JEE Main & Advanced Physics Thermodynamical Processes Question Bank Topic Test - Thermodynamical Processes

  • question_answer
    A carnot’s reversible engine converts \[\frac{1}{6}th\] of heat input into work. When the temperature of sink is reduced 62 K, the efficiency of carnot’ss cycle, becomes \[\frac{1}{3}.\] Calculate temperature of source and sink

    A) 372K, 310K       

    B) 772K, 312K

    C) 672K, 610K

    D) None of these

    Correct Answer: A

    Solution :

    [a] \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] where \[{{T}_{1}}\And {{T}_{2}}\] are source & sink temperatures respectively.
    \[\frac{1}{6}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\]               ……(1)
    Modified case,
    \[\frac{1}{3}=1-\frac{{{T}_{2}}-62}{{{T}_{1}}}\]         …..(2)
    Solving (1) and (2), we get

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