Skip to main content
Ch 19: Work, Heat, and the First Law of Thermodynamics

Chapter 19, Problem 19

Most stars are main-sequence stars, a group of stars for which size, mass, surface temperature, and radiated power are closely related. The sun, for instance, is a yellow main-sequence star with a surface temperature of 5800 K. For a main-sequence star whose mass M is more than twice that of the sun, the total radiated power, relative to the sun, is approximately P/Pₛᵤₙ=1.5(M/Mₛᵤₙ)^3.5 . The star Regulus A is a bluish main-sequence star with mass 3.8Mₛᵤₙ and radius 3.1Rₛᵤₙ. What is the surface temperature of Regulus A?

Verified Solution
Video duration:
7m
This video solution was recommended by our tutors as helpful for the problem above.
544
views
Was this helpful?

Video transcript

Hey, everyone. This problem is dealing with the Stefan Boltzmann theorem. Let's see what it's asking us. So we're told that the energy from a main sequence star is a function of its mass and radius. The power radiated by a main sequence star of a given mass M is given by the equation P equals 1.5 multiplied by M divided by MS to the 3.5 power, all multiplied by P where M sub S and P sub S is the mass and power of the sun. We are asked to consider a hypothetical star V that has a mass of 2.1 times the mass of the sun and a radius of 2.4 times the radius of the sun. And we are asked to determine the surface temperature of this star. They give us the sun's surface temperature as approximately 5780 Kelvin. And we're told that both of the, both the star V and our son have the same emissivity. Our multiple choice answers here are a 65 Kelvin B 7200 Kelvin C 7900 Kelvin or D 8300 Kelvin So the key to this problem is we're calling the Stefan Boltzmann equation which tells us that power is equal to area multiplied by the emissivity multiplied by the Stefen Boman constant sigma multiplied by the temperature to the fourth power. And so I made it look or the power of star V given using the equation for power given to us in the equation or sorry, give it to us. In the problem statement, we can write that the power of star V is equal to 1.5 times or multiplied by the mass of B divided by the mass of the sun to the 3.5 power multiplied by the power of the sun. And so the mass of star V was given as 2.1 multiplied by the mass of the sun. And we can use the Stefen Boldman theory or Stefen Boldman formula to solve her temperature. So what that looks like is you have the area of V then multiplied by the mi multiplied by the step in both constant multiplied by the temperature of V to the fourth power. And that equals 1.5 multiplied by 2.1, the mass multiplied by the mass of the sun divided by the mass of the sun to the 3.5 power multiplied by the area, the sun multiplied by the emissivity multiplied by the sun con bolt and constant multiply by the temperature of the sun to the fourth hour. And so as we look to simplify things, we emi are the same. So those cancel stuff involvement, constant cancels the mask, the sun cancel and then area is given by IRS square. And so the area of V is going to be pi multiplied by the radius of V which is 2.4 multiplied by the radius of the sun square. And so the area of the sun is simply going to be pi multiplied by the radius of the sun squared. And so we can plug that in. So we'll have pi multiplied by two, sorry, 2. multiplied by the radius of the sun. That quantity squared multiplied by TV to the fourth times or sorry equals 1.5 multiplied by 2.1 to the 3. multiplied by pi R S squared multiplied by A T S to the fourth. So now our, our S squares cancel P cancel and we are left with 5. multiplied by TV. Four equals 20.1 multiplied by T S four. And so solving for TV will take the fourth route of 20.1 times or sorry, 5780 Kelvin to the fourth divided by 5.76. We plug that into our calculator and we get 7900 K. So looking at our multiple choice answers that aligns with answer choice C so that's all we have for this one. We'll see you in the next video.
Related Practice
Textbook Question
The beaker in FIGURE P19.45, with a thin metal bottom, is filled with 20 g of water at 20°C. It is brought into good thermal contact with a 4000 cm^3 container holding 0.40 mol of a monatomic gas at 10 atm pressure. Both containers are well insulated from their surroundings. What is the gas pressure after a long time has elapsed? You can assume that the containers themselves are nearly massless and do not affect the outcome.
319
views
Textbook Question
A typical nuclear reactor generates 1000 MW (1000 MJ/s) of electric energy. In doing so, it produces 2000 MW of 'waste heat' that must be removed from the reactor to keep it from melting down. Many reactors are sited next to large bodies of water so that they can use the water for cooling. Consider a reactor where the intake water is at 18°C. State regulations limit the temperature of the output water to 30°C so as not to harm aquatic organisms. How many liters of cooling water have to be pumped through the reactor each minute?
629
views
Textbook Question
Liquid helium, with a boiling point of 4.2 K, is used in ultralow-temperature experiments and also for cooling the superconducting magnets used in MRI imaging in medicine. Storing liquid helium so far below room temperature is a challenge because even a small 'heat leak' will boil the helium away. A standard helium dewar, shown in FIGURE P19.67, has an inner stainless-steel cylinder filled with liquid helium surrounded by an outer cylindrical shell filled with liquid nitrogen at –196°C. The space between is a vacuum. The small structural supports have very low thermal conductivity, so you can assume that radiation is the only heat transfer between the helium and its surroundings. Suppose the helium cylinder is 16 cm in diameter and 30 cm tall and that all walls have an emissivity of 0.25. The density of liquid helium is 125 kg/m^3 and its heat of vaporization is 2.1×10^4 J/kg. a. What is the mass of helium in the filled cylinder?
654
views
1
rank
Textbook Question
A 750 g aluminum pan is removed from the stove and plunged into a sink filled with 10.0 L of water at 20.0°C . The water temperature quickly rises to 24.0°C. What was the initial temperature of the pan in °C and in °F?
390
views
Textbook Question
10 g of aluminum at 200°C and 20 g of copper are dropped into 50 cm^3 of ethyl alcohol at 15°C. The temperature quickly comes to 25°C . What was the initial temperature of the copper?
445
views
Textbook Question
A lava flow is threatening to engulf a small town. A 400-m-wide, 35-cm-thick tongue of 1200°C lava is advancing at the rate of 1.0 m per minute. The mayor devises a plan to stop the lava in its tracks by flying in large quantities of 20°C water and dousing it. The lava has density 2500 kg/m^3, specific heat 1100 J/kg K, melting temperature 800°C, and heat of fusion 4.0×10^5 J/kg. How many liters of water per minute, at a minimum, will be needed to save the town?
502
views