Thank you for reporting, we will resolve it shortly
Q.
An engine is working. It takes $100$ calories of heat from source and leaves $80$ calories of heat to sink. If the temperature of source is $127^{\circ} C$, then temperature of sink is
Heat taken from source $Q_{1} = 100$ cal
Heat left to sink $Q_{2} = 80$ cal
$\therefore $ efficiency of the engine $\eta = 1 - \frac{Q_{2}}{Q_{1} }$
$ = 1-\frac{80}{100} = 20\%$
Temperature of the source $T_{1} = 127ºC = 400\, K$
Temperature of the sink $T_{2} = ?$
We know that
$\eta = 1-\frac{T_{2}}{T_{1}}$
$\Rightarrow 0.2 = 1-\frac{T_{2}}{400}$
$\Rightarrow \frac{T_{2}}{400} = 1-0.2 = 0.8$
$\Rightarrow T_{2} = 320\, K = 47^{\circ}C$