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 $5802\, K$, then
(1) the temperature of $B$ is $1934\, K$
(2) $\lambda_{B}=1.5 \, \mu m$
(3) the temperature of $B$ is $11604\, K$
(4) the temperature of $B$ is $2901 \, K$

• 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}=3 T_{B} $
$T_{B}=\frac{1}{3} T_{A}=\frac{1}{3}(5802\, K) $
$T_{B}=1934 \,K $
$\lambda_{A}=\frac{1}{2} \lambda_{B}$ (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 $
$\lambda_{B}=1.5\, \mu m$