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  • A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be :&

A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be :         
                 

Option: 1

300 K


Option: 2

900 K


Option: 3

1000 K


Option: 4

360 K


Answers (1)

best_answer

\begin{aligned} & \eta=1-\frac{T_L}{T_H} \\ & 0.5=1-\frac{T_L}{600} \Rightarrow T_L=(1-0.5) \times 600 \mathrm{~K}=300 \mathrm{~K} \\ & \text { Now } 0.7=1-\frac{300}{T_2} \\ & \frac{300}{T_2}=0.3 \Rightarrow T_2=\frac{300}{0.3}=1000 \mathrm{~K} \end{aligned}

Posted by

Gautam harsolia

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