Question

Two bodies of same shape, same size and same radiating power have emissivitys 0.2 and 0.8. The ratio of their temperatures is:

• A. $ \sqrt{3}:1 $
• B. $ \sqrt{2}:1 $
• C. $ 1:\sqrt{5} $
• D. $ 1:\sqrt{8} $

Answer 1

Answer: (B)

Radiating power of a body of area $ {{\text{A}}_{1}}, $ emissivity $ {{e}_{1}} $ and surface temperature $ {{T}_{1}} $ is $ {{P}_{1}}=\sigma {{e}_{1}}T_{1}^{4}{{T}_{1}} $ ?(i) Similarly radiating power of a body of area $ {{A}_{2}}, $ emissivity $ {{e}_{2}} $ and surface temperature $ {{T}_{2}} $ is $ {{P}_{2}}=\sigma {{e}_{2}}T_{2}^{4}{{A}_{2}} $ ?(ii) Given: $ {{P}_{1}}={{P}_{2}} $ and $ {{A}_{1}}={{A}_{2}}, $ $ \sigma {{e}_{1}}T_{1}^{4}{{A}_{1}}=\sigma {{e}_{2}}T_{2}^{4}{{A}_{2}} $ $ \Rightarrow $ $ {{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}=\left( \frac{{{e}_{2}}}{{{e}_{1}}} \right)\left( \frac{{{A}_{2}}}{{{A}_{1}}} \right) $ $ \frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{e}_{2}}}{{{e}_{1}}} \right)}^{1/4}}.1 $ $ (\because {{A}_{1}}={{A}_{2}}) $ Here $ {{e}_{1}}=0.2,\,{{e}_{2}}=0.8 $ Hence, $ \frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{0.8}{0.2} \right)}^{1/4}}={{\left( \frac{4}{1} \right)}^{1/4}}={{\left( \frac{{{2}^{2}}}{{{1}^{1}}} \right)}^{1/4}} $ $ \frac{{{T}_{1}}}{{{T}_{1}}}={{\left( \frac{2}{1} \right)}^{1/2}}=\sqrt{\frac{2}{1}} $ $ {{T}_{1}}:{{T}_{2}}=\sqrt{2}:1 $