JEE Main & Advanced Sample Paper JEE Main Sample Paper-47

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

    A)  \[4.2\times {{10}^{6}}\,J\]             

    B)   \[8.4\times {{10}^{6}}J\]

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

    D)  zero

    Correct Answer: B

    Solution :

     Given, \[{{T}_{1}}=627+273=900\,K\] \[{{Q}_{1}}=3\times {{10}^{6}}\,cal\] \[{{T}_{1}}=27+273=300\,K\]             At,        \[\frac{{{Q}_{1}}}{{{T}_{1}}}=\frac{{{Q}_{2}}}{{{T}_{2}}}\] \[\Rightarrow \]            \[{{Q}_{2}}=\frac{{{T}_{2}}}{{{T}_{1}}}\times {{Q}_{1}}\] \[=\frac{300}{900}\times 3\times {{10}^{6}}\] \[=1\times {{10}^{6}}\,cal\] 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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