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NEET Physics Thermodynamical Processes NEET PYQ-Thermodynamical Processes

  • question_answer
    A Carnot engine having an efficiency of \[\frac{1}{10}\]  as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is                                                                                         [NEET-2017]

    A)  100 J              

    B)  1 J

    C)  90 J    

    D)       99 J

    Correct Answer: C

    Solution :

    [c]        \[\beta =\frac{1-\eta }{\eta }\]
    \[=\frac{1-\frac{1}{10}}{\frac{1}{10}}=\frac{\frac{9}{10}}{\frac{1}{10}}\]
    \[\beta =9\]
    \[\beta =\frac{{{Q}_{2}}}{W}\]
    \[{{Q}_{2}}=9\times 10=90\,J\]


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