CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    An ideal gas heat engine operates in a Carnots cycle between \[{{227}^{o}}C\] and \[{{127}^{o}}C\]. It absorbs \[6\times {{10}^{4}}J\] at high temperature. The amount of heat converted into work is :

    A)  \[1.6\times {{10}^{4}}J\]             

    B)  \[1.2\times {{10}^{4}}J\]

    C)  \[4.8\times {{10}^{4}}J\]                             

    D)  \[3.5\times {{10}^{4}}J\]

    Correct Answer: B

    Solution :

    Using the relation                 \[\frac{W}{{{Q}_{1}}}=\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}\] or            \[\frac{W}{{{Q}_{1}}}=1-\frac{{{Q}_{2}}}{{{Q}_{1}}}\] or            \[\frac{W}{{{Q}_{1}}}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\]                        \[\left( \because \frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{{{T}_{1}}}{{{T}_{2}}} \right)\] or            \[W={{Q}_{1}}\,\left( 1-\frac{{{T}_{2}}}{{{T}_{1}}} \right)\] \[\therefore \]  \[W=6\times {{10}^{-4}}\left[ 1-\frac{(127+273)}{(227+273)} \right]\] or            \[W=6\times {{10}^{4}}\left( 1-\frac{400}{500} \right)\]                 \[=6\times {{10}^{4}}\times \frac{100}{500}\]                 \[=1.2\times {{10}^{4}}J\]


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