J & K CET Medical J & K - CET Medical Solved Paper-2011

  • question_answer
    The temperature of the sink of a Carnot engine is \[27{}^\circ C\] and its efficiency is 25%. The temperature of the source is

    A) \[227{}^\circ C\]

    B) \[27{}^\circ C\]

    C) \[327{}^\circ C\]

    D) \[127{}^\circ C\]

    Correct Answer: D

    Solution :

                     \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\frac{25}{100}=1-\frac{300}{{{T}_{1}}}\] \[\frac{300}{{{T}_{1}}}=1-\frac{25}{100}\] \[\frac{300}{{{T}_{1}}}=\frac{100-25}{100}\] \[\frac{300}{{{T}_{1}}}=\frac{75}{100}\] \[{{T}_{1}}=\frac{300\times 100}{75}\] \[{{T}_{1}}=400\] \[\therefore \]  \[{{T}_{1}}=400-273\] \[={{127}^{o}}C\]


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