Power of the drilling machine, $P=10 kW =10 \times 10^{3} W$

Mass of the aluminum block, $m=8.0 kg =8 \times 10^{3} g$

Time for which the machine is used, $t=2.5 min =2.5 \times 60=150 s$

Specific heat of aluminium, $c=0.91 J g ^{-1} K ^{-1}$

Rise in the temperature of the block after drilling $=\delta T$

Total energy of the drilling machine $=P t$

$=10 \times 10^{3} \times 150$

$=1.5 \times 10^{6} J$

It is given that only $50 \%$ of the power is useful. Useful energy, $\Delta Q=\frac{50}{100} \times 1.5 \times 10^{6}=7.5 \times 10^{5} J$

But $\Delta Q=m c \Delta T$

$\therefore \Delta T=\frac{\Delta Q}{m c}$

$=\frac{7.5 \times 10^{5}}{8 \times 10^{3} \times 0.91}$

$=103\,^{\circ} C$

Therefore, in $2.5$ minutes of drilling, the rise in the temperature of the block is $103\,^{\circ} C$