JEE Main & Advanced Physics Thermodynamical Processes JEE PYQ-Thermodynamical Processes

  • question_answer
    A Carnot engine has an efficiency of 1/6. When the temperature of the sink is reduced by \[62{}^\text{o}C\], its efficiency is doubled. The temperatures of the source and the sink are, respectively                       [JEE Main 12-4-2019 Afternoon]

    A)  \[124{}^\text{o}C,62{}^\text{o}C\]

    B)       \[37{}^\text{o}C,99{}^\text{o}C\]

    C)  \[62{}^\text{o}C,124{}^\text{o}C\]

    D)       \[99{}^\text{o}C,37{}^\text{o}C\]

    Correct Answer: B

    Solution :

    [b] Efficiency of Carnot engine \[=1-\frac{{{T}_{\sin k}}}{{{T}_{source}}}\]
    Given,
    \[\frac{1}{6}=1-\frac{{{T}_{\sin k}}}{{{T}_{source}}}\Rightarrow \frac{{{T}_{\sin k}}}{{{T}_{source}}}=\frac{5}{6}\]              .....(1)
    Also,
    \[\frac{2}{6}=1-\frac{{{T}_{\sin k}}-62}{{{T}_{source}}}\Rightarrow \frac{62}{{{T}_{source}}}=\frac{1}{6}\]                    …..(2)
    \[\therefore \]\[{{T}_{source}}=372K={{99}^{o}}C\]
    Also, \[{{T}_{\operatorname{sink}}}=\frac{5}{6}\times 372=310K={{37}^{o}}C\]
    (Note :- Temperature of source is more than
    temperature of sink)


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