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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Self Evaluation Test - Thermodynamics

  • 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 :

    [b] \[\eta =\frac{\left( 627 \right)+\left( 273 \right)-\left( 273+27 \right)}{627+273}\]           \[=\frac{900-300}{900}=\frac{600}{900}=\frac{2}{3}\] \[\text{work=}\left( \eta  \right)\times \text{Heat =}\frac{2}{3}\times 3\times {{10}^{6}}\times 4.2J\]         \[\,=8.4\times {{10}^{6}}J\]


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