Heat and Thermodynamics 3 Question 26
26. 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 $\lambda _B$ corresponding to maximum spectral radiancy in the radiation from $B$ shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from $A$, by $1.00 \mu m$. If the temperature of $A$ is $5802 K$
(1994, 2M)
(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 $2901 K$
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Answer:
Correct Answer: 26. (a,b)
Solution:
- Power radiated and surface area is same for both $A$ and $B$. Therefore, $\quad e _A \sigma T _A^{4} A=e _B \sigma T _B^{4} A$
$$ \begin{aligned} & \therefore \quad \frac{T _A}{T _B}=\left(\frac{e _B}{e _A}\right)^{1 / 4}=\left(\frac{0.81}{0.01}\right)^{1 / 4}=3 \\ & \therefore \quad T _B=\frac{T _A}{3}=\frac{5802}{3} \\ & =1934 K \\ & T _B=1934 K \end{aligned} $$
According to Wien’s displacement law,
$$ \begin{array}{rlrl} & & \lambda _m T & =\text { constant } \\ \therefore & \lambda _A T _A & =\lambda _B T _B \\ \text { or } \quad & \lambda _A & =\lambda _B\left(\frac{T _B}{T _A}\right)=\frac{\lambda _B}{3} \end{array} $$
Given, $\quad \lambda _B-\lambda _A=1 \mu m$
$\Rightarrow \quad \lambda _B-\frac{\lambda _B^{A}}{3}=1 \mu m$
$$ \begin{aligned} & \text { or } \quad \frac{2}{3} \lambda _B=1 \mu m \\ & \Rightarrow \quad \lambda _B=1.5 \mu m \end{aligned} $$
NOTE
$\lambda _m T=b=$ Wien’s constant value of this constant for perfectly black body is $2.89 \times 10^{-3} m-K$. For other bodies this constant will have some different value. In the opinion of author option (b) has been framed by assuming $b$ to be constant for all bodies. If we take $b$ different for different bodies. Option (b) is incorrect.
