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JEE Main & Advanced Physics Thermodynamical Processes JEE PYQ-Thermodynamical Processes

  • 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 :

    [a] 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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