Question

A 10 kW drilling machine is used to drill a bore in a small aluminum 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 = 0.91 J g–1 K–1

Answer 1

Power of the drilling machine, P = 10 kW = 10 × 103 W
Mass of the aluminum block, m = 8.0 kg = 8 × 103 g
Time for which the machine is used, t = 2.5 min = 2.5 × 60 = 150 s
Specific heat of aluminum, c = 0.91 J g–1 K–1
Rise in the temperature of the block after drilling = δT
The total energy of the drilling machine = Pt
= 10 × 103 × 150
= 1.5 × 106 J
It is given that only 50% of the power is useful.
Useful energy, △Q = $\frac{50}{100}$ x 1.5 x 106 = 7.5 x 105 J
But △Q = mc△T
∴ △T = $\frac{\Delta Q}{mc}$
=$\frac{ 7.5 \times 10^5}{8 \times 10^3 \times 0.91}$
= 103°C
Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.