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I have a doubt, kindly clarify. - Thermodynamics - JEE Main

A Carnot engine has as efficiency 0f 1/6. When the temperature of the sink is reduced by 62^{\circ}C, its efficiency is doubled. The temperatures of the source and the sink are, respectively,

 

  • Option 1)

    62^{\circ}C124^{\circ}C

  • Option 2)

    99^{\circ}C, 37^{\circ}C

  • Option 3)

    124^{\circ}C, 62^{\circ}C

  • Option 4)

    37^{\circ}C, 99^{\circ}C

 
Answers (1)
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given 

\frac{1}{6}= 1- \frac{Tsink}{Tsourse}

\frac{Tsink}{Tsourse}=\frac{5}{6}\cdot \cdot \cdot (1)

also 

\frac{2 }{6}=1-\frac{Tsink-62}{Tsourse}

\frac{1}{3}=1-\frac{5}{6}+\frac{62}{Tsourse}           (from (1))

\frac{1}{6}=\frac{62}{Tsourse}

Tsourse=372K=99^{\circ}C

Tsink=\frac{5}{6}\times 372=310 K=37^{\circ}C


Option 1)

62^{\circ}C124^{\circ}C

Option 2)

99^{\circ}C, 37^{\circ}C

Option 3)

124^{\circ}C, 62^{\circ}C

Option 4)

37^{\circ}C, 99^{\circ}C

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