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A carnot engine separates between two reservoirs of temperature 900 K and 300 K. The engine performs 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir in a cycle is:-  
Option: 1 600
Option: 2 550
Option: 3 475
Option: 4 700

Answers (1)

best_answer

Carnot Engine -

      So \eta =\frac{W}{Q_{1}}=1-\frac{T_{2}}{T_{1}}

      where 

            T_{1}= Source temperature, T_{2}= Sink Temperature   and    \left ( T_{1} > T_{2}\right )

           and    T_{1}\, and\, T_{2}  are in kelvin

 

 

given, W=1200 J

\eta _{carnot}=1-\frac{T_L}{T_H}=1-\frac{300}{900}=\frac{6}{9}

\eta _{carnot}=\frac{W}{Q_1}

So, \frac{W}{Q_1}=\frac{6}{9}\Rightarrow Q_1=1800 J

So, W=Q_1-Q_2\Rightarrow Q_2=Q_1-W=1800-1200=600J

 

So, the Correct answer is option 1. 

Posted by

Ritika Jonwal

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