The spectrum of a black body at two temperatu | vedclass

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(D) According to Stefan-Boltzmann law,the total energy radiated per unit area per unit time (emissive power $E$) by a black body is directly proportio

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$16 : 1$

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(D) According to Stefan-Boltzmann law,the total energy radiated per unit area per unit time (emissive power $E$) by a black body is directly proportional to the fourth power of its absolute temperature $(T)$.$E \propto T^4$Since the area under the intensity-wavelength curve represents the total emissive power,we have $A \propto T^4$.Given temperatures are $T_1 = 27\,^oC = (27 + 273)\,K = 300\,K$ and $T_2 = 327\,^oC = (327 + 273)\,K = 600\,K$.Therefore,the ratio of the areas is:$\frac{A_2}{A_1} = \frac{T_2^4}{T_1^4} = \left(\frac{600}{300}\right)^4 = (2)^4 = 16$.Thus,the value of $\frac{A_2}{A_1}$ is $16 : 1$.

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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