An engine has an efficiency of 1/6. When the temperature of sink is reduced by 50°C, its efficiency is doubled. Temperature of the source is :
Answer & explanation
Correct answer: option 1
Since efficiency of engine is : \(\eta = 1 - \frac{T_2}{T_1}\)= $\frac{1}{6} \Rightarrow \frac{T_2}{T_1}=\frac{5}{6}$
When the temperature of the sink is reduced by 50°C, its efficiency is doubled : \(2\eta = 1 - \frac{T_2-50}{T_1} = \frac{1}{3}\)
$\Rightarrow \frac{T_2-50}{T_1} = \frac{2}{3} \Rightarrow \frac{T_2}{T_1} - \frac{50}{T_1} = \frac{2}{3}$
$\Rightarrow \frac{5}{6} - \frac{50}{T_1}= \frac{2}{3} \Rightarrow \frac{50}{T_1} = \frac{1}{6} $
$\Rightarrow T_1 = 300 = 27^oC $