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JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    A Carnot's engine works as a refrigerator between \[250K\] and \[300K\]. It receives \[500cal\] heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is:                   [JEE Online 15-04-2018]

    A) \[420J\]                       

    B) \[2100J\]         

    C) \[772J\]                       

    D) \[2520J\]         

    Correct Answer: A

    Solution :

    COP = \[\frac{{{T}_{1}}}{{{T}_{2}}-{{T}_{1}}}=\frac{250}{300-250}=5\] \[COP=5=\frac{{{Q}_{L}}}{W}\] \[W=\frac{500\times 4.184}{5}=420J\]


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