JEE Main & Advanced Physics Thermodynamical Processes Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    In Carnot engine efficiency is 40% at hot reservoir temperature T. For efficiency 50% what will be temperature of hot reservoir?

    A) \[\frac{T}{5}\]

    B) \[\frac{2T}{5}\]

    C) \[6T\]

    D) \[\frac{6T}{5}\]

    Correct Answer: D

    Solution :

    [d] \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] \[{{T}_{1}}=T\] (Temperature of hot reservoir) For \[\eta =40%\] \[\frac{40}{100}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\Rightarrow \frac{{{T}_{2}}}{T}=\frac{3}{5}\Rightarrow {{T}_{2}}=\frac{3}{5}T\] For \[\eta =50%\] \[\frac{500}{1000}=1-\frac{\frac{3}{5}T}{{{T}_{1}}}\Rightarrow {{T}_{1}}=\frac{6T}{5}\]


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