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  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-29

  • question_answer
    A carnot engine takes \[840\,J\]. heat from the source. If its efficiency is\[20%\]. How much heat will be rejected to sink.

    A) \[200\,cal\]                        

    B) \[180\,cal\]

    C) \[42\,cal\]                          

    D) \[160\,cal\]

    Correct Answer: D

    Solution :

    efficiency = 20/100 = Work / heat absorbed = W/840 W= 168J Heat rejected = heat absorb - W = 840-168=672J=672/4.2 cal=160 cal


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