Question

For a black body at temperature $ 727{}^\circ C $ its radiating power is 60 W and temperature of surrounding is $ 227{}^\circ C $ . If the temperature of the black body is changed to $ 1227{}^\circ C, $ then its radiating power will be

• A. 120 W
• B. 240 W
• C. 304 W
• D. 320 W

Answer 1

Answer: (D)

From Stefans-Boltzmann law $ E=\sigma ({{T}^{4}}-T_{0}^{4}) $ $ \frac{{{E}_{2}}}{{{E}_{1}}}=\frac{(T_{2}^{4}-T_{0}^{4})}{(T_{1}^{4}-T_{0}^{4})} $ $ \Rightarrow $ $ {{E}_{2}}=\left( \frac{T_{2}^{4}-T_{0}^{4}}{T_{1}^{4}-T_{0}^{4}} \right){{E}_{1}} $ ?(i) Here $ {{E}_{1}}=60\,\,W, $ $ {{T}_{0}}=227{}^\circ C=500\,K, $ $ {{T}_{1}}=727{}^\circ C=1000\,K, $ $ {{T}_{2}}=1227{}^\circ C=1500\,K $ From Eq. (i), we get $ \therefore $ $ {{E}_{2}}=\frac{{{(1500)}^{4}}-{{(500)}^{4}}}{{{(1000)}^{4}}-{{(500)}^{4}}}\times 60 $ $ =\frac{{{(500)}^{4}}[{{3}^{4}}-1]}{{{(500)}^{4}}[{{2}^{4}}-1]}\times 60 $ $ =\frac{80}{15}\times 60=320\,W $