The value of $∆$E when the system absorbs 1000 kJ of heat and does 650 kJ of work on the surroundings is ------kJ

The heat absorbed by the system (q) is 1000 kJ. As heat absorbed by the system, therefore it is taken as positive. Work (W) is done by the system, therefore it is taken as negative.

$\begin{array}{l}\mathrm{\Delta E}=\mathrm{q}+\mathrm{W}\\ \mathrm{\Delta E}=1000-650=350\mathrm{kJ}\end{array}$