Question

Two bodies $A$ and $B$ have thermal emissivities of $0.01$ and $0.81$, respectively. The outer surface areas of two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength ${\lambda }_B$ corresponding to maximum spherical spectral radiance in the radiation from $B$ is shifted from the wavelength corresponding to the maximum spectral radiance in radiation $A$ by $1.00\mu m$. If the temperature of $A$ is $5802K$, then
(1) the temperature of $B$ is $1934K$
(2) ${\lambda }_B=1.5\mu m$
(3) the temperature of $B$ is $11604K$
(4) the temperature of $B$ is $2901K$

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (B)

Since both bodies emit total radiant power at the same rate.
$e_A\sigma \left(T_A^4\right)=e_B\sigma \left(T_B^4\right)$
$0.01\left(T_A^4\right)=0.81\left(T_B^4\right)$
$T_A=3T_B$
$T_B=\frac{1}{3}{T}_A=\frac{1}{3}(5802K)$
$T_B=1934K$
${\lambda }_A=\frac{1}{2}{\lambda }_B$ (due to Wien's law) Since
${\lambda }_B-{\lambda }_A=1\mu m$
${\lambda }_B-\frac{{\lambda }_B}{3}=1\mu m$
$\frac{2}{3}{\lambda }_B=1\mu m$
${\lambda }_B=1.5\mu m$