CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2010

  • question_answer
    The efficiency of Carnot's heat heat engine is 0.5 when the temperature of the source is \[{{T}_{1}}\]that of sink is \[{{T}_{2}}\]. The efficiency of source. Carnot's heat engine is also 0.5. The temperatures of source and second engine are respectively

    A) \[2{{T}_{1}},2{{T}_{2}}\]                               

    B) \[2{{T}_{1}},\frac{{{T}_{2}}}{2}\]

    C) \[{{T}_{1}}+5,\,{{T}_{2}}-5\]       

    D) \[{{T}_{1}}+10,\,{{T}_{2}}-10\]  

    Correct Answer: A

    Solution :

    Efficiency of Carnot's heat engine, \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\]. Efficiency remains same when both \[{{T}_{1}}\] and \[{{T}_{2}}\]are increased by same factor.


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