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 = 0.91 J g^–1 K^–1.

Given,

Power of the drilling machine, P = 10 kW = 10 × 10³ W

Mass of the aluminium block, m = 8.0 kg = 8 × 10³ g

Machine operation time, t = 2.5 min = 2.5×60=150 s Specific heat of aluminium, c = 0.91 J g^–1 K^–1

Let the rise in the temperature of the block after drilling be ΔT.

Total energy of the drilling machine, E = Pt

⇒ E = 10 × 10³ × 150

⇒ E = 1.5 × 10⁶ J

It is given that only 50% of the power is useful.

So, Useful energy, ∆Q = (50/100) × E

⇒ ΔQ = 7.5 × 10⁵ J

But ∆Q = mc∆T

∴ ∆T = ∆Q/mc = (7.5 × 10⁵ J)/(8×10³ g × 0.91 J g^–1 K^–1) = 103 K Hence, the rise in the temperature of the block is 103 K in 2.5 minutes of drilling.

NOTE: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.