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 and radiate energy at the same rate. The wavelength ${\lambda }_B$, corresponding to the maximum spectral radiancy in the radiation from $B$, is shifted from the wavelength corresponding to the maximum spectral radiancy in the radiation from $A$ by $1.00\mu m$. If the temperature of $A$ is $5802$ K,

• A. The temperature of $B$ is 1934 K
• B. ${\lambda }_B=1.5\mu m$
• C. The temperature of $B$ is 11604 K
• D. The temperature of $B$ is 2901 K

Answer 1

Answer: (A), (B)

The correct options are
A The temperature of $B$ is 1934 K
B ${\lambda }_B=1.5\mu m$

Power radiated: $P=eA\sigma T^4$

$\frac{P_A}{P_B}=\frac{e_A}{e_B}\frac{T_A^4}{T_B^4}=\frac{0.01}{0.81}\frac{T_A^4}{T_B^4}$

Given, $P_A=P_B$

$\Rightarrow {\left(\frac{T_A}{T_B}\right)}^4=81=3^4$

$\frac{T_A}{T_B}=3$

Given, $T_A=5802K$

$\Rightarrow T_B=\frac{T_A}{3}=\frac{5802}{3}={1934}^{0}K$

$({\lambda }_m{)}_A{T}_A=({\lambda }_m{)}_B{T}_B=\left(\right({\lambda }_m{)}_A+1)\frac{T_A}{3}$

$\frac{\left(\right({\lambda }_m{)}_A+1)}{({\lambda }_m{)}_A}=3$

$\Rightarrow \frac{1}{({\lambda }_m{)}_A}=2$

$\Rightarrow ({\lambda }_m{)}_B=({\lambda }_m{)}_A+1=1.5\mu m$