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CMC Medical CMC-Medical VELLORE Solved Paper-2009

  • question_answer
    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

    Correct Answer: D

    Solution :

                    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 \[\therefore \]  \[{{E}_{2}}=\frac{{{(1500)}^{4}}-{{(500)}^{4}}}{{{(1000)}^{4}}-{{(500)}^{4}}}\times 60\] \[=\frac{{{(500)}^{4}}[{{3}^{4}}-1]}{{{(500)}^{4}}[{{2}^{4}}-1]}\times 60\] \[=\frac{80}{15}\times 60=320\,W\]


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