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Manipal Medical Manipal Medical Solved Paper-2005

  • 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 :

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


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