Question

The power (energy per unit time) radiated by a blackbody per unit area of surface expressed in units of \(\mathrm{W} \mathrm{m}^{-2}\) is given by \(P=\sigma T^{4}\) with \(\sigma=5.67 \times 10^{-8} \mathrm{W} \mathrm{m}^{-2} \mathrm{K}^{-4} .\) The radius of the sun is approximately \(7.00 \times 10^{5} \mathrm{km}\) and the surface temperature is \(5800 .\) K. Calculate the total energy radiated per second by the sun. Assume ideal blackbody behavior.

Short answer

The total energy radiated per second by the sun is approximately \(3.8 \times 10^{26} \ \mathrm{W}\).

Step-by-step solution

Calculate the surface area of the sun.

The surface area of a sphere is given by the formula \(A = 4 \pi r^2\). Therefore, using the given radius of the sun, \(r = 7.00 \times 10^5 \ \mathrm{km}\), we can calculate its surface area. Note that we have to convert the radius to meters: \(7.00 \times 10^5 \ \mathrm{km} = 7.00 \times 10^5 \times 10^3 \ \mathrm{m} = 7.00 \times 10^8 \ \mathrm{m}\). Now, we can find the surface area: \(A = 4 \pi (7.00 \times 10^8 \ \mathrm{m})^2\)

Compute the power radiated per unit area of surface.

We are given the formula for power (energy per unit time) radiated by a blackbody per unit area of surface, \(P = \sigma T^4\). Using the provided values for \(\sigma = 5.67 \times 10^{-8} \ \mathrm{W} \ \mathrm{m}^{-2} \ \mathrm{K}^{-4}\) and the surface temperature, \(T = 5800 \ \mathrm{K}\), we can calculate the power radiated per unit area: \(P = (5.67 \times 10^{-8} \ \mathrm{W} \ \mathrm{m}^{-2} \ \mathrm{K}^{-4}) (5800 \ \mathrm{K})^4\)

Compute the total energy radiated per second by the sun.

To compute the total energy radiated per second by the sun, we have to multiply the power radiated per unit area of surface (from step 2) by the surface area of the sun (from step 1): \(E = P \times A = (5.67 \times 10^{-8} \ \mathrm{W} \ \mathrm{m}^{-2} \ \mathrm{K}^{-4}) (5800 \ \mathrm{K})^4 \times 4 \pi (7.00 \times 10^8 \ \mathrm{m})^2\) Now, let's compute this value. \(E = (5.67 \times 10^{-8}) (5800)^4 \times 4 \pi (7.00 \times 10^8)^2 \ \mathrm{W} \approx 3.8 \times 10^{26} \ \mathrm{W}\) So, the total energy radiated per second by the sun is approximately \(3.8 \times 10^{26} \ \mathrm{W}\).

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann law is a fundamental principle in the field of thermodynamics that describes how much radiant energy a blackbody emits. A blackbody is an idealized object that perfectly absorbs and radiates energy at every frequency. According to the law, the power radiated by a blackbody per unit area is directly proportional to the fourth power of the blackbody's absolute temperature.

Mathematically, this is expressed as \( P = \.sigma \.T^4 \), where \( P \) is the power radiated per unit area, \( \.sigma \) (sigma) is the Stefan-Boltzmann constant, and \( \.T \) is the absolute temperature measured in Kelvins (K). The Stefan-Boltzmann constant \( \.sigma \) has a value of approximately \( 5.67 \times 10^{-8} \.mathrm{W} \.mathrm{m}^{-2} \.mathrm{K}^{-4} \), which is a universal constant. It reveals the powerful influence that temperature has on the energy output of a stellar body, as even small changes in temperature can result in significant variations in energy emission.

Surface Area of a Sphere

Understanding the surface area of a sphere is crucial when calculating how much energy a spherical body, such as a star or planet, emits into space. The surface area determines the 'skin' from which all the radiation is emitted. The formula to calculate the surface area of a sphere is given by \( A = 4 \.pi \.r^2 \), where \( A \) represents the surface area and \( r \) is the radius of the sphere.

For celestial bodies like the sun, understanding the surface area helps scientists estimate the amount of energy it emits. The radius of the sun is approximately \( 7.00 \times 10^5 \.mathrm{km} \), and converting this to meters (since the constant \( \.sigma \) is in terms of meters), we use \( 7.00 \times 10^8 \.mathrm{m} \) for calculations. Applying these values to the formula reveals the vast expanse of the sun's radiative 'skin,' which plays a key role in determining its energy output.

Solar Energy Output

The solar energy output refers to the amount of energy the sun emits into space. This is a significant quantity because it influences the earth's climate and the potential for solar energy harvesting. The amount of energy produced by the sun, and indeed any star, can be modeled using the Stefan-Boltzmann law.

When estimating the energy output, scientists rely on the sun's surface temperature and its surface area. With the sun's temperature at around 5800 K and its vast surface area, the total energy emitted is astonishingly high. To put it in perspective, the energy radiated by the sun every second is more than sufficient to meet the global energy demands of humanity for over a million years.

Temperature to the Fourth Power

The concept of 'temperature to the fourth power' is a cornerstone of the Stefan-Boltzmann law and plays a pivotal role in understanding blackbody radiation. The law states that the radiant energy emitted by a blackbody is proportional to the fourth power of its absolute temperature.

When we raise the temperature to the fourth power, it exemplifies the exponential relationship between temperature and radiated energy: a small increase in temperature leads to a disproportionate increase in emitted energy. This principle allows us to understand why temperature differences across various celestial bodies result in vastly different radiation levels. As a real-world application, looking at our sun, raising its surface temperature of 5800 K to the fourth power results in a substantial number that, when multiplied by the Stefan-Boltzmann constant and the sun's huge surface area, demonstrates how the sun is such a powerful energy source.