J & K CET Engineering J and K - CET Engineering Solved Paper-2011

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

    A)  \[{{227}^{o}}C\] 

    B)  \[{{27}^{o}}C\]

    C)  \[{{327}^{o}}C\]         

    D)  \[{{127}^{o}}C\]

    Correct Answer: D

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner