For a body emitting blackbody radiation, the total power emitted is proportional to the 4th power of the body's absolute temperature: ( T in kelvins)
And the wavelength of the emitted EM radiation that has the highest intensity is inversely proportional to the body's absolute temperature according to: ( in meters, T in kelvins)
Assume an object is emitting blackbody radiation.
As the temperature of a body increases,
A) the body gets brighter.
B) the color of the body shifts from red to white.
C) the wavelength of the most intense light decreases.
D) the frequency of the most intense light increases.
E) all the above.

Question

For a body emitting blackbody radiation, the total power emitted is proportional to the 4th power of the body's absolute temperature:  ( T in kelvins)
And the wavelength of the emitted EM radiation that has the highest intensity is inversely proportional to the body's absolute temperature according to:  (  in meters, T in kelvins)
Assume an object is emitting blackbody radiation.
As the temperature of a body increases,
A) the body gets brighter. 
B) the color of the body shifts from red to white. 
C) the wavelength of the most intense light decreases. 
D) the frequency of the most intense light increases. 
E) all the above.

Quizplus · Accepted Answer

As per the first law, the total power emitted is proportional to T^4. This means that as the temperature increases, the body will get brighter. As per the second law, the wavelength of the most intense light is inversely proportional to T. This means that as the temperature increases, the wavelength of the most intense light decreases, which corresponds to a shift from red to white. The frequency of the most intense light is directly proportional to the temperature, but this is not explicitly given as an option. Thus, the correct answer is E as all the above statements are true.

For a Body Emitting Blackbody Radiation, the Total Power Emitted