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JCECE Engineering JCECE Engineering Solved Paper-2011

  • question_answer
    An ideal gas heat engine operates in a Carnot cycle between \[{{227}^{o}}C\] and\[{{127}^{o}}C\]. It absorbs \[6.0\times {{10}^{4}}\,\,cal\] at the higher temperature. The amount of heat converted into work is equal to

    A) \[4.8\times {{10}^{4}}cal\]                          

    B) \[3.5\times {{10}^{4}}cal\]

    C)  \[1.6\times {{10}^{4}}cal\]                         

    D)  \[1.2\times {{10}^{4}}cal\]

    Correct Answer: D

    Solution :

    \[\eta =\frac{W}{Q}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] Work,    \[W=\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}{{Q}_{1}}\] \[W=\frac{(227+273)-(127+273)}{(227+273)}\times 6.0\times {{10}^{4}}cal\]                 \[=\frac{100}{500}\times 6.0\times {{10}^{4}}cal\]


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