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  • Help me understand! - Chemical Thermodynamics - JEE Main-2

A Carnot engine operating between temperatures T_{1}\: and \: T_{2} has efficiency \frac{1}{6} When T_{2} is lowered by 62 K , its efficiency increases to \frac{1}{3}.Then T_{1}\: and \: T_{2} are, respectively

  • Option 1)

    372 \: K\: and \: 310\: K

  • Option 2)

    372 \: K\: and \: 330\: K

  • Option 3)

    330 \: K and \: 268 \: K

  • Option 4)

    310 K and \: 248 K

 

Answers (1)

best_answer

As we have learnt,

 

The efficiency is given by 

\\*\eta _1 = 1 - \frac{T_2}{T_1} \Rightarrow \frac{1}{}6 = 1 - \frac{T_2}{T_1} \Rightarrow \frac{T_2}{T_1} = \frac{5}{6} \;\;\; -(1) \\* \eta_2 = 1 - \frac{T_2 - 62}{T_1} \Rightarrow \frac{1}{3} = 1 - \frac{T_2 - 62}{T_1} \;\;\; -(2)

On solving (1) and (2) we get,

T_1 = 372K \;\&\; T_2 = 310 K

 


Option 1)

372 \: K\: and \: 310\: K

Option 2)

372 \: K\: and \: 330\: K

Option 3)

330 \: K and \: 268 \: K

Option 4)

310 K and \: 248 K

Posted by

SudhirSol

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