Question

A 10 kw drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium $=0.91J{g}^{-1}{K}^{-1}$

Answer 1

Power of the drilling machine, $P=10kW$
Mass of the aluminum block, $m=8.0kg$
Time for which the machine is used,$t=2.5min=2.5\times 60=150s$

Specific heat of aluminium,$c=0.91J/gK$

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

Total energy of the drilling machine$=Pt=10\times {10}^3\times 150=1.5\times {10}^{6}J$
It is given that only 50% of the power is useful.
Useful energy, $Q=\left(50/100\right)\times 1.5\times {10}^6=7.5\times {10}^{5}J$

BUT $Q=mc\Delta T$

$\Delta T=Q/mc=(7.5\times {10}^5)/(8\times {10}^3\times 0.91)={103}^{o}C$
Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is ${103}^{o}C$.