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Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    A reversible engine working between the temperature limits of 600 K and 1200 K receives 50 kJ of heat. The work done by the engine will be

    A)  50 kJ           

    B)  100 kJ

    C)  25 kJ            

    D)  \[-25\text{ }kJ\]

    Correct Answer: C

    Solution :

     Work done by the engine is given by \[\frac{W}{Q}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\frac{W}{50\times {{10}^{3}}}=1-\frac{600}{1200}\] \[\frac{W}{50\times {{10}^{3}}}=1-\frac{1}{2}\] \[\Rightarrow \] \[W=25\times {{10}^{3}}J=25\,kJ\]


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