Problem 51 - Chemistry The Central Science In Si Units

Question

The energy from solar radiation falling on Earth is \(1.07 \times 10^{16} \mathrm{~kJ} / \mathrm{min} .(\mathbf{a})\) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (b) If the energy released in the reaction $$ { }^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n} $$ \(\left(\begin{array}{c}235 \\ \mathrm{U} \text { nuclear mass, } 234.9935 \mathrm{u} ;{ }^{141} \mathrm{Ba} \text { nuclear mass, }\end{array}\right.\) \(140.8833 \mathrm{u} ;{ }^{92} \mathrm{Kr}\) nuclear mass, \(91.9021 \mathrm{u}\) ) 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

(a) 1.71 x 10^5 kg mass loss from the Sun. (b) 1.94 x 10^5 kg of uranium-235 required.

Step 1: Calculate Total Energy Received by Earth in One Day

First, determine the total energy received by Earth in one day. Given that Earth receives \(1.07 \times 10^{16}\) kJ per minute, we need to convert this amount into one day's worth of energy. Calculate the number of minutes in a day: \[ 1 \text{ day} = 24 \text{ hours} \times 60 \text{ minutes/hour} = 1440 \text{ minutes} \]Now multiply the energy per minute by the total number of minutes in a day:\[ 1.07 \times 10^{16} \text{ kJ/min} \times 1440 \text{ min} = 1.5408 \times 10^{19} \text{ kJ} \].

Step 2: Use Mass-Energy Equivalence to Find Mass Loss

Using Einstein's mass-energy equivalence principle, \(E = mc^2\), we can find the mass loss equivalent to this energy. Here, \(c\) is the speed of light, approximately \(3.00 \times 10^8 \text{ m/s}\), and \(E\) is the energy, which we just calculated.Convert the energy from kJ to joules:\[ 1.5408 \times 10^{19} \text{ kJ} = 1.5408 \times 10^{22} \text{ J} \]Next, rearrange the equation to solve for mass \(m\):\[ m = \frac{E}{c^2} = \frac{1.5408 \times 10^{22} \text{ J}}{(3.00 \times 10^8 \text{ m/s})^2} = 1.71 \times 10^5 \text{ kg} \].

Step 3: Determine Required Uranium-235 Mass

To find the mass of \(^{235}U\) required, we first need the energy of the nuclear reaction. Using the given nuclear masses, compute the mass defect:\[\text{Mass defect} = \left(234.9935 \text{ u} - (140.8833 + 91.9021) \text{ u}\right) = 0.2081 \text{ u}\]Convert this mass defect from atomic mass units (u) to energy in MeV using the conversion \(1 \text{ u} = 931.5 \text{ MeV}/c^2\): \[0.2081 \text{ u} \times 931.5 \frac{\text{MeV}}{c^2} = 193.75 \text{ MeV}\]Convert MeV to joules: \[193.75 \text{ MeV} = 193.75 \times 1.602 \times 10^{-13} \text{ J} \approx 3.1041 \times 10^{-11} \text{ J}\]To match \(0.10\%\) of solar energy falling on Earth, calculate the energy:\[0.001 \times 1.5408 \times 10^{22} \text{ J} = 1.5408 \times 10^{19} \text{ J}\]Find the number of uranium reactions required:\[\frac{1.5408 \times 10^{19} \text{ J}}{3.1041 \times 10^{-11} \text{ J/reaction}} \approx 4.96 \times 10^{29} \text{ reactions}\]Since each reaction consumes one \(^{235}U\) nucleus and 1 mole \( \approx 6.022 \times 10^{23} \) nuclei:\[\text{Moles of } ^{235}U = \frac{4.96 \times 10^{29}}{6.022 \times 10^{23}} \approx 8.24 \times 10^{5} \text{ moles}\]The molar mass of \(^{235}U \) is 235 g/mol, so:\[\text{Mass of } ^{235}U = 8.24 \times 10^{5} \times 235 \text{ g/mol} = 1.936 \times 10^{8} \text{ g} = 1.94 \times 10^{5} \text{ kg}\]

Step 4: Results Interpretation

Part (a) results in a solar mass loss of \(1.71 \times 10^5\) kg per day. For Part (b), \(1.94 \times 10^5\) kg of \(^{235}U\) is required to produce 0.10% of the solar energy received by Earth in one day.

Key Concepts

Solar Radiation Energy

The energy we receive from the sun is colossal.Each minute, the Earth absorbs around \(1.07 \times 10^{16}\) kJ from solar radiation.This energy is critical as it drives countless processes on Earth, including weather patterns, plant growth, and even human technology through solar panels.

Solar radiation energy is a result of nuclear reactions happening in the sun.These reactions convert mass into energy, as described by Einstein’s famous equation \(E = mc^2\).The energy permeating through space as sunlight gives life to our planet and keeps it warm enough to sustain life.

• Solar energy affects Earth's climate and ecosystems.
• It is a major source of renewable energy.
• Human applications include solar panels and solar water heaters.

Understanding how much energy the Earth receives is vital for assessing solar energy's role in our environment and potential uses in technology.

Nuclear Reaction Energetics

At the heart of nuclear energetics are the reactions that release or absorb energy.When nuclei come together or split apart, they release energy due to the conversion of mass into energy.

One such nuclear process is the basis for the energy computations in nuclear reactors and nuclear weapons.This energy, far greater than chemical reactions, stems from the conversion of just a tiny amount of mass due to the large multiplicative factor \(c^2\).These reactions transform the binding energy of nuclei into usable energy.

• Nuclear reactions can either fission (split) or fusion (combine) nuclei.
• The amount of energy released depends on the mass defect – the change in mass from reactants to products.
• A single nuclear reaction releases energy measured in MeV (Mega-electronvolts), which is then converted to Joules for practical applications.

Nuclear energetics is vital for understanding both the potential and the risks associated with nuclear technology.

Uranium-235 Fission

Uranium-235 is a prominent fuel in nuclear reactors due to its ability to undergo fission.When hit by a neutron, it splits into smaller nuclei, releasing a significant amount of energy.

The fission of \(^{235}U\) kicks off a chain reaction.Each fission event releases energy and more neutrons, sustaining the reaction by splitting additional uranium nuclei. This is the basis for both nuclear power generation and nuclear explosions.

• The fission of \(^{235}U\) leads to products like \(^{141}Ba\), \(^{92}Kr\), and additional neutrons.
• The mass defect in these reactions converts to energy, releasing about 193.75 MeV per fission event.
• Nuclear reactors control the rate of these fissions to safely generate usable electricity.

Understanding uranium fission is essential for safe nuclear energy utilization and handling nuclear materials responsibly.