Discussion Forum

Consider a reversible engine of efficiency  $\frac{1}{6}$. When the temperature of the sink is reduced by $62^{0}C$ , its efficiency gets doubled. The temperature of the source and sink  respectively are

Answer:

Explanation:

 Given

           $\eta_{1}=\frac{1}{6}$

 According to the question,

$\eta_{1}=\frac{T_{1}-T_{2}}{T_{1}}$

 $\frac{T_{1}-T_{2}}{T_{1}}=\frac{1}{6}$......(i)

$\eta_{2}=\frac{T_{1}-(T_{2}-62)}{T_{1}}\Rightarrow 2 \times \eta_{1}=\frac{T_{1}-T_{2}+62}{T_{1}}$

$2\times \frac{1}{6}=\frac{T_{1}-T_{2}-62}{T_{1}}\Rightarrow \frac{1}{3}=\frac{T_{1}-T_{2}+62}{T_{1}}$ .....(ii)

 from  Eq.(i) we ,get,

 $T_{1}-T_{2}=\frac{T_{1}}{6}$

 Substituting this value in Eq(11), we get

   $\frac{1}{3}=\frac{\frac{T_{1}}{6}+62}{T_{1}}\Rightarrow \frac{1}{3}= \frac{T_{1}+372}{6T_{1}}$

  $6T_{1}=3T_{1}+1116\Rightarrow 3T_{1}=1116 \Rightarrow T_{1}=372 K$

 From Eq.(i) , we get

      $\frac{372-T_{2}}{372}= \frac{1}{6}$

 $6(372- T_{2})=372$

 $2232-6T_{2}=372$

   $6T_{2}=2232-372$

  $6T_{2}=1860$

  $T_{2}=\frac{1860}{6}$

    $T_{2}=310 K$