An ideal heat engine is exhausting heat energy to a reservoir at ${27}^0 C$. If the initial efficiency of 50% is decreased to 40% by changing the temperature of sink, the new temperature of sink is

$efficiency of heat engine=\eta =1-\frac{T_2}{T_1}---\left(1\right)$

$T_1,T_2 are temperatures of source and \sin k$

$given T_2=27+273=300K;$

$efficiency=50\%=\frac{50}{100}$

$\eta =1-\frac{T_2}{T_1}$

$\Rightarrow \frac{50}{100}=1-\frac{300}{T_1}$

$\Rightarrow \frac{1}{2}=\frac{300}{T_1}$

$\Rightarrow T_1=600K---\left(2\right)$

$new temperature of \sin k=T_2^1;$

$new efficiency=40\%=\frac{40}{100};and eqn\left(2\right) substitute in following equation,$

${\eta }^1=1-\frac{{T^1}_2}{T_1}$

$\begin{array}{l}\frac{40}{100}=1-\frac{T_2^1}{600}\Rightarrow \frac{T_2^1}{600}=1-\frac{2}{5}=\frac{3}{5}\\ \end{array}$

$\Rightarrow \frac{T_2^1}{600}=1-\frac{2}{5}$

$\begin{array}{l}\frac{T_2^1}{600}=\frac{3}{5}\\ T_2^1=\frac{3}{5}X600=360K.\end{array}$