Problem 51 - Chemistry The Central Science

Question

The energy from solar radiation falling on Earth is \(1.07 \times 10^{16} \mathrm{kJ} / \mathrm{min.}\) (a) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (\mathbf{b} )If the energy released in the reaction \begin{equation}^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \longrightarrow_{56}^{141} \mathrm{Ba}+_{36}^{92} \mathrm{Kr}+3_{0}^{1} \mathrm{n} \end{equation}\(\left(^{235} \mathrm{U}\right.\)nuclear mass,234.9935 amu; \(^{235} \mathrm{Ba}.\) nuclear mass, 140.8833 amu; \(^{92} \mathrm{Kr}\) nuclear mass, 91.9021 amu) is taken as typical of that occurring in a nuclear reactor, what mass of uranium-235 is required to equal 0.10\(\%\) of the solar energy that falls on Earth in 1.0 day?

Step-by-Step Solution

Answer

The loss of mass from the Sun in one day from just the energy falling on Earth is \(1.71 * 10^7\) kg. The required mass of uranium-235 to equal 0.10% of the solar energy that falls on Earth in one day is 130000 kg.

Step 1: Calculate the energy from solar radiation in one day

To find the total energy from solar radiation in one day, we simply multiply the given energy per minute by the number of minutes in a day (1 day = 1440 minutes): Energy per day = 1440 (minutes in a day) * 1.07 * 10^16 (kJ/minute) = \(1.54 * 10^{19}\) kJ

Step 2: Calculate loss of mass from the Sun (part a)

We'll use the equation E=mc^2 to find the mass loss: Here, E = Energy (in Joules), m is the mass loss (in kg), and c is the speed of light (\(3 * 10^8\) m/s). Since we need to convert energy from kJ to J, let's do that first: E = \(1.54 * 10^{19}\) kJ * 1000 (J/kJ) = \(1.54 * 10^{22}\) J Now, rearrange the equation to solve for mass loss: m = E / c^2 m = \(\frac{1.54 * 10^{22}}{(3 * 10^8)^2}\) kg = \(1.71 * 10^7\) kg So, the loss of mass from the Sun in one day from just the energy falling on Earth is \(1.71 * 10^7\) kg.

Step 3: Find the energy released in the nuclear reaction (part b)

First, we need to find the total mass change during the reaction. We'll do that by comparing the mass of reactants and products: Mass change = (Mass of uranium-235 + Mass of neutron) - (Mass of Ba + Mass of Kr) Mass change = (234.9935 + 1) - (140.8833 + 91.9021) = 3.2081 amu Since 1 amu = 1.66054 * 10^-27 kg, our mass change in kg is: Mass change = 3.2081 amu * 1.66054 * 10^-27 kg/amu = \(5.33 * 10^{-27}\) kg Now, we can use E=mc^2 to find the energy released: E = (\(5.33 * 10^{-27}\) kg) * (\(3 * 10^8\) m/s)^2 = \(4.79 * 10^{-11}\) J

Step 4: Calculate mass of uranium-235 required for 0.10% of solar energy to Earth (part b)

Now, let us find 0.10% of the energy received from the Sun in one day: 0.10% of solar energy = \(1.54 * 10^{19}\) kJ * 0.001 * 1000 J/kJ = \(1.54 * 10^{16}\) J We can then find the number of uranium-235 reactions needed to produce this energy: Number of reactions = \(\frac{1.54 * 10^{16} J}{4.79 * 10^{-11} J}\) = \(3.21 * 10^{26}\) reactions Finally, we can find the mass of uranium-235 required for the desired energy: Mass of uranium-235 = (Number of reactions) * (Mass of uranium-235 per reaction) Mass of uranium-235 = (\(3.21 * 10^{26}\) reactions) * (234.9935 amu * 1.66054 * 10^-27 kg/amu) = 130000 kg So, the required mass of uranium-235 to equal 0.10% of the solar energy that falls on Earth in one day is 130000 kg.

Key Concepts

Solar Radiation Energy

Solar radiation energy is the power that we receive from the Sun, which provides the Earth with the light and warmth essential for life. This immense energy originates from nuclear reactions, specifically fusion, occurring within the Sun's core, where hydrogen atoms combine to form helium, releasing energy in the process. The amount of solar radiation energy that reaches the Earth is quantified by measuring the radiant energy per minute. For example, it's reported that Earth receives about 1.07 x 10¹⁶ kilojoules (kJ) per minute. This energy is vital for numerous natural processes, including climate and weather systems, plant photosynthesis, and it's also harnessed through solar panels for electricity.

E=mc²

The equation E=mc² is one of the most recognized equations in physics, formulated by Albert Einstein as part of his theory of special relativity. It describes the relationship between energy (E), mass (m), and the speed of light (c). This profound equation reveals that mass can be converted into energy and vice versa, implying that a small amount of mass can be transformed into a large amount of energy. In the context of solar radiation, E=mc² helps us calculate how much mass the Sun loses when it emits solar radiation. Despite the seemingly large mass loss, it is actually minuscule in comparison to the Sun's total mass, indicating that the Sun will continue to radiate energy for billions of years.

Nuclear Reaction

A nuclear reaction involves a change in an atom's nucleus and often includes the transformation of elements. In the core of stars like the Sun, nuclear fusion is the predominant process, where lighter elements merge to create heavier ones. On Earth, in nuclear reactors, a different kind of nuclear reaction takes place, known as fission. Uranium-235 is often the fuel for these reactions. When a neutron hits the nucleus of a Uranium-235 atom, it splits into smaller elements, releasing additional neutrons and a significant amount of energy. This energy can then be harnessed for electricity generation. The reaction's byproducts are lighter elements and more neutrons, which can propagate the reaction further if conditions allow.

Uranium-235

Uranium-235 (U-235) is an isotope of uranium that is crucial for both nuclear reactors and nuclear weapons because of its ability to sustain a chain reaction. Natural uranium contains about 0.7% of Uranium-235, with the rest being Uranium-238, which is less fissionable. To be used in reactors, the U-235 content must be increased through a process called enrichment. During a fission reaction, U-235 is bombarded with neutrons, causing the nucleus to split and release more neutrons, as well as a tremendous amount of energy. This energy release is the foundation of nuclear power. For example, to match a tiny fraction of the solar energy Earth receives in a day, a massive amount of U-235 would be required, highlighting the concentrated power of nuclear fuel.