MGIMS WARDHA MGIMS WARDHA Solved Paper-2010

  • question_answer
    An ideal gas heat engine operates in Carnot cycle between \[227{}^\circ C\] and \[127{}^\circ C\]. It absorbs \[6\times {{10}^{4}}cal\]of heat at higher temperature. Amount of heat converted into work is

    A) \[1.2\times {{10}^{4}}1\]                             

    B)  \[2.2\times {{10}^{4}}cal\]

    C) \[6.0\times {{10}^{4}}cal\]                          

    D)  \[4.8\times {{10}^{4}}cal\]

    Correct Answer: A

    Solution :

    As          \[\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\therefore \]  \[\frac{{{Q}_{2}}}{6\times {{10}^{4}}}=\frac{127+273}{227+273}=\frac{400}{500}\] \[\therefore \]  \[{{Q}_{2}}=\frac{4}{5}\times 6\times {{10}^{4}}\] \[=4.8\times {{10}^{4}}cal\] \[\therefore \]  \[W={{Q}_{1}}-{{Q}_{2}}\]                 \[=6\times {{10}^{4}}-4.8\times {{10}^{4}}\]                 \[=1.2\times {{10}^{4}}cal\]


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