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Question

Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are the same . 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


A

the temperature of B is 1934 K

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B

λB=1.5μm

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C

the temperature of B is 11604 K

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D

the temperature of B is 2901 K

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Solution

The correct options are
A

the temperature of B is 1934 K


B

λB=1.5μm


Power radiated and surface area is same for both A and B .
Therefore, eAσT4AA=eBσT4BA
TATB=(eBeB)14=(0.810.01)14=3
thereforeTB=TA3=58023
= 1934 K
TB=1934K
According to Wien’ s displacement law,
λmT=constant
λATA=λBTB
or λATB=λB(TBTA)=λB3
Given, λBλA=1μm
λBλB3=1μm
or 23λB=1μm
λB=1.5μm

Note: λmT=B = Wien’s constant value of this constant for perfectly black body is 2.89×103 m-K. For other bodies this constant will have some different value. In the option (b) has different for different bodies. Option (b) is incorrect.


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