The Stefan-Boltzmann law says that hot objects emit energy proportional to the fourth power of their temperature.One star has a temperature of 30,000 K and another star has a temperature of 6,000 K.Compared to the cooler star,how much more energy per second will the hotter star radiate from each square metre of its surface?

Question

Quizplus · Accepted Answer

According to the Stefan-Boltzmann law, the energy radiated per second per unit area, or the radiant flux density, is proportional to the fourth power of the temperature. 
Thus, for the two stars in question, the ratio of their radiant flux densities would be:
$$(\frac{F_1}{F_2}) = (\frac{T_1}{T_2})^4$$
Substituting the temperatures,
$$(\frac{F_1}{F_2}) = (\frac{30,000}{6,000})^4 = 5^4 = 625$$
Thus, the hotter star will radiate 625 times more energy per second from each square metre of its surface compared to the cooler star. 
Therefore, the correct answer is choice C) 625 times.

The Stefan-Boltzmann Law Says That Hot Objects Emit Energy Proportional