JEE Main & Advanced Physics Thermodynamical Processes Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    A refrigerator with coefficient of performance releases 200 J of heat to a hot reservoir. Then the work done on the working substance is

    A) \[\frac{100}{3}J\]

    B) \[100J\]

    C) \[\frac{200}{3}J\]

    D) \[150J\]

    Correct Answer: D

    Solution :

    [d] The coefficient of performance of a refrigerator is given by \[\alpha =\frac{{{Q}_{2}}}{W}=\frac{{{Q}_{2}}}{{{Q}_{1}}-{{Q}_{2}}}\] Substituting the given values, we get \[\frac{1}{3}=\frac{{{Q}_{2}}}{200-{{Q}_{2}}}\] \[\Rightarrow 200-{{Q}_{2}}=3{{Q}_{2}}\Rightarrow 4{{Q}_{2}}=200\] \[\text{or }{{Q}_{2}}=\frac{200}{4}J=50J\] \[\therefore W={{Q}_{1}}-{{Q}_{2}}=200J-50J=150J\]


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