SATHEE: Heat and Thermodynamics 3 Question 26

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 $λ_{B}$ 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 $1.00 μ m$ . If the temperature of $A$ is $5802 K$

(1994, 2M)

(a) the temperature of $B$ is $1934 K$

(b) $λ_{B} = 1.5 μ m$

(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, $e_{A} σ T_{A}^{4} A = e_{B} σ T_{B}^{4} A$

$\begin{aligned} ∴ \frac{T_{A}}{T_{B}} = {(\frac{e_{B}}{e_{A}})}^{1 / 4} = {(\frac{0.81}{0.01})}^{1 / 4} = 3 \\ ∴ T_{B} = \frac{T_{A}}{3} = \frac{5802}{3} \\ = 1934 K \\ T_{B} = 1934 K \end{aligned}$

According to Wien’s displacement law,

$\begin{aligned} λ_{m} T & = constant \\ ∴ & λ_{A} T_{A} & = λ_{B} T_{B} \\ or & λ_{A} & = λ_{B} (\frac{T_{B}}{T_{A}}) = \frac{λ_{B}}{3} \end{aligned}$

Given, $λ_{B} - λ_{A} = 1 μ m$

$\Rightarrow λ_{B} - \frac{λ_{B}^{A}}{3} = 1 μ m$

$\begin{aligned} or \frac{2}{3} λ_{B} = 1 μ m \\ \Rightarrow λ_{B} = 1.5 μ m \end{aligned}$

NOTE

$λ_{m} T = b =$ 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.

Heat and Thermodynamics 3 Question 26

Answer:

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Heat and Thermodynamics 3 Question 26

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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