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  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2001

  • question_answer
    In Carnot cycle, the temperature of source is 500 K. If engine lakes in 300 cal of heat from a source and gives 150 cal of heat to a sink. The temperature of sink will be

    A)  500 K           

    B)  250 K

    C)  750 K           

    D)  125 K

    Correct Answer: B

    Solution :

     \[1-\frac{{{Q}_{2}}}{{{Q}_{1}}}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\]or \[\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\Rightarrow \] \[\frac{150}{300}=\frac{{{T}_{2}}}{500}\] \[\Rightarrow \] \[{{T}_{2}}=250\,K\]


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