JEE Main & Advanced Physics Thermodynamical Processes Question Bank Topic Test - Thermodynamical Processes

  • question_answer
    An ideal refrigerator has a freezer at a temperature of \[-13{}^\circ C\]. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) is.

    A) \[320{}^\circ C\]

    B) \[39{}^\circ C\]

    C) 325 K

    D) \[325{}^\circ C\]

    Correct Answer: B

    Solution :

    [b]Given \[\beta =5\]
                We have, \[\beta =\frac{{{T}_{2}}}{{{T}_{1}}-{{T}_{2}}}\]
                where \[{{T}_{1}}=\] temp. of surrounding,
                \[{{T}_{2}}=\] temp. of cold body
                or         \[5=\frac{260}{{{T}_{1}}-260}\]
                \[5{{T}_{1}}=260+1300\] or \[{{T}_{1}}=312\,K={{39}^{o}}C\].

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