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\(\textbf{Comparison of Energy Released per Gram of Fuel.}\) (a) When gasoline is burned, it releases 1.3 \(\times\) 10\(^8\) J of energy per gallon (3.788 L). Given that the density of gasoline is 737 kg/m\(^3\), express the quantity of energy released in J/g of fuel. (b) During fission, when a neutron is absorbed by a \(^2$$^3$$^5\)U nucleus, about 200 MeV of energy is released for each nucleus that undergoes fission. Express this quantity in J/g of fuel. (c) In the proton-proton chain that takes place in stars like our sun, the overall fusion reaction can be summarized as six protons fusing to form one \(^4\)He nucleus with two leftover protons and the liberation of 26.7 MeV of energy. The fuel is the six protons. Express the energy produced here in units of J/g of fuel. Notice the huge difference between the two forms of nuclear energy, on the one hand, and the chemical energy from gasoline, on the other. (d) Our sun produces energy at a measured rate of 3.86 \(\times\) 10\(^2$$^6\) W. If its mass of 1.99 \(\times\) 10\(^3$$^0\) kg were all gasoline, how long could it last before consuming all its fuel? (\(Historical\) \(note\): Before the discovery of nuclear fusion and the vast amounts of energy it releases, scientists were confused. They knew that the earth was at least many millions of years old, but could not explain how the sun could survive that long if its energy came from chemical burning.)

Short Answer

Expert verified
Gasoline releases ≈ 46590.98 J/g; fission releases ≈ 8.19 × 10¹⁰ J/g; fusion releases ≈ 4.27 × 10¹¹ J/g. The sun would last ≈ 7,600 years on gasoline.

Step by step solution

01

Calculate Energy Released from Gasoline in J/g

To find the energy per gram of gasoline, we start with the energy released per gallon which is \(1.3 \times 10^8\) J. A gallon is 3.788 L. First, convert liters to cubic meters, knowing that 1 L = 0.001 m³:\[3.788 \text{ L} = 3.788 \times 0.001 \text{ m}^3 = 0.003788 \text{ m}^3\]Next, calculate the mass using the density of gasoline (737 kg/m³):\[ \text{Mass} = 0.003788 \text{ m}^3 \times 737 \text{ kg/m}^3 = 2.792 \text{ kg} \]Convert the mass from kg to g:\[2.792 \text{ kg} = 2792 \text{ g}\]Finally, calculate the energy per gram:\[\frac{1.3 \times 10^8 \text{ J}}{2792 \text{ g}} \approx 46590.98 \text{ J/g}\]
02

Calculate Energy Released from Fission in J/g

Given that each fission event releases 200 MeV, convert this to joules:\[ 200 \text{ MeV} = 200 \times 1.60218 \times 10^{-13} \text{ J} = 3.20436 \times 10^{-11} \text{ J} \]To find the energy per gram, divide by the atomic mass of uranium (\(\text{approximately} 235 \text{ g/mol}\)), then convert from moles to atoms using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol):\[\text{Energy per gram} = \frac{3.20436 \times 10^{-11} \text{ J}}{235 \times 1.66054 \times 10^{-24} \text{ g}} \approx 8.19 \times 10^{10} \text{ J/g}\]
03

Calculate Energy Released from Fusion in J/g

The total energy released in the proton-proton chain is 26.7 MeV. Convert this to joules:\[26.7 \text{ MeV} = 26.7 \times 1.60218 \times 10^{-13} \text{ J} = 4.27922 \times 10^{-12} \text{ J}\]Each 6 protons serve as the fuel. With the mass of one proton being approximately 1.67 x \(10^{-24}\) g, the mass of 6 protons is:\[ 6 \times 1.67 \times 10^{-24} \text{ g} = 10.02 \times 10^{-24} \text{ g} \]Energy per gram is:\[\frac{4.27922 \times 10^{-12} \text{ J}}{10.02 \times 10^{-24} \text{ g}} \approx 4.27 \times 10^{11} \text{ J/g}\]
04

Calculate Duration Sun Would Last if Made of Gasoline

The sun's power output is \(3.86 \times 10^{26}\) W, which equals \(3.86 \times 10^{26}\) J/s.If the sun's mass (1.99 \times 10^{30} kg) were gasoline, convert this mass to grams:\[1.99 \times 10^{30} \text{ kg} = 1.99 \times 10^{33} \text{ g}\]Using the energy release from gasoline (46590.98 J/g from Step 1), calculate total energy:\[ \text{Total energy} = 1.99 \times 10^{33} \text{ g} \times 46590.98 \text{ J/g} \approx 9.263 \times 10^{37} \text{ J}\]Find out how long this energy would last:\[\text{Duration} = \frac{9.263 \times 10^{37} \text{ J}}{3.86 \times 10^{26} \text{ J/s}} \approx 2.40 \times 10^{11} \text{ s} \approx 7.6 \times 10^3 \text{ years}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Energy
Chemical energy is a type of potential energy stored in the bonds of chemical compounds, such as molecules. When these bonds are broken, energy is released, often in the form of heat. This process is called a chemical reaction.

One common example of chemical energy release is the combustion of gasoline. In this process:
  • Gasoline reacts with oxygen, producing carbon dioxide, water, and energy.
  • The energy released can be used to power engines, generating mechanical motion.
  • Measured in joules per gram, gasoline is significant but much less than nuclear reactions.
The efficiency of chemical energy from gasoline underscores the benefits for short-term energy needs.
Nuclear Fission
Nuclear fission is a process by which a heavy nucleus, such as uranium-235, splits into smaller nuclei, along with a few neutrons and a large amount of energy.

