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 same. The two bodies 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 $5802K$, then:

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

Answer 1

Answer: (A), (B)

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

From Stefan's Law:
$\sigma A{ϵ}_A{T}_A^4=\sigma A{ϵ}_B{T}_B^4$ ....(1)
where, $T_A=5802K$ is temp of A and $T_B$ is temp of B,

${ϵ}_A=0.01$ is emissivity of A,
${ϵ}_B=0.81$ is emissivity of B,
$\sigma$ is Stefan's constant,
$A$ is the surface area of the bodies A and B

Substituting the values in (1)

$0.01\times {5802}^4=0.81T_B^4$

or, ${\left(\frac{T_B}{5802}\right)}^4=\frac{0.01}{0.81}={\left(\frac{1}{3}\right)}^4$

$\therefore T_B=\frac{5802}{3}=1934K$

From Wien's displacement Law
$({\lambda }_A{)}_m{T}_A=({\lambda }_B{)}_m{T}_B$ ......(2)

Given, $({\lambda }_B{)}_m=({\lambda }_A{)}_m+1\times {10}^{-6}$ ....(3)

Substituting $({\lambda }_B{)}_m$ from (3) in (2)
$({\lambda }_A{)}_m{T}_A=(({\lambda }_A{)}_m+1\times {10}^{-6})T_B$
$\therefore ({\lambda }_A{)}_m\times 3=({\lambda }_A{)}_m+1\times {10}^{-6}$ since $\frac{T_A}{T_B}=3$
$\therefore 2({\lambda }_A{)}_m={10}^{-6}$
$\therefore ({\lambda }_A{)}_m=0.5\times {10}^{-6}$
$\therefore ({\lambda }_B{)}_m=0.5\times {10}^{-6}+1\times {10}^{-6}=1.5\times {10}^{-6}=1.5\mu m$