Thermal Radiation And Stefan's Law

Question

Two stars behave as perfect black bodies, with the second having twice the surface temperature and twice the radius of the first. What is the ratio of the total radiant power emitted by the second star to that emitted by the first?

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A black body radiates energy at a rate governed by the Stefan–Boltzmann law, $P = \sigma A T^4$, where the emitting area $A = 4\pi R^2$ for a sphere and $T$ is the absolute surface temperature. Combining these, the total power is $P = \sigma\,(4\pi R^2)\,T^4$, so $P \propto R^2 T^4$. Forming the ratio for the two stars, $\frac{P_2}{P_1} = \left(\frac{R_2}{R_1}ight)^2\left(\frac{T_2}{T_1}ight)^4 = (2)^2	imes(2)^4 = 4	imes16 = 64$. The value 8 ignores the temperature exponent and uses $R^2T$. The value 16 keeps only the $T^4$ factor and forgets the area. The value 32 mixes an incorrect $R^2T^3$ combination. This strong dependence, especially the fourth power of absolute temperature, is the essence of the NCERT treatment of black-body radiation, and it must use the Kelvin scale because $T$ measures the true thermal energy content of the radiator. Wien's displacement law accompanies Stefan's law and tells us the hotter star also radiates its peak at a shorter wavelength, appearing bluer than its cooler companion. As a plausibility check, doubling the temperature alone already multiplies power sixteenfold, and the larger surface adds another factor of four, so a total ratio of 64 is the physically expected, sharply amplified result for a hotter, bigger star.

Thermal Radiation And Stefan's Law

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