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

  • question_answer
    In a Carnot engine, when \[{{T}_{2}}={{0}^{o}}C\] and \[{{T}_{1}}={{200}^{o}}C,\] its efficiency is \[{{\eta }_{1}}\]and when \[{{T}_{1}}=0{{\,}^{o}}C\] and \[{{T}_{2}}=-200{{\,}^{o}}C\], Its efficiency is \[{{\eta }_{2}}\], then what is \[{{\eta }_{1}}/{{\eta }_{2}}\]                    [DCE 2004]

    A)            0.577                                       

    B)            0.733

    C)            0.638                                       

    D)            Cannot be calculated

    Correct Answer: A

    Solution :

                       \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}\] \[\Rightarrow \,{{\eta }_{1}}=\frac{(473-273)}{473}=\frac{200}{473}\]            and \[{{\eta }_{2}}=\frac{273-73}{273}=\frac{200}{273}\] So required ratio \[\frac{{{\eta }_{1}}}{{{\eta }_{2}}}=\frac{273}{473}=0.577\]


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