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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2006

  • question_answer
    A Carnot engine takes heat from a reservoir at \[{{627}^{o}}C\] and rejects heat to a sink at \[{{27}^{o}}C\]. Its efficiency will be:

    A)  3/5                                       

    B)  1/3

    C)  2/3                                       

    D)  200/209

    Correct Answer: C

    Solution :

                    \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\therefore \]  \[\eta =1-\frac{(27+273)}{(273+627)}\]                 \[=1-\frac{300}{900}=\frac{600}{900}=\frac{2}{3}\]


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