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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Heat Engine Refrigerator and Second Law of Thermodynamics

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

    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 :

                       \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{W}{Q}\]Þ\[W=\left( 1-\frac{{{T}_{1}}}{{{T}_{2}}} \right)\ Q=\left\{ 1-\frac{(273+27)}{(273+627)} \right\}\] Þ\[W=\left( 1-\frac{300}{900} \right)\times 3\times {{10}^{6}}\]\[=2\times {{10}^{6}}\times 4.2\ J=8.4\times {{10}^{6}}J\]


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