JEE Main & Advanced Physics Thermodynamical Processes Question Bank Topic Test - Thermodynamical Process (2021)

  • 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

    A) 1000 K

    B) 960 K

    C) 846 K

    D) 754 K

    Correct Answer: B

    Solution :

    [b] 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\[x=\frac{600\times 8}{5}=960K\]

You need to login to perform this action.
You will be redirected in 3 sec spinner