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Measurements on two stars indicate that Star X has a surface temperature of \(5727^{\circ} \mathrm{C}\) and Star Y has a surface temperature of \(11727^{\circ} \mathrm{C}\). If both stars have the same radius, what is the ratio of the luminosity (total power output) of Star Y to the luminosity of Star X? Both stars can be considered to have an emissivity of \(1.0\).

Short Answer

Expert verified
The ratio of the luminosity of Star Y to the luminosity of Star X is 16.

Step by step solution

01

Convert Temperatures to Kelvin

First, we have to convert the given temperatures from Celsius to Kelvin, since the Stefan-Boltzmann law requires temperatures in Kelvin. This can be done by adding 273.15 to each Celsius temperature. Therefore, the temperature of Star X in Kelvin is \(5727^{\circ} \mathrm{C} + 273.15 = 6000 \mathrm{K}\) and that of Star Y is \(11727^{\circ} \mathrm{C} + 273.15 = 12000 \mathrm{K}\).
02

Calculate the Ratio of Temperatures Raised to the Fourth Power

The luminosity of a star is directly proportional to the fourth power of its temperature (according to the Stefan-Boltzmann law). Hence, the luminosity of Star Y to the luminosity of Star X is the same as the fourth power of the ratio of their temperatures. Therefore, the required ratio is \( \left( \frac{12000}{6000} \right) ^4 = 16\).

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Key Concepts

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

Luminosity of Stars
In astrophysics, the luminosity of a star is a key concept that represents the amount of energy a star emits into space per unit time, typically measured in watts. It's an intrinsic property that doesn't depend on how far away the observer is from the star. A star’s luminosity is closely related to two of its basic physical properties: its radius and surface temperature.

The energy radiated by a star is generated in its core through nuclear fusion processes and then emitted from the surface. This energy emission is crucial in understanding a star's life cycle and classifying it's type within the Hertzsprung-Russell diagram, which is pivotal in astrophysics.

Relation to Surface Temperature and Radius

One of the vital insights from the field of stellar physics is that a star's luminosity is proportional to its radius squared and the fourth power of its surface temperature. Understanding this relationship helps astronomers to determine crucial properties of distant stars just by studying their light.

For example, if two stars have the same radius but quite different surface temperatures, we can compare their luminosity by considering the temperatures alone. This is simplified using the Stefan-Boltzmann law, which asserts that luminosity (\( L \)) is proportional to the fourth power of the star’s temperature (\( T \)) and given by the formula \( L = 4\pi R^2\sigma T^4 \) where \( \sigma \) is the Stefan-Boltzmann constant and \( R \) is the radius of the star.
Temperature Conversion
Temperature plays a critical role in many areas of physics and astronomy, particularly when dealing with the Stefan-Boltzmann law. However, temperature can be measured in different scales, such as Celsius, Fahrenheit, and Kelvin. For scientific purposes, especially in physics, Kelvin is the preferred scale because it is the standard unit of thermodynamic temperature and begins at absolute zero, unlike Celsius or Fahrenheit.

The conversion is relatively straightforward but essential for accurate calculations. To convert from Celsius to Kelvin, one simply adds 273.15 to the Celsius temperature. Remember, the size of one degree on both the Kelvin and Celsius scales is the same, so they only differ in their starting points: 0 degrees Celsius is equivalent to 273.15 Kelvin.

Why Convert to Kelvin in Astrophysics?

For many astronomical equations, including those used to calculate a star's luminosity, the input temperatures must be in Kelvin. This is because these equations are often derived from the laws of thermodynamics, which are based on the Kelvin scale. In the step by step solutions, this conversion was demonstrated in determining the luminosity ratio between two stars.
Proportionality in Physics
Proportionality is an essential concept in physics used to describe the relationship between two quantities that change together in a predictable way. When two variables are proportional, as one increases or decreases, the other does so in response, at a consistent rate. This can be a direct proportionality, such as the case with the luminosity of stars being directly proportional to the fourth power of their temperature, or it could be an inverse proportionality where the increase in one variable results in the decrease of the other.

The understanding of proportionality allows scientists to create models and formulas that describe the behavior of physical systems. The Stefan-Boltzmann law utilizes the concept of direct proportionality, expressing that luminosity is directly proportional to the fourth power of the temperature when the radius is constant. In physics equations, proportionality is often denoted by either the symbols \( \propto \) in qualitative discussions, or precise constants of proportionality in quantitative formulas.

