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Two bodies A and B have thermal emissivities of 0.81 respectively. The outer surface areas of the two bodies are the same. The two bodies radiate energy at the same rate. The wavelength λB, corresponding to the maximum spectral radiancy in the radiation from B, is shifted from the wavelength corresponding to the maximum spectral radiancy in the radiation from A by 1.00 µm. If the temperature of A is 5802 K,

(A) The temperature of B is 1934 K

(B) λB = µ 1.5 m

(C) The temperature of B is 11604 K

(D) The temperature of B is 2901 K

1 Answer

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Final answer:

To find the temperature of body B, one must use Wien's displacement law in conjunction with the given temperature of body A and the shift in peak wavelength. The resulting calculation does not match any of the provided answer choices, suggesting an issue with the given answers or the question's information.

Step-by-step explanation:

The question involves the concept of blackbody radiation in physics, especially focusing on Wien's displacement law and the Stefan-Boltzmann law. Using Wien's displacement law, which is given as λmax T = 2.898 × 10-3 m·K, we can calculate the temperature of body B given that the peak wavelength shift from body A is 1.00 µm and the temperature of A is 5802 K.

First, we find the wavelength of maximum spectral radiancy for body A using Wien's displacement law: λA = λmax = (2.898 × 10-3) / 5802 K. Then, we get λB = λA + 1.00 µm. Knowing that body B radiates energy at the same rate and using the equation for Wien's displacement law, we can find the temperature of B by rearranging Wien's law to T = (2.898 × 10-3) / λB.

After performing the calculation, we can see which option correctly represents the temperature of body B. The correct answer is that the temperature of B is not listed in the options provided (2901 K, 1934 K, or 11604 K), indicating there may be an issue either with the provided answers or the question's information being incomplete.

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