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JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank 2nd and 3rd law of thermodynamics and Entropy

  • question_answer
    An engine operating between \[{{150}^{o}}C\] and \[{{25}^{o}}C\] takes 500 J heat from a higher temperature reservoir if there are no frictional losses, then work done by engine is [MH CET 1999]

    A)                 147.7 J  

    B)                 157.75 J

    C)                 165.85 J

    D)                 169.95 J

    Correct Answer: A

    Solution :

           \[{{T}_{2}}=150+273=423\,K\]                     \[{{T}_{1}}=25+273=298\,K\]            \[Q=500\,K\]                 \[\frac{W}{Q}=\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{2}}}\]; \[W=500\left( \frac{423-298}{423} \right)=147.7\,J\].


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