Thermal Radiation And Stefan's Law
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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Solution
64
A black body radiates energy at a rate governed by the Stefan–Boltzmann law, P=σAT4, where the emitting area A=4πR2 for a sphere and T is the absolute surface temperature. Combining these, the total power is P=σ(4πR2)T4, so P∝R2T4. Forming the ratio for the two stars, P1P2=(R1R2)2(T1T2)4=(2)2×(2)4=4×16=64. The value 8 ignores the temperature exponent and uses R2T. The value 16 keeps only the T4 factor and forgets the area. The value 32 mixes an incorrect R2T3 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.
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About This Question
- Subject
- physics
- Chapter
- properties of solids and liquids
- Topic
- thermal radiation and stefan's law
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
64
A black body radiates energy at a rate governed by the Stefan–Boltzmann law, P=σAT4, where the emitting area A=4πR2 for a sphere and T is the absolute surface temperature. Combining these, the total power is P=σ(4πR2)T4, so P∝R2T4. Forming the ratio for the two stars, P1P2=(R1R2)2(T1T2)4=(2)2×(2)4=4×16=64. The value 8 ignores the temperature exponent and uses R2T. The value 16 keeps only the T4 factor and forgets the area. The value 32 mixes an incorrect R2T3 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.
This hard difficulty physics question is from the chapter properties of solids and liquids, covering the topic of thermal radiation and stefan's law. It appeared in the 2025 exam.
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