Those with a higher luminosity or a greater absolute magnitude.
That question is a little tricky. What is certain is that if the light is coming from a hot object like a glowing heater element, yellow light indicates that the element is hotter. It is also true that a single photon of yellow light carries more energy than a single photon of red light and is "hotter" in that sense. So in general, yellow light is hotter, but if you are simply talking about narrowband filtered light from a single source, a human will not ordinarily be able to sense the difference, and which color is more intense will depend on the spectrum of the light source.
Yes. Since it has a temperature, it is giving off infra-red radiation. Some parts of it (molten lavea) are so hot that they give off visible light as well.
Blue photons are higher-energy than red photons. The equations governing the emission of light based solely on temperature state that an object giving off primarily blue light is hotter than one giving off primarily red light.
Through a comparison of solar luminosity and temperature seen with the H-R diagram. On this diagram, stars sharing the same temperature but different luminosities show measurable differences in radius as well as mass. So if two stars both have the same visible surface temperature but one is more luminous, it has to be larger. This is further explained by Stefan Boltzmann's law,L=(4πR^2 x sigma(constant) x T^4)A Star with the the same surface temperature with larger surface area results in more Luminosity.
That's done by analyzing the star's spectrum.
No, it is not necessarily giving off more energy. A stars brightness is connected to both it's distance from Earth, it's size, and the energy being produced. Therefore a distant star giving off lots of energy may not appear to be as bright as a closer star, that doesn't give off as much energy. This effect is similar to standing next to a 40w bulb and comparing it's brightness with a street light in the distance. The street light may be 200w and gives off more energy, but because we're standing next to the 40w bulb it appears brighter.
Giving off is emitting it and reflecting light is reflecting it
It would take 65 years for us on Earth to notice that the star has stopped giving off light energy, since light travels at the speed of light. This means that the light emitted by the star today will take 65 years to reach us, so we would only be able to detect the absence of its light after that time has passed.
That question is a little tricky. What is certain is that if the light is coming from a hot object like a glowing heater element, yellow light indicates that the element is hotter. It is also true that a single photon of yellow light carries more energy than a single photon of red light and is "hotter" in that sense. So in general, yellow light is hotter, but if you are simply talking about narrowband filtered light from a single source, a human will not ordinarily be able to sense the difference, and which color is more intense will depend on the spectrum of the light source.
Basically the more intense the heat the star is giving off the brighter the star will appear to be. For example, let's say there is a red colored star and a white colored star in the vicinity of the planet you're currently on to observe them. The red colored star is giving off light at the visible spectrum that reveals red most prominently. I'm not 100% sure how the colors line up but I do know that red is a lower frequency wavelength of light then pure white. This means that the higher the temperature of the star the brighter light it will give off, by bright of course I mean the higher frequency colors slowly approaching pure white light.
The amount of light a star gives off is the intensity or luminosity. Absolute Magnitude measures the amount of light given off.
Stars begin their life cycles in a nebula.
Magnitude
A off
Yes. Since it has a temperature, it is giving off infra-red radiation. Some parts of it (molten lavea) are so hot that they give off visible light as well.
A star
Blue photons are higher-energy than red photons. The equations governing the emission of light based solely on temperature state that an object giving off primarily blue light is hotter than one giving off primarily red light.