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The sun radiates \(3.9 \times 10^{23} \mathrm{~J}\) of energy into space every \(\mathrm{sec}-\) ond. What is the rate at which mass is lost from the sun?

Short Answer

Expert verified
The rate at which mass is lost from the sun is approximately \(4.33 \times 10^6 kg/s\).

Step by step solution

01

Write down the given values

We are given the energy radiated into space per second by the sun: \(E = 3.9 \times 10^{23} J\).
02

Write down the equation relating energy and mass

We will use Einstein's famous equation relating energy and mass: \(E = mc^2\), where \(E\) is energy, \(m\) is mass, and \(c\) is the speed of light, \( 3 \times 10^8 m/s\).
03

Solve the equation for mass

To find the mass lost per second, we need to rearrange the equation \(E = mc^2\) to solve for mass, \(m\). Divide both sides of the equation by the speed of light squared, \(c^2\): \(m = \frac{E}{c^2}\)
04

Plug in the given energy value and the speed of light

Now, plug in the energy value given in the exercise \(E = 3.9 \times 10^{23} J\) and the speed of light \(c = 3 \times 10^8 m/s\) into the rearranged equation from Step 3: \(m = \frac{3.9 \times 10^{23} J}{(3 \times 10^8 m/s)^2}\)
05

Calculate the mass

Carry out the calculation: \(m = \frac{3.9 \times 10^{23} J}{(9 \times 10^{16} m^2/s^2)}\) \(m = \frac{3.9 \times 10^{23} J}{9 \times 10^{16} kg}\) \(m = 4.33 \times 10^6 kg\)
06

Determine the mass loss rate

Finally, we have found that the mass lost per second is approximately \(4.33 \times 10^6 kg/s\). This is the rate of mass lost from the sun due to the energy radiated into space every second.

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Key Concepts

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

Einstein's Equation
Einstein's equation, \(E = mc^2\), is a profound revelation that connects mass (\(m\)) and energy (\(E\)) in a way that was unprecedented before its discovery. This formula tells us that mass can be converted into energy and vice versa, with the speed of light squared (\(c^2\)) acting as the conversion factor. In simple terms, a small amount of mass can be transformed into a huge amount of energy, which is the cornerstone principle behind both nuclear reactors and thermonuclear explosions.

The equation itself is deceptively simple, but it unpacks into a world of meaning. It implies that mass and energy are two sides of the same coin and one can be considered a different form of the other. Considering the speed of light is a large number (\(3 \times 10^8 m/s\)), when squared, it becomes colossal, which is why even a tiny amount of mass can lead to a vast outlay of energy.
Energy Conversion
Energy conversion is the process of changing one form of energy into another. In the context of Einstein's equation, the conversion we are interested in is that of mass into energy. While this may sound like science fiction, it's a reality that occurs in various natural and human-made processes.

For example, in nuclear reactors, atoms are split during fission to release energy. Conversely, in the sun, the process is fusion where atoms combine; yet, in both cases, some matter is converted into pure energy according to \(E = mc^2\). In the exercise, we calculated how this principle explains the sun's radiated energy and the corresponding mass loss. The sun is essentially a giant engine that converts mass into the light and heat we receive on Earth. This mass-energy conversion is critical for understanding not just astrophysics, but also for harnessing nuclear power safely and efficiently on Earth.
Nuclear Fusion in the Sun
Nuclear fusion is the process that powers stars, including our sun. At the core of the sun, intense pressure and heat cause hydrogen atoms to fuse into helium, releasing vast amounts of energy in the process.

This fusion reaction is the sun's way of converting mass into energy, evidenced by the solar radiation we experience. The sun loses mass as it emits energy, a calculation we've explored in the textbook exercise using Einstein's equation. What makes nuclear fusion in the sun so fascinating is that it serves as the universe's primary energy producer, creating light and warmth that has allowed life to flourish on Earth. It's also a clean and potent form of energy conversion that scientists are working to replicate on Earth for a sustainable, long-term energy source that could revolutionize our energy infrastructure.

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Most popular questions from this chapter

Each of the following isotopes has been used medically for the purpose indicated. Suggest reasons why the particular element might have been chosen for this purpose. a. cobalt- 57, for study of the body's use of vitamin \(B_{12}\) b. calcium- 47 , for study of bone metabolism c. iron-59, for study of red blood cell function

The mass ratios of \({ }^{40} \mathrm{Ar}\) to \({ }^{40} \mathrm{~K}\) also can be used to date geologic materials. Potassium-40 decays by two processes: \({ }_{19}^{40} \mathrm{~K}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{18}^{40} \mathrm{Ar}(10.7 \%) \quad t_{1 / 2}=1.27 \times 10^{9}\) years \({ }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{20}^{40} \mathrm{Ca}+{ }_{-1}^{0} \mathrm{e}(89.3 \%)\) a. Why are \({ }^{40} \mathrm{Ar} /{ }^{40} \mathrm{~K}\) ratios used to date materials rather than \({ }^{40} \mathrm{Ca} /{ }^{40} \mathrm{~K}\) ratios? b. What assumptions must be made using this technique? c. A sedimentary rock has an \({ }^{40} \mathrm{Ar} /{ }^{40} \mathrm{~K}\) ratio of \(0.95\). Calculate the age of the rock. d. How will the measured age of a rock compare to the actual age if some \({ }^{40} \mathrm{Ar}\) escaped from the sample?

Calculate the binding energy per nucleon for \({ }_{1}^{2} \mathrm{H}\) and \({ }_{1}^{3} \mathrm{H}\). The atomic masses are \({ }_{1}^{2} \mathrm{H}, 2.01410 \mathrm{u} ;\) and \({ }_{1}^{3} \mathrm{H}, 3.01605 \mathrm{u}\).

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A chemist wishing to do an experiment requiring \({ }^{47} \mathrm{Ca}^{2+}\) (half-life \(=4.5\) days) needs \(5.0 \mu \mathrm{g}\) of the nuclide. What mass of \({ }^{47} \mathrm{CaCO}_{3}\) must be ordered if it takes \(48 \mathrm{~h}\) for delivery from the supplier? Assume that the atomic mass of \({ }^{47} \mathrm{Ca}\) is \(47.0 \mathrm{u}\).

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