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The efficiency of a Carnot cycle is 1/6. If on reducing the temperature of the sink by 65^{\circ} \mathrm{C}, the efficiency becomes 1/3, find the initial and final temperatures between which the cycle is working.
 

Option: 1

225 \mathrm{~K}


 


Option: 2

280 \mathrm{~K}
 


Option: 3

300 \mathrm{~K}
 


Option: 4

325 \mathrm{~K}


Answers (1)

best_answer

Given \mathrm{\eta_1=\frac{1}{6}, \eta_2=\frac{1}{3}}

If the temperatures of the source and the sink between which the cycle is working are \mathrm{T_1\: and \: T_2}, then the efficiency in the first case will be

\mathrm{ \eta_1=1-\frac{T_2}{T_1}=\frac{1}{6} }

In the second case \mathrm{ \eta_2=1-\frac{T_2-65}{T_1} \quad=\frac{1}{3} }

Solving \mathrm{ \mathrm{T}_1=390 \mathrm{~K} \: and \: \mathrm{T}_2=325 \mathrm{~K} \text {. } }

Posted by

Rakesh

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