Transmission Of Heat Question 110
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. 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 $ is 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 $
[IIT 1994; DCE 1996]
Options:
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 $
Show Answer
Answer:
Correct Answer: A
Solution:
According to Stefan’s law $ E=eA\sigma T^{4}\Rightarrow E _{1}=e _{1}A\sigma T _{1}^{4} $ and $ E _{2}=e _{2}A\sigma T _{2}^{4} $
$ \because $
$ E _{1}=E _{2} $
$ \
Therefore $ $ e _{1}T _{1}^{4}=e _{2}T _{2}^{4} $
$ \Rightarrow $ $ T _{2}={{( \frac{e _{1}}{e _{2}}T _{1}^{4} )}^{\frac{1}{4}}}={{( \frac{1}{81}\times {{(5802)}^{4}} )}^{\frac{1}{4}}} $
$ \Rightarrow $ $ T _{B}=1934\ K $ And, from Wein’s law $ {\lambda _{A}}\times T _{A}={\lambda _{B}}\times T _{B} $
$ \Rightarrow \frac{{\lambda _{A}}}{{\lambda _{B}}}=\frac{T _{B}}{T _{A}} $
$ \Rightarrow $ $ \frac{{\lambda _{B}}-{\lambda _{A}}}{{\lambda _{B}}}=\frac{T _{A}-T _{B}}{T _{A}} $
$ \Rightarrow $ $ \frac{1}{{\lambda _{B}}}=\frac{5802-1934}{5802}=\frac{3968}{5802}\Rightarrow {\lambda _{B}}=1.5\ \mu m $
BETA
