The spectrum of a black body at two temperatu | vedclass

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(D) The area under the intensity versus wavelength curve for a black body represents the total emissive power $(E)$ of the body.According to the Stefa

Vedclass · Accepted Answer

$16:1$

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(D) The area under the intensity versus wavelength curve for a black body represents the total emissive power $(E)$ of the body.According to the Stefan-Boltzmann law,the total emissive power is proportional to the fourth power of the absolute temperature: $E \propto T^4$.Therefore,the ratio of the areas under the curves is equal to the ratio of the fourth powers of their absolute temperatures: $\frac{A_2}{A_1} = \frac{T_2^4}{T_1^4}$.Given temperatures are $T_1 = 27^oC = (27 + 273) K = 300 K$ and $T_2 = 327^oC = (327 + 273) K = 600 K$.Substituting these values into the ratio: $\frac{A_2}{A_1} = \left( \frac{600}{300} \right)^4 = (2)^4 = 16$.Thus,the ratio $\frac{A_2}{A_1} = 16:1$.

The spectrum of a black body at two temperatures $27^oC$ and $327^oC$ is shown in the figure. Let $A_1$ and $A_2$ be the areas under the two curves respectively. The value of $\frac{A_2}{A_1}$ is

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