Practical Use of Proportionality

Using proportionality in equations like the Stefan-Boltzmann law allows scientists to predict the luminosity of stars accurately. It also simplifies the understanding of complex relationships, as seen in the exercise, where we calculated the luminosity ratio by raising the temperature ratio to the fourth power, due to their direct proportionality.

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Most popular questions from this chapter

Three liquids are at temperatures of \(10^{\circ} \mathrm{C}, 20^{\circ} \mathrm{C}\), and \(30^{\circ} \mathrm{C}\), respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is \(17^{\circ} \mathrm{C}\). Equal masses of the second and third are then mixed, and the equilibrium temperature is \(28^{\circ} \mathrm{C}\). Find the equilibrium temperature when equal masses of the first and third are mixed.

A \(1.5-\mathrm{kg}\) copper block is given an initial speed of \(3.0 \mathrm{~m} / \mathrm{s}\) on a rough horizontal surface. Because of friction, the block finally comes to rest. (a) If the block absorbs \(85 \%\) of its initial kinetic energy as internal energy, calculate its increase in temperature. (b) What happens to the remaining energy?

In the summer of 1958 in St. Petersburg, Florida, a new sidewalk was poured near the childhood home of one of the authors. No expansion joints were supplied, and by mid-July the sidewalk had been completely destroyed by thermal expansion and had to be replaced, this time with the important addition of expansion joints! This event is modeled here. A slab of concrete \(4.00 \mathrm{~cm}\) thick, \(1.00 \mathrm{~m}\) long, and \(1.00 \mathrm{~m}\) wide is poured for a sidewalk at an ambient temperature of \(25.0^{\circ} \mathrm{C}\) and allowed to set. The slab is exposed to direct sunlight and placed in a series of such slabs without proper expansion joints, so linear expansion is prevented. (a) Using the linear expansion equation (Eq. 10.4), eliminate \(\Delta L\) from the equation for compressive stress and strain (Eq. 9.3). (b) Use the expression found in part (a) to eliminate \(\Delta T\) from Equation \(11.3\), obtaining a symbolic equation for thermal energy transfer \(Q\). (c) Compute the mass of the concrete slab given that its density is \(2.40 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). (d) Concrete has an ultimate compressive strength of \(2.00 \times 10^{7} \mathrm{~Pa}\), specific heat of \(880 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and Young's modulus of \(2.1 \times 10^{10} \mathrm{~Pa}\). How much thermal energy must be transferred to the slab to reach this compressive stress? (e) What temperature change is required? (f) If the Sun delivers \(1.00 \times 10^{3} \mathrm{~W}\) of power to the top surface of the slab and if half the energy, on the average, is absorbed and retained, how long does it take the slab to reach the point at which it is in danger of cracking due to compressive stress?

The thermal conductivities of human tissues vary greatly. Fat and skin have conductivities of about \(0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(0.020 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively, while other tissues inside the body have conductivities of about \(0.50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assume that between the core region of the body and the skin surface lies a skin layer of \(1.0 \mathrm{~mm}\), fat layer of \(0.50 \mathrm{~cm}\), and \(3.2 \mathrm{~cm}\) of other tissues. (a) Find the \(R\)-factor for each of these layers, and the equivalent \(R\)-factor for all layers taken together, retaining two digits. (b) Find the rate of energy loss when the core temperature both a protective layer of clothing and an insulating layer of unmoving air are absent, and a body area of \(2.0 \mathrm{~m}^{2}\).

When you jog, most of the food energy you burn above your basal metabolic rate (BMR) ends up as internal energy that would raise your body temperature if it were not eliminated. The evaporation of perspiration is the primary mechanism for eliminating this energy. Determine the amount of water you lose to evaporation when running for 30 minutes at a rate that uses \(400 \mathrm{kcal} / \mathrm{h}\) above your BMR. (That amount is often considered to be the "maximum fat-burning" energy output.) The metabolism of 1 gram of fat gen- erates approximately \(9.0 \mathrm{kcal}\) of energy and produces approximately 1 gram of water. (The hydrogen atoms in the fat molecule are transferred to oxygen to form water.) What fraction of your need for water will be provided by fat metabolism? (The latent heat of vaporization of water at room temperature is \(2.5 \times 10^{6} \mathrm{~J} / \mathrm{kg}\).)

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