if the temperature of star b is twice the temperature of star a, what can we say about the energy emitted by the surface of star b compared to the energy emitted by star a?



Answer :

The blackbody radiation energy emitted by star a is 1/16 factor of star b energy.

We need to know about black body radiation to solve this problem. The energy radiated by a black body object is proportional to the area and the fourth power of temperature. It can be determined as

P = A . e . σ . T⁴

where P is power, A is surface area, e is emissivity, σ is Stefan Boltzmann's constant ( 5.67 x 10¯⁸ W/m²K⁴) and T is temperature.

From the question above, we know that

Tb = 2 Ta

By using the black body radiation formula, the ratio of temperature is

Ta⁴ / Tb⁴ = Pa / Pb

Ta⁴ / Tb⁴ = Ea / Eb

Hence,

Ta⁴ / Tb⁴ = Ea / Eb

Ta⁴ / (2Ta)⁴ = Ea / Eb

Ta⁴ / 16Ta⁴ = Ea / Eb

1/16 . Eb = Ea

Thus, the energy of star a will have 1/16 factor of star b energy

Find more on black body radiation at: https://brainly.com/question/18273651

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