Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Two bodies A and B have thermal emissivities of 0.01 and 0.81, respectively. The outer surface areas of two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength\[{{\lambda }_{B}}\]corresponding to maximum spherical spectral radiance in the radiation from B is shifted from the wavelength corresponding to the maximum spectral radiance in radiation A by\[1.00\,\mu m\]. If the temperature of A is 5802 K, then (1) the temperature of B is 1934 K (2) \[{{\lambda }_{B}}=1.5\,\mu m\] (3) the temperature of B is 11604 K (4) the temperature of B is 2901 KA) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: B
Solution :
Since both bodies emit total radiant power at the same rate. \[{{e}_{A}}\sigma (T_{A}^{4})={{e}_{B}}\sigma (T_{B}^{4})\] \[0.01\,(T_{A}^{4})=0.81(T_{B}^{4})\] \[{{T}_{A}}=3{{T}_{B}}\] \[{{T}_{B}}=\frac{1}{3}{{T}_{A}}=\frac{1}{3}(5802K)\] \[{{T}_{B}}=1934K\] \[{{\lambda }_{A}}=\frac{1}{2}{{\lambda }_{B}}\](due to Wien's law) Since \[{{\lambda }_{B}}-{{\lambda }_{A}}=1\,\mu m\] \[{{\lambda }_{B}}-\frac{{{\lambda }_{B}}}{3}=1\,\mu m\] \[\frac{2}{3}{{\lambda }_{B}}=1\,\mu m\] \[{{\lambda }_{B}}=1.5\,\mu m\]
You need to login to perform this action.
You will be redirected in
3 sec
