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Haryana PMT Haryana PMT Solved Paper-2006

  • question_answer
    Container 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 :

                    \[\beta =\frac{{{Q}_{2}}}{W}=\frac{{{T}_{L}}}{{{T}_{H}}-{{T}_{L}}}\] \[{{Q}_{2}}=\frac{273\times 1}{303-273}=\frac{273}{30}=90J\] Heat delivered to the surrounding \[{{Q}_{1}}={{Q}_{2}}+W=9+1=10J\]


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