JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    A Carnot engine takes \[3\times {{10}^{6}}\] cal of heat from a reservoir at \[{{627}^{o}}C\] and gives it to a sink at \[{{27}^{o}}C\]. The work done by the engine is     AIEEE  Solved  Paper-2003

    A) \[42\times {{10}^{6}}\]J     

    B)                       \[8.4\times {{10}^{6}}\]J              

    C) \[16.8\times {{10}^{6}}\] J     

    D)                       Zero

    Correct Answer: B

    Solution :

                    Given,       \[{{T}_{1}}=627+273=900\,K\]                     \[{{Q}_{1}}=3\times {{10}^{6}}\] cal                     \[{{T}_{2}}=27+273=300\,K\] \[\therefore \]      \[\frac{{{Q}_{1}}}{{{T}_{1}}}=\frac{{{Q}_{2}}}{{{T}_{2}}}\] \[\Rightarrow \]   \[{{Q}_{2}}=\frac{{{T}_{2}}}{{{T}_{1}}}\times {{Q}_{1}}=\frac{300}{900}\times 3\times {{10}^{6}}\]                     \[=1\times {{10}^{6}}\] cal Now, work done \[={{Q}_{1}}-{{Q}_{2}}\]                     \[=3\times {{10}^{6}}-1\times {{10}^{6}}=2\times {{10}^{6}}\,cal\]                     \[=2\times 4.2\times {{10}^{6}}J\]                 \[=8.4\times {{10}^{6}}J\]


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