Answer
$2.625\frac{J}{K}$
Work Step by Step
The change in entropy in hot reservoir is
$dS_h=\frac{Q_h}{T_h}$
We plug in the known values to obtain:
$dS_h=\frac{6400}{610}=10.5\frac{J}{K}$
The change in entropy in cold reservoir is
$dS_c=\frac{Q_c}{T_c}$
We plug in the known values to obtain:
$dS_c=\frac{4200}{320}=13.125\frac{J}{K}$
Now, the total change in entropy is
$dS_c-dS_h=13.125-10.5=2.625\frac{J}{K}$
