# Brightness change with temperature

• Krushnaraj Pandya
In summary, the black body will be initially brighter than the yellow body, but eventually both will have the same brightness.f

Gold Member

## Homework Statement

A black and yellow body at room temperature are thrown into a furnace at very high temperature. How will the initial and final brightness of both compare?

λT=constant

## The Attempt at a Solution

The rise in the black body's temperature will be higher initially, finally both will have the same temperature so the wavelength of light they emit will be initially lower for the black body and the same finally but the brightness is the number of photons not their energy so how do we deduce the brightness from this?

## Homework Statement

A black and yellow body at room temperature are thrown into a furnace at very high temperature. How will the initial and final brightness of both compare?

λT=constant

## The Attempt at a Solution

The rise in the black body's temperature will be higher initially, finally both will have the same temperature so the wavelength of light they emit will be initially lower for the black body and the same finally but the brightness is the number of photons not their energy so how do we deduce the brightness from this?
The question is a bit ambiguous but I think you are to asked to compare the relative brightness of the bodies at the following times: 1. before being put into the furnace and 2. after they have been placed in the furnace and have reached thermal equilibrium with the furnace. At both times, the bodies would have the same temperature.

What is the relationship between temperature and rate at which a black body radiates energy? How does emissivity of a body affect a body's rate of radiation emission compared to the rate of emission for a black body at the same temperature? How does the emissivity of the yellow body compare to that of a black body?

AM

CWatters
The question is a bit ambiguous but I think you are to asked to compare the relative brightness of the bodies at the following times: 1. before being put into the furnace and 2. after they have been placed in the furnace and have reached thermal equilibrium with the furnace. At both times, the bodies would have the same temperature.

What is the relationship between temperature and rate at which a black body radiates energy? How does emissivity of a body affect a body's rate of radiation emission compared to the rate of emission for a black body at the same temperature? How does the emissivity of the yellow body compare to that of a black body?

AM
I thought we were being asked to compare it just after they've been thrown into the furnace and after they've achieved thermal equilibrium (judging from the options in my book).
The formula is σεAT^4. ε for black body is 1 and for yellow body is lower. Since the black body gains temperature faster and also has a higher emissivity it should always be brighter but the answer given is initially yellow is brighter and finally both have same brightness

someone there?

I thought we were being asked to compare it just after they've been thrown into the furnace and after they've achieved thermal equilibrium (judging from the options in my book).
The formula is σεAT^4. ε for black body is 1 and for yellow body is lower. Since the black body gains temperature faster and also has a higher emissivity it should always be brighter but the answer given is initially yellow is brighter and finally both have same brightness
I am confused by the term "yellow body". How is it defined? Either that or I am confused by the term "brightness". It does not appear to be same as radiation power.

AM

Last edited:
CWatters
I am confused by the term "yellow body". How is it defined? Either that or I am confused by the term "brightness". It does not appear to be same as radiation power.

AM
I really have no clue. These are the very reasons I can't solve this problem

Oh, well. No point to keep discussing an ambiguous question- but I still learned a lot from what @Andrew Mason said, thank you for that :D