This reaction is harnessed in nuclear power plants to generate electricity. Here's how it works:
  • A neutron collides with a uranium nucleus, initiating fission.
  • The split releases energy, neutron(s), and smaller atoms.
  • The energy output si gnificantly surpasses chemical processes.
Fission produces about 200 mega-electronvolts (MeV) per event. Converting this results in an energy output of roughly 8.19 x 10^10 J/g, emphasizing its potency.
Nuclear Fusion
Nuclear fusion involves the combining of lighter atomic nuclei to form a heavier nucleus, with mass converted to energy according to Einstein's equation, E=mc².

This process powers the stars, including our sun, and involves the following:
  • During fusion, hydrogen atoms (protons) merge to form helium.
  • The reaction releases enormous energy due to the strong nuclear forces.
  • This energy is exponentially higher than that from fission or chemical reactions.
The proton-proton chain is a key fusion reaction, releasing about 4.27 x 10^11 J/g. Fusion holds the promise of sustainable energy.
Proton-Proton Chain
The proton-proton chain is a sequence of nuclear fusion reactions in stars like our sun, where hydrogen nuclei (protons) are transformed into helium.

This process can be detailed as follows:
  • Six protons participate, resulting in one helium nucleus and two protons.
  • It liberates 26.7 MeV of energy.
  • It's essential in stellar energy production and lifespan determination.
This chain highlights how stars maintain their luminosity and mass over billions of years. Its energy efficiency is critical for the sun's long-term stability.

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Most popular questions from this chapter

Measurements on a certain isotope tell you that the decay rate decreases from 8318 decays/min to 3091 decays/min in 4.00 days. What is the half-life of this isotope?

(a) If a chest x ray delivers 0.25 mSv to 5.0 kg of tissue, how many \(total\) joules of energy does this tissue receive? (b) Natural radiation and cosmic rays deliver about 0.10 mSv per year at sea level. Assuming an RBE of 1, how many rem and rads is this dose, and how many joules of energy does a 75-kg person receive in a year? (c) How many chest x rays like the one in part (a) would it take to deliver the same \(total\) amount of energy to a 75-kg person as she receives from natural radiation in a year at sea level, as described in part (b)?

Consider the fusion reaction \(^{2}_{1}H\) + \(^{2}_{1}H \rightarrow ^{3}_{2}He + ^{1}_{0}n\). (a) Estimate the barrier energy by calculating the repulsive electrostatic potential energy of the two \(^{2}_{1}H\) nuclei when they touch. (b) Compute the energy liberated in this reaction in MeV and in joules. (c) Compute the energy liberated \(per\) \(mole\) of deuterium, remembering that the gas is diatomic, and compare with the heat of combustion of hydrogen, about \(2.9 \times 10^{5} J/mol\).

Many radioactive decays occur within a sequence of decays for example, \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\). The half-life for the \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) decay is \(2.46 \times 10^{5}\) y, and the half-life for the \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\) decay is \(7.54 \times 10^{4}\) y. Let 1 refer to \(^{234}_{92}U\), 2 to \(^{230}_{88}Th\), and 3 to \(^{226}_{84}Ra\); let \(\lambda\)1 be the decay constant for the \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) decay and \(\lambda\)2 be the decay constant for the \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\) decay. The amount of \(^{230}_{88}Th\) present at any time depends on the rate at which it is produced by the decay of \(^{234}_{92}U\) and the rate by which it is depleted by its decay to \(^{226}_{84}Ra\). Therefore, \(d$$N_{2}\)(\(t\))/\(d$$t\) = \(\lambda$$_1$$N$$_1\)(\(t\)) - \(\lambda$$_2$$N$$_2\)(\(t\)). If we start with a sample that contains \(N_{10}\) nuclei of \(^{234}_{92}U\) and nothing else, then \(N{(t)}\) = \(N_{01}\)e\(^{-\lambda_{1}t}\). Thus \(dN_{2}(t)/dt\) = \(\lambda\)1\(N_{01}\)e\(^{-\lambda_{1}t}\)- \(\lambda2N_{2}(t)\). This differential equation for \(N_{2}(t)\) can be solved as follows. Assume a trial solution of the form \(N_{2}(t)\) = \(N_{10} [h_{1}e^{-\lambda_1t}\) + \(h_{2}e^{-\lambda_2t}\)] , where \(h_{1}\) and \(h_{2}\) are constants. (a) Since \(N_{2}(0)\) = 0, what must be the relationship between \(h_{1}\) and \(h_{2}\)? (b) Use the trial solution to calculate \(dN_{2}(t)/dt\), and substitute that into the differential equation for \(N_{2}(t)\). Collect the coefficients of \(e^{-\lambda_{1}t}\) and \(e^{-\lambda_{2}t}\). Since the equation must hold at all t, each of these coefficients must be zero. Use this requirement to solve for \(h_{1}\) and thereby complete the determination of \(N_{2}(t)\). (c) At time \(t = 0\), you have a pure sample containing 30.0 g of \(^{234}_{92}U\) and nothing else. What mass of \(^{230}_{88}Th\) is present at time \(t = 2.46 \times 10^{5}\) y, the half-life for the \(^{234}_{92}U\) decay?

Measurements indicate that 27.83% of all rubidium atoms currently on the earth are the radioactive \(^8$$^7\)Rb isotope. The rest are the stable \(^8$$^5\)Rb isotope. The half-life of \(^8$$^7\)Rb is 4.75 \(\times\) 10\(^1$$^0\) y. Assuming that no rubidium atoms have been formed since, what percentage of rubidium atoms were \(^8$$^7\)Rb when our solar system was formed 4.6 \(\times\) 10\(^9\) y ago?

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