Question

If the surface temperature of hot body is doubled, then rate of radiation energy emitted will be enhanced by a factor of

• A. 2
• B. 4
• C. 16
• D. 32

Hint:

Always remember that radiation heat transfer scales with the fourth power of absolute temperature ($T^4$): - Temperature doubled ($2T$) $\implies$ Radiation increases by $2^4 = 16$ times. - Temperature tripled ($3T$) $\implies$ Radiation increases by $3^4 = 81$ times. - Temperature quadrupled ($4T$) $\implies$ Radiation increases by $4^4 = 256$ times. This non-linear scaling makes radiation the dominant mode of heat transfer at high temperatures.

Answer 1

The Correct Option is C

Step 1: Pick a convenient starting temperature.
Since only the ratio matters, take a sample value \(T_1 = 300 \text{ K}\). Doubling the surface temperature gives \(T_2 = 600 \text{ K}\). By the Stefan-Boltzmann law, emissive power scales as \(E \propto T^4\).

Step 2: Compute both fourth powers directly.
\[ T_1^4 = 300^4 = 8.1 \times 10^{9} \] \[ T_2^4 = 600^4 = 1.296 \times 10^{11} \]

Step 3: Take the ratio of emitted energy.
\[ \frac{E_2}{E_1} = \frac{T_2^4}{T_1^4} = \frac{1.296 \times 10^{11}}{8.1 \times 10^{9}} = 16 \] \[ \boxed{E_2 = 16 \, E_1} \] Since any starting temperature would give the same ratio, the emitted radiation rises by a factor of 16 whenever the absolute temperature is doubled, matching option 3.