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Chapter 16: Problem 70

A blue giant star whose surface temperature is \(23 \mathrm{kK}\) radiates energy at the rate of \(3.4 \times 10^{30}\) W. Find the star's radius, assuming it behaves like a blackbody.

Short Answer

Expert verified
The radius of the blue giant star is found to be approximately \( 6.964 \times 10^8 \) m

Step by step solution

01

Convert the temperature

First, make sure the temperature of the star is in Kelvin, as it's the units used in this formula. In this case, it's given that the temperature is \(23 \mathrm{kK}\), meaning 23,000 Kelvin.
02

Apply Stefan's Law

Next, impose Stefan's Law to find the radius of the star. According to Stefan's Law, the power being radiated by a star can be described as \( P = \sigma * A * T^4 \), where \( P = 3.4 \times 10^{30} \) W, \( \sigma = 5.67 \times 10^{-8} \, \mathrm{W} \, \mathrm{m}^{-2} \, \mathrm{K}^{-4} \) (the Stefan-Boltzmann constant), and \( T = 23000 \, \mathrm{K} \). To find 'A', the surface area of the star, substitute the given values into Stefan's Law and solve for A. We get: \( A = P / (\sigma * T^4) \).
03

Find the radius from Surface Area

Now that 'A' is known, the radius of the star can be found by substituting the calculated 'A' for the area in the formula for the surface area of a sphere, which is \( A = 4 * \pi * r^2 \). By rearranging the formula and solving for 'r', we get \( r = \sqrt{A/(4\pi)} \).

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Key Concepts

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

Blackbody Radiation
When studying stars or any celestial object, understanding blackbody radiation is crucial. This is a type of electromagnetic radiation that all objects emit, provided they are not reflecting any light and are not transparent. In astrophysics, stars are often approximated as perfect blackbodies because they emit a spectrum of radiation that depends solely on their temperature.

A key aspect of a blackbody is that it absorbs all radiation incident upon it and, in theory, does not reflect or transmit any light. This makes it a perfect emitter as well. The radiation emitted covers a range of wavelengths which, when plotted, gives the famous 'blackbody curve'. This curve varies based on the temperature of the object - hotter objects shift towards the blue end of the spectrum, while cooler objects emit more towards the red.

The blackbody model is what enables scientists to calculate various properties of stars, such as radius or luminosity, by characterizing the radiation emitted from their surfaces using laws like Stefan's Law.
Surface Area of a Sphere
When it comes to geometry, one fundamental formula relates to finding the surface area of a sphere. This formula, expressed as \( A = 4\pi r^2 \) where \( A \) is the surface area and \( r \) is the radius, is incredibly important in astronomy and physics.

Understanding the concept of surface area is especially important when dealing with celestial bodies like stars, planets, and moons. Their spherical shapes mean that we can use this formula to deduce many other properties, such as their volume, mass (if the density is known), and in our specific application, calculating the star's radius based on its emitted radiation. This geometry concept is simple yet powerful, lending itself to a variety of applications in the natural world.
Surface Temperature of Stars
Estimating the surface temperature of stars is a pivotal approach to learn about their size, life stage, and the type of light they emit. The surface temperature can also be indicative of a star's color: the hotter the star, the bluer it appears, while cooler stars appear red or orange.

In astronomy, the temperature of a star is typically measured in Kelvin. This temperature is a direct indicator of the amount of heat energy being produced by the nuclear reactions at its core. To assess the radius or other characteristics of a star, scientists deploy laws of thermodynamics and electromagnetism, such as the Stefan-Boltzmann Law, which gives a relationship between the power radiated by an object (such as a star), its surface area, and its absolute temperature to the fourth power.

By understanding this relationship, we can solve for unknowns like the radius in the given exercise, provided we have the other values. This marriage of physics and celestial mechanics opens the door to further discoveries about the cosmos, simply starting from understanding the surface temperature of stars.

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Most popular questions from this chapter

At low temperatures, the specific heats of solids are approximately cH proportional to the cube of the temperature: \(c(T)=a\left(T / T_{0}\right)^{3}\). For copper, \(a=31 \mathrm{~J} / \mathrm{g} \cdot \mathrm{K}\) and \(T_{0}=343 \mathrm{~K}\). Find the heat required to bring \(33 \mathrm{~g}\) of copper from \(13.0 \mathrm{~K}\) to \(29.0 \mathrm{~K}\).

In 2014 the European Space Agency's Rosetta spacecraft was \(5000 \mathrm{~km}\) from the comet 67 P/Churyumov-Gerasimenko. Rosetta turned its infrared sensors toward the comet and measured a flux of \(96.3 \mathrm{~W}\) per square meter of cometary surface. Assuming the dark, dusty comet radiated like a blackbody, what was its temperature?

The outdoor temperature rises by \(27^{\circ} \mathrm{F}\). What's that rise in Celsius?

A washing machine's "warm" setting calls for water at \(34.0^{\circ} \mathrm{C}\). If the cold-water supply is at \(12.4^{\circ} \mathrm{C}\) and the hot-water supply is at \(51.7^{\circ} \mathrm{C}\), what ratio of hot to cold water should the washing machine's fill valves admit to the machine?

In a low-temperature physics experiment, a metal block is sur\(\mathrm{CH}\) rounded on five faces by near-perfect insulation that prevents any conductive heat loss, and it's coated on those faces with a perfect reflector that prevents radiation. The remaining face is painted black so it behaves like a blackbody of emissivity 1 , and it's covered with a slab of material with themal conductivity \(k\) and thickness \(d\) that's transparent to radiation. The other side of the slab is in contact with liquid helium at nearly \(0 \mathrm{~K}\). (a) Find an expression for the temperature of the metal block if it loses energy equally by radiation and conduction. You can assume that heat flows straight through the slab, perpendicular to its interface with the metal block, with no heat loss out its sides. (b) Evaluate your expression when the slab is \(2.85\) \(\mathrm{cm}\) thick and is made of insulating foam with \(k=0.0166 .\)
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