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.