Turnbull Stephens
01/28/2023 · Senior High School
For a body emitting blackbody radiation, the total power emitted is proportional to the 4 th power of the body's absolute temperature: \( P \propto T^{A} \) ( \( T \) 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: \( \lambda_{\max }=\frac{0.0029}{T} \) ( \( \lambda_{\max } \) in meters, \( T \) in kelvins) Assume an object is emitting blackbody radiation. If the Kelvin temperature of an object is doubled, the amount of radiant energy emitted each second is the original amount.
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If the Kelvin temperature of an object is doubled, the radiant energy emitted increases by a factor of 16, and the wavelength of maximum intensity is halved.
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