According to the Stefan-Boltzmann Law, is the total energy emitted by a blackbody linear with respect to temperature?

According to the Stefan-Boltzmann Law, is the total energy emitted by a blackbody linear with respect to temperature?

• A. Yes
• B. No - fourth power
• C. It is proportional to temperature squared
• D. It depends on emissivity

Answer: (B)

The Stefan-Boltzmann law shows that the energy radiated by a blackbody scales with the fourth power of its absolute temperature, not linearly. Specifically, the power per area is P/A = σ T^4, and for a blackbody with area A, P = ε σ A T^4 with ε = 1. This means as temperature increases, the emitted energy rises much faster than a straight line, following T^4 rather than T. The idea of linear dependence would violate this law. The note about emissivity only matters for non-ideal surfaces; for a true blackbody emissivity is 1, so the T^4 relationship is the defining factor. The quadratic option is also incorrect because the law uses the fourth power, not the square.

The Stefan-Boltzmann law shows that the energy radiated by a blackbody scales with the fourth power of its absolute temperature, not linearly. Specifically, the power per area is P/A = σ T^4, and for a blackbody with area A, P = ε σ A T^4 with ε = 1. This means as temperature increases, the emitted energy rises much faster than a straight line, following T^4 rather than T. The idea of linear dependence would violate this law. The note about emissivity only matters for non-ideal surfaces; for a true blackbody emissivity is 1, so the T^4 relationship is the defining factor. The quadratic option is also incorrect because the law uses the fourth power, not the square.