Heat TransferEasy
Question
Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface area of the two bodies are the sme. The two bodies emit total radiant power at the same rate. The wavelength λB corresponding to maximum spectral radiancy in the radiation from B shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A, by 1.00 μm. If the temperature of A is 5802 K
Options
A.the temperature of B is 1934 K
B.λB = 1.5 μm
C.the temperature of B is 11604 K
D.the temperature of B is 2901 K
Solution
Power radiate and surface area is same for both A and B
Therefore, eAσTA4 A = eB σTB4A
∴
∴
TB = 1934K
According to Wien’s displacement law,
λmT = constant
∴ λATA = λBTB or λA = λB
Given, λB - λA = 1 μm
⇒
or
Therefore, eAσTA4 A = eB σTB4A
∴
∴
TB = 1934K
According to Wien’s displacement law,
λmT = constant
∴ λATA = λBTB or λA = λB
Given, λB - λA = 1 μm
⇒
or
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