If a star has twice the temperature, how much more luminous would it be?

Question

BlackTom AI · Accepted Answer

The Stefan–Boltzmann law states that a star’s luminosity L (total power radiated) is L = σ A T^4, and for a spherical star A = 4πR^2, so L = 4πσR^2T^4. Key concepts: blackbody radiation, proportionality of luminosity to surface area and the fourth power of temperature.

Step-by-step reasoning:
1. Assume the star’s radius R (surface area) does not change; only the surface temperature T changes. This is the standard interpretation when asking how luminosity scales with temperature alone.
2. Write the luminosity dependence as L ∝ T^4 (with R constant).
3. If the temperature doubles, new temperature T2 = 2T1.
4. Compute the new luminosity factor: L2/L1 = (T2/T1)^4 = (2T1/T1)^4 = 2^4.
5. Evaluate 2^4 = 16.

All intermediate arithmetic: (2)^4 = 2 × 2 × 2 × 2 = 16, so L2 = 16 L1.

Why this is correct: Stefan–Boltzmann’s T^4 dependence means small changes in temperature produce large changes in emitted power per unit area; doubling temperature increases emitted power per unit area by 16×, and with the same area the total luminosity increases by the same factor.

Why other options are incorrect:
- "8 times more luminous": would require an L ∝ T^3 dependence (2^3 = 8), which is not how blackbody radiation behaves.
- "it would have the same luminosity": contradicts the T^4 dependence; changing temperature changes luminosity.
- "2 times more luminous": corresponds to a linear dependence L ∝ T, not applicable for blackbody emitters.

Therefore, with radius held constant, doubling the star’s temperature makes it 16 times more luminous. This matches the provided correct answer: 16 times more luminous.

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