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

  • 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) will be [BHU 2000; CPMT 2002]

    A)            325°C                                       

    B)             325K

    C)            39°C                                         

    D)             320°C

    Correct Answer: C

    Solution :

                       Coefficient of performance                    \[K=\frac{{{T}_{2}}}{{{T}_{1}}-{{T}_{2}}}\]Þ\[5=\frac{(273-13)}{{{T}_{1}}-(273-13)}=\frac{260}{{{T}_{1}}-260}\]                    Þ \[5{{T}_{1}}-1300=260\]Þ \[5{{T}_{1}}=1560\]            Þ \[{{T}_{1}}=312K\to 39{}^\circ C\]


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