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

  • question_answer
    Efficiency of a Carnot engine is 50% when temperature of outlet is 500 K. In order to increase efficiency up to 60% keeping temperature of intake the same what is temperature of outlet                [CBSE PMT 2002]

    A)            200 K                                       

    B)            400 K

    C)            600 K                                       

    D)            800 K

    Correct Answer: B

    Solution :

                       \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\]Þ\[\frac{1}{2}=1-\frac{500}{{{T}_{1}}}\]Þ \[\frac{500}{{{T}_{1}}}=\frac{1}{2}\]                                ?..(i)                    \[\frac{60}{100}=1-\frac{{{T}_{2}}'}{{{T}_{1}}}\]Þ\[\frac{{{T}_{2}}'}{{{T}_{1}}}=\frac{2}{5}\]                                          ?..(ii)            Dividing equation (i) by (ii), \[\frac{500}{{{T}_{2}}'}=\frac{5}{4}\]Þ\[{{T}_{2}}=400K\]


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