Answer
$ Q_H' = 1692 J$ or $1.70 \space kJ$
Work Step by Step
From the figure given, we get $Q_H = 4000J$ when $T_H = 325 K$
Then we find the work done first
$\frac{W}{Q_H} = 1 - \frac{T_C}{T_H} $
$W = (1 - \frac{250K}{325K}) 4000J $
$W = 923.08 J$
Now, find the $Q_H$ when $T_H$ is set to 550K is
$\frac{W}{Q_H'} = 1 - \frac{T_C}{T_H} $
$\frac{923.08J}{Q_H'} = 1 - \frac{250K}{550K} $
$ Q_H' = 1692 J$ or $1.70 \space kJ$
