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AMU Medical AMU Solved Paper-2001

  • question_answer
    A Camot engine working between 300 K and 600 K has work output of 800 J/cycle. The amount of heat energy supplied to the engine from source per cycle will be

    A)  800                                       

    B)  1600

    C)  1200                                     

    D)  900

    Correct Answer: B

    Solution :

                     Efficiency of Carnot engine \[(\eta )\] is given by                 \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{W}{Q}\] Given,   \[{{T}_{2}}=300K,\,\,{{T}_{1}}=600\,K,\,\,W=800\,J\] \[\therefore \]  \[1-\frac{300}{600}=\frac{800}{Q}\] \[\Rightarrow \]               \[Q=\frac{800}{0.5}=1600\]J/cycle


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