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BVP Medical BVP Medical Solved Paper-2012

  • 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)  \[2m/s\]                            

    B)  \[1m/s\]

    C)  \[10m/s\]                                          

    D)  \[7m/s\]

    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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