CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    In a Carnot engine, the temperature of reservoir is\[{{927}^{o}}C\]and that of sink is\[{{27}^{o}}C\]. If the work done by the engine when it transfers heat from reservoir to sink is\[12.6\times {{10}^{,}}^{6}J,\]the quantity of heat absorbed by the engine from the reservoir is

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

    B)         \[4\times {{10}^{6}}J\]

    C)  \[7.6\times {{10}^{6}}J\]             

    D)         \[4.25\times {{10}^{6}}J\]

    E)  \[20.8\times {{10}^{6}}J\]

    Correct Answer: A

    Solution :

    \[\frac{Q}{W}=\frac{{{T}_{1}}}{{{T}_{2}}}\] \[\therefore \]  \[\frac{Q}{12.6\times {{10}^{6}}}=\frac{927+273}{27+273}\] Or           \[Q=16.8\times {{10}^{6}}J\]


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