Question

Energy is being emitted from the surface of a black body at $ 127{}^\circ C $ at the rate of $ (1.0\times {{10}^{6}})/s{{m}^{2}} $ . The temperature of a black body at which the rate of energy emission is $ (16.0\times {{10}^{6}})/s{{m}^{2}} $ will be

• A. $ 727{}^\circ C $
• B. $ 527{}^\circ C $
• C. $ 508{}^\circ C $
• D. $ 254{}^\circ C $

Answer 1

Answer: (B)

Here, $ {{T}_{1}}=127{}^\circ C=400K $
$ {{E}_{2}}=16\times {{10}^{6}}J/s{{m}^{2}} $
$ {{E}_{1}}=1\times {{10}^{6}}J/s{{m}^{2}} $
Using the relation. $ \frac{{{E}_{2}}}{{{E}_{1}}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{4}} $
$ \frac{{{T}_{2}}}{{{T}_{1}}}=\frac{{{E}_{2}}}{{{E}_{1}}}={{\left( \frac{16.0\times {{10}^{6}}}{1\times {{10}^{6}}} \right)}^{1/4}}=2 $
$ {{T}_{2}}=2\times {{T}_{1}}=2\times 400=800K $
$ {{T}_{2}}=527{}^\circ C $