Problem 84 - Chemistry The Central Science In Si Units

Question

The Sun radiates energy into space at the rate of \(3.9 \times 10^{26} \mathrm{~J} / \mathrm{s} .\) (a) Calculate the rate of mass loss from the Sun in \(\mathrm{kg} /\) s. (b) How does this mass loss arise? (c) It is esti- mated that the Sun contains \(9 \times 10^{56}\) free protons. How many protons per second are consumed in nuclear reactions in the Sun?

Step-by-Step Solution

Answer

(a) The Sun's mass loss rate is \(4.33 \times 10^9 \text{ kg/s}\). (b) Mass loss arises from nuclear fusion reactions. (c) \(3.9 \times 10^{38}\) protons are consumed per second.

Step 1: Understanding Energy-Mass Equivalence

According to Einstein's mass-energy equivalence principle, energy and mass are related by the equation \( E = mc^2 \), where \( E \) is energy, \( m \) is mass, and \( c \) is the speed of light (\(3 \times 10^8 \text{ m/s}\)). This relationship allows us to calculate the mass loss from the Sun from its energy output.

Step 2: Calculate the Rate of Mass Loss

We begin by equating the Sun’s rate of energy radiation to its rate of mass loss: \( E = mc^2 \). Rearranging the formula gives \( m = \frac{E}{c^2} \). Substituting the given energy \( E = 3.9 \times 10^{26} \text{ J/s} \) and the speed of light \( c = 3 \times 10^8 \text{ m/s} \), we find the mass loss rate as:\[m = \frac{3.9 \times 10^{26} \text{ J}}{(3 \times 10^8 \text{ m/s})^2} = 4.33 \times 10^9 \text{ kg/s}\]

Step 3: Explaining the Source of Mass Loss

The Sun loses mass primarily due to nuclear fusion reactions in its core. Hydrogen nuclei (protons) fuse to form helium, and during this process a small fraction of mass is converted to energy, which is radiated into space.

Step 4: Calculate the Number of Protons Consumed Per Second

Each fusion reaction in the Sun primarily involves the conversion of hydrogen nuclei (protons) to helium. The process emits energy equivalent to the mass difference. Knowing the Sun radiates \(3.9 \times 10^{26} \text{ J/s}\), we reverse calculate the number of protons:The energy released in a single fusion event converting four protons to helium is approximately \(4 \times 10^{-12} \text{ J}\). The number of reactions per second is:\[n = \frac{3.9 \times 10^{26} \text{ J/s}}{4 \times 10^{-12} \text{ J/reaction}} \approx 9.75 \times 10^{37} \text{ reactions/sec}\]Each reaction consumes 4 protons, so the number of protons consumed per second is:\[4 \times n = 3.9 \times 10^{38} \text{ protons/s}\]

Key Concepts

Nuclear Fusion

Nuclear fusion is the fundamental process that powers our Sun and other stars in the universe. It's a type of nuclear reaction where two light atomic nuclei combine to form a heavier nucleus. In the Sun, hydrogen nuclei, known as protons, fuse together to create helium. During this fusion process, there is a conversion of a small amount of mass into energy.

This transformation of mass into energy is described by Einstein's equation, \( E = mc^2 \). Here, \( E \) represents energy, \( m \) is the mass, and \( c \) is the speed of light. It's this mass-to-energy conversion during fusion reactions that is responsible for the incredible energy output of the Sun.

In summary:

• Fusion combines light nuclei to form heavier nuclei.
• A tiny amount of mass is lost but converted into a significant amount of energy.
• The energy emitted by the Sun comes from these nuclear reactions.

Protons in the Sun

Protons are one of the key ingredients in the Sun's core nuclear reactions. The Sun contains an abundance of free protons, roughly estimated at \(9 \times 10^{56}\). A proton is a positively charged particle, equivalent to a hydrogen nucleus in atomic terms.

In the Sun, protons undergo various fusion reactions. The primary reaction in the Sun is the proton-proton chain reaction. Through a series of steps, protons fuse to form helium nuclei, releasing energy in the process. This energy is then radiated out of the Sun, providing light and heat.

Each second, a massive number of protons are consumed in these nuclear reactions. Given the Sun's radiant energy of \(3.9 \times 10^{26} \text{ J/s}\), about \(3.9 \times 10^{38}\) protons are consumed every second. Despite this enormous consumption rate, the Sun still has enough protons to sustain its energy output for billions of years into the future.

• Protons are the raw fuel for the Sun’s nuclear fusion.
• Each second, huge numbers of protons fuse to form helium.
• The consumption rate is maintained over billions of years.

Mass Loss in Stars

Stars, like the Sun, experience mass loss due to the process of nuclear fusion. A seemingly small fraction of mass is transformed into energy. Although small at the atomic level, this mass loss accumulates substantially over time as energy is emitted into space.

For the Sun, this conversion can account for a mass loss of approximately \(4.33 \times 10^9 \text{ kg/s}\) based on its energy output. Nuclear fusion, while converting hydrogen to helium, is the main source of this mass-energy transformation.

The accumulated loss of mass does not immediately affect the Sun’s size or stability significantly. However, over billions of years, it can lead to changes in the Sun’s structure and lifecycle. Understanding this concept is essential for astrophysics and helps explain the lifecycle of stars.

• Mass is lost during nuclear fusion and converted into energy.
• The Sun loses around \(4.33 \times 10^9 \text{ kg/s}\) due to energy radiation.
• Mass loss impacts a star’s lifecycle over long periods.