First slide

Radiation

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 $λ_{B}$ corresponding to maximum spectral radiancy in the radiation from B is greater than the wavelength I, corresponding to maximum spectral radiancy in the radiation from A by 1.00 $μ$ m. If the temperature of A is 5802 K

Moderate

A

the temperature of B is 1934 K
B

$λ_{B} = 1 \cdot 5 μm$
C

the temperature of B is 11604 K
D

the temperature of B is 2901 K

Solution

Power radiated $P = EA = {eσT}^{4} A$
Now    $e_{A} T_{A}^{4} = e_{B} T_{B}^{4} (∵ P_{1} = P_{2})$
or $T_{B} = {(\frac{e_{A}}{e_{B}})}^{1 / 4} \times T_{A} = {(\frac{0 \cdot 01}{0 \cdot 81})}^{1 / 4} \times 5802$
   $= \frac{1}{3} \times 5802 or T_{B} = 1934 K$
According to Wien's displacement law,
$λ_{m} T = constant$
So    $λ_{A} T_{A} = λ_{B} T_{B} or (λ_{B} / λ_{A}) = (T_{A} / T_{B})$
or    $\frac{λ_{B}}{λ_{A}} = \frac{5802}{1934} or λ_{A} = (λ_{B} / 3)$
Further, $λ_{B} - λ_{A} = 1 μm$
or $λ_{B} - (\frac{λ_{B}}{3}) = 1 μm or λ_{B} = 1 \cdot 5 μm$

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