NEET Sample Paper NEET Sample Test Paper-19

  • question_answer The efficiency of a Carnot engine operating between temperatures of \[100{}^\circ C\] and \[-23{}^\circ C\] will be

    A) \[\frac{100-23}{273}\]                      

    B) \[\frac{100+23}{273}\]

    C) \[\frac{100+23}{100}\]                    

    D) \[\frac{100-23}{100}\]   

    Correct Answer: B

    Solution :

    Efficiency of Carnot engine is given by \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}\] Given, \[{{T}_{1}}=\]temperature of reservoir \[=100+273=373\,K\] \[{{T}_{2}}=\]temperature of sink \[=-\,23\,+\,273\] = 250 K Substituting in Eq.(i), we get \[\therefore \]\[\eta =\frac{373-250}{373}=\frac{123}{373}\] \[=\frac{100+23}{373}\] Hence, the correction option is [b].


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