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NEET Physics Thermodynamical Processes NEET PYQ-Thermodynamical Processes

  • question_answer
    For a black body at temperature \[727{}^\circ C,\] its radiating power is 60W 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:                                                                                                                                   [AIPMT 2002]

    A)  123 W   

    B)                   240 W

    C)  304 W             

    D)       320 W

    Correct Answer: D

    Solution :

    Boltzmann corrected Stefan’s law and stated that the amount of radiations emitted by the body, not only depend upon the temperature of the body but also on the temperature of the surrounding. The power by the body is given by radiated \[P=\sigma \,\,({{T}^{4}}-T_{0}^{4})\]                            ...(i)
    where \[{{T}_{0}}\] is the absolute temperature of surrounding.
    \[\therefore \]      \[\frac{{{P}_{2}}}{{{P}_{1}}}=\left( \frac{T_{2}^{4}-T_{0}^{4}}{T_{1}^{4}-T_{0}^{4}} \right)\]                        ....(ii)
    Here,     \[{{p}_{1}}=60\,\,W,\,{{T}_{1}}={{727}^{o}}C=1000\,K\]
    \[{{T}_{0}}={{227}^{o}}C=500\,K,\,{{T}_{2}}={{1227}^{o}}C=1500\,K\]
    Substituting in relation Eq. (ii), we get
    \[{{P}_{2}}=\frac{{{(1500)}^{4}}-{{(500)}^{4}}}{{{(1000)}^{4}}-{{(500)}^{4}}}\times 60\]
    \[=\frac{{{(500)}^{4}}}{{{(500)}^{4}}}\times \left[ \frac{{{3}^{4}}-1}{{{2}^{4}}-1} \right]\times 60\]
    \[=\frac{80}{15}\times 60=320\,W\]


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