Discussion Forum

A carnot  engine  with efficiency $\eta$ operates between two heat  reservoirs with temperatures $T_{1}$ and $T_{2}$  , where $T_{1} >T_{2}$ . If only $T_{1}$ is changed by 0.4% , the  change in efficiency is $\triangle\eta_{1}$ , whereas if only  $T_{2}$  is changed by 0.2% , the efficiency is changed by $\triangle\eta_{2}$ . The ratio  $\frac{\triangle\eta_{1}}{\triangle\eta_{2}}$ is approximately

Answer:

Explanation:

Key idea  Efficiency of carnot engine

   $\eta=1-\frac{T_{2}}{T_{1}}$

where, $T_{1}$= source temperature and $T_{2}$= sink temperature

If $T_{1}$ is changed by 0.4%, then

 $\frac{\triangle \eta_{1}}{\eta}= \frac{ \triangle T_{1}}{T_{1}}+\frac{\triangle T_{2}}{T_{2}}$

   ( from combination of the error)

$\Rightarrow$      $\frac{\triangle \eta_{1}}{\eta}= \frac{0.4}{100}+0$  .........(i)

Similarly , $T_{2}$ is changed  by 0.2% then

  $\frac{\triangle \eta_{2}}{\eta}= 0+\frac{0.2}{100}$  .........(ii)

 So, from Eq.(i) and (ii) , we get

 $\frac{\triangle \eta_{1}}{\triangle \eta_{2}}=\frac{0.4}{0.2}=2$

Hence, option (a) is correct