A Carnot engine with sink temperature at ${17}^{0}C$ has 50% efficiency. By how much should its source temperature be changed to increase its efficiency to 60%?

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

Initially,          $\frac{50}{100}=1-\frac{273+17}{T_1}$

or                   $\frac{290}{T_1}=\frac{1}{2}$  or $T_1=580K$

Finally,            $\frac{60}{100}=1-\frac{273+17}{T_1\text{'}}$  or  $\frac{290}{T_1\text{'}}$ $=\frac{2}{5}$

or                        $T\text{'}=725K$  

$\therefore$       Change in source temperature

                              $=\left(725-580\right)K=145K$