ThermodynamicsHard
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
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
In the following $p - V$ diagram the equation of state along the curved path is given by $(V - 2)^{2} = 4ap$ where $a$ i...One mole of a monatomic ideal gas is taken along two cyclic processes E → F → G → E and E → F &#...On a cold morning, a metal surface on touching is felt colder than a wooden surface,because the metal has:...Four curves A, B, C and D are drawn in the adjoining figure for a given amount of gas. The curves which represent adiaba...A black body is heated from 27oC to 127oC. The ratio of their energies of radiation emitted will be:...