JEE Main & Advanced Physics Thermodynamical Processes Question Bank Heat Engine Refrigerator and Second Law of Thermodynamics

  • question_answer
    A carnot engine has the same efficiency between  800 K to 500 K and x K to 600 K. The value of x is [Pb. PMT 1996; CPMT 1996]

    A)                                                      1000 K                                     

    B)            960 K

    C)            846 K                                       

    D)            754 K

    Correct Answer: B

    Solution :

                       In first case, \[({{\eta }_{1}})=1-\frac{500}{800}=\frac{3}{8}\]                    and in second case, \[({{\eta }_{2}})=1-\frac{600}{x}\]                    Since \[{{\eta }_{1}}={{\eta }_{2}},\] therefore \[\frac{3}{8}=1-\frac{600}{x}\]            or \[\frac{600}{x}=1-\frac{3}{8}=\frac{5}{8}\]or\[x=\frac{600\times 8}{5}=960K\]


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