{ }}kg\).
b. The values we have:
So, the mass obtained is:
The ration then will be:
\(\begin{aligned}\dfrac{{\Delta m}}{m} = \dfrac{{6.5{\rm{ }}\times{\rm{ }}{{10}^{ - 8}}{\rm{ }}kg}}{{150kg}}\\ = 4.4{\rm{ }}\times{\rm{ }}{10^{ - 10}}\end{aligned}\)
Therefore, the ratio is: \(4.4{\rm{ }}\times{\rm{ }}{10^{ - 10}}\).