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.