Problem 60 - 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 encrgy falling on Farth? (b) If the energy released in the reaction $$ { }^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n} $$ \(\left({ }^{235} \mathrm{U}\right.\) nuclear mass, \(234.9935 \mathrm{amu} ;{ }^{141} \mathrm{Ba}\) nuclear mass, \(140.8833 \mathrm{amu} ;{ }^{92} \mathrm{Kr}\) 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 mass loss from the Sun in one day due to the energy falling on Earth is calculated using mass-energy equivalence, and is found to be \( \frac{\text{Total energy in a day}}{c^2} \). To find the mass of uranium-235 required to produce 0.10% of the solar energy in a day, first calculate 0.10% of the total solar energy, then find the energy released per uranium-235 reaction, and determine the number of reactions needed to produce that energy. Finally, obtain the mass of uranium-235 required for these reactions by multiplying the number of reactions by the mass of one uranium-235 atom.

Step 1: Part (a): Mass loss from the Sun in one day

We are given the energy falling on Earth per minute due to solar radiation. We need to find the mass loss from the Sun in one day, which accounts for that energy. First, we will find the total energy in a day. Then, we can use the mass-energy equivalence to find the corresponding mass loss. 1. Calculate the total energy falling on Earth in one day: Energy rate = 1.07 x 10^16 kJ/min Convert the energy rate to joules: Energy rate (in Joules) = 1.07 x 10^16 x 1000 J/min There are 60 minutes in an hour and 24 hours in a day, so: Total energy in a day = Energy rate (in Joules) x 60 min/hour x 24 hour/day 2. Find the mass loss from the Sun using mass-energy equivalence (E=mc^2): Mass loss = \( \frac{E}{c^2} \) Where c is the speed of light = \(3 \times 10^8 ms^{-1} \) Mass loss from the Sun in one day = \( \frac{\text{Total energy in a day}}{c^2} \)

Step 2: Part (b): Mass of uranium-235 required to produce 0.10% of the solar energy

Next, we need to determine the mass of uranium-235 that would need to undergo the given nuclear reaction in order to produce 0.10% of the solar energy falling on Earth in one day. 1. Calculate 0.10% of the total solar energy in a day: Solar energy = Total energy in a day x 0.001 2. Calculate the energy released per uranium-235 reaction: Find the mass change (Δm) for the reaction: Δm = [mass of uranium-235 - (mass of barium-141 + mass of krypton-92)] x 1 amu (amu = atomic mass unit) Energy released per reaction (E_reaction) = Δm x c^2 3. Find the number of reactions required to produce 0.10% of the solar energy: Number of reactions = \( \frac{\text{Solar energy}}{\text{E_reaction}} \) 4. Calculate the mass of uranium-235 required for these reactions: Mass of uranium-235 required = Number of reactions x mass of one uranium-235 atom Finally, we have found the mass of uranium-235 needed to produce 0.10% of the solar energy that falls on Earth in 1.0 day.

Key Concepts

Solar Energy

The Sun is a massive source of energy for Earth, primarily due to the solar radiation it emits. Every minute, Earth receives an incredible amount of energy from the Sun, approximately \(1.07 \times 10^{16}\) kilojoules. This energy plays a vital role in maintaining life on our planet. However, have you ever wondered how the Sun loses mass because of this energy emission?

According to Einstein's mass-energy equivalence principle, expressed as \(E = mc^2\), energy (\(E\)) and mass (\(m\)) are interconvertible. Here, \(c\) is the speed of light \(3 \times 10^8 \text{ m/s}\). This means that the Sun loses a tiny amount of mass each time it emits solar energy.

Let's think about it daily. The energy Earth receives in one day from the Sun totals to a large number when calculated by multiplying the minute rate by 60 (to get hours) and then by 24 (to get days). This daily energy when used in \(E = mc^2\) can calculate the Sun's mass loss. While significant over millennia, the daily mass loss is minuscule compared to the Sun's total mass.

Nuclear Reactions

Nuclear reactions are processes where nuclei of atoms are transformed, leading to the release or absorption of energy. These reactions often involve heavy elements, such as Uranium-235, undergoing fission. Fission is a form of a nuclear reaction where a heavy nucleus splits into smaller nuclei with the emission of neutrons and a large amount of energy.

In the given nuclear reaction, Uranium-235 absorbs a neutron and splits into Barium-141, Krypton-92, and three free neutrons. This process releases energy due to the conversion of a small amount of mass into energy, again echoing Einstein's mass-energy equivalence \(E = mc^2\).

• The original mass of Uranium-235 and a neutron is slightly more than the mass of the resultant products (Barium, Krypton, neutrons).
• This difference in mass (known as mass defect) becomes the energy released.

Nuclear reactors exploit such reactions to generate electricity due to their high energy output relative to the mass of fuel used.

Uranium-235 Calculations

Calculations involving Uranium-235 often revolve around determining how much is needed to produce a specified amount of energy. This is especially important in comparing nuclear energy to other sources, such as solar.

If we wish to match just a fraction (say 0.10%) of Earth's daily solar energy through Uranium-235, we must first determine how much energy is released in a typical Uranium-235 fission reaction. Each reaction has a mass defect, calculated from the atomic masses of the reactants and products. This mass defect, when put through \(E = mc^2\), gives us the energy per fission event.

• First, calculate 0.10% of the solar energy received.
• Then, apply the reaction energy formula to determine the number of fission reactions necessary to produce that energy.
• Finally, we derive the total mass of Uranium-235 required by multiplying the number of reactions by the mass of a single Uranium-235 atom.

This process outlines not only the calculation but highlights the potency of nuclear energy compared to other energy forms.