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VMMC VMMC Medical Solved Paper-2006

  • question_answer
    The inside and outside temperatures of a refrigerator are 273 K and 303 K respectively. Assuming that refrigerator cycle is reversible, for every joule of work done, the heat delivered to the surrounding will be :

    A) 10 J        

    B)                        20 J                        

    C) 30 J                        

    D)        50 J

    Correct Answer: A

    Solution :

    \[\beta =\frac{{{Q}_{2}}}{W}=\frac{{{T}_{L}}}{{{T}_{H}}-{{T}_{L}}}\] \[{{Q}_{2}}=\frac{273\times 1}{303\times 273}=\frac{273}{30}=9\,J\] Heat delivered to the surrounding \[{{Q}_{1}}={{Q}_{2}}+W=9+1=10\,J\]


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