JEE Main & Advanced Physics Thermodynamical Processes Question Bank Heat Engine Refrigerator and Second Law of Thermodynamics

  • question_answer
    The temperature of sink of Carnot engine is \[{{27}^{o}}C\]. Efficiency of engine is 25%. Then temperature of source is [DCE 2002; CPMT 2002]

    A)            \[{{227}^{o}}C\]                  

    B)            \[{{327}^{o}}C\]

    C)            \[{{127}^{o}}C\]                  

    D)            \[{{27}^{o}}C\]

    Correct Answer: C

    Solution :

                       \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] Þ \[\frac{25}{100}=1-\frac{300}{{{T}_{1}}}\] Þ \[\frac{1}{4}=1-\frac{300}{{{T}_{1}}}\] \[{{T}_{1}}=400\,K=127{}^\circ C\]


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