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  • A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by <img alt="40^{\circ} \mathrm{C}" src="https://entrancecorner.oncodecogs.com/gif.latex?40%5E%7B%5Ccirc%7D%20%5Cmathrm%

A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40^{\circ} \mathrm{C}, its efficiency increases by 30%. The temperature of the source will be :
 

Option: 1

166.7 \mathrm{~K}


Option: 2

255.1 \mathrm{~K}


Option: 3

266.7 \mathrm{~K}


Option: 4

367.7 \mathrm{~K}


Answers (1)

best_answer

\mathrm{\eta=0.5=1-\frac{T_1}{T_2} \rightarrow 1 }

\mathrm{\eta^{\prime}=0.5+\frac{30}{100}\: \eta=0.65 }

\mathrm{\eta^{\prime}=0.65=1-\left(\frac{T_1-40}{T_2}\right) }

\mathrm{0.65=1-\frac{T_1}{T_2}+\frac{40}{T_2} }

\mathrm{0.15=\frac{40}{T_2} }

\mathrm{T_2=\frac{4000}{15}=\frac{800}{3} }

\mathrm{T_2=266.7 \mathrm{~K}}

The temperature of the source is \mathrm{266.7 \mathrm{~K}}

Hence 3 is correct option.






 

Posted by

Shailly goel

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