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 = .
Given: Power of the drilling machine, P = 10 kW = ³ W = 10,000 W Mass of the aluminium block, m = 8.0 kg = 8000 g Time, t = 2.5 min = = 150 s Specific heat of aluminium, s = 0.91 J g⁻¹ K⁻¹
Since 50% of the power is lost to the surroundings, only the remaining 50% is used to heat the block.
Power used for heating: P_used = 50% of 10 kW = 5 kW = 5000 W
Energy supplied to the block (Q): Q = Power × Time Q = 5000 W × 150 s = 750,000 J
Using the heat formula from the context (Q = msΔT): 750,000 = 8000 g × 0.91 J g⁻¹ K⁻¹ × ΔT 750,000 = 7280 × ΔT
Rise in temperature (ΔT): ΔT = 750,000 / 7280 ΔT 103 °C
The rise in temperature of the block is 103 °C.
Explanation
The question tests the application of the specific heat formula Q = msΔT in a real-world scenario involving power and energy conversion. The key is to first determine the effective power contributing to the heating process. Since 50% is lost, the effective power is 5 kW. Multiplying this by time gives the heat energy (Q) transferred to the aluminium block. Using the specific heat value provided in the context (0.91 J g⁻¹ K⁻¹), one must ensure the mass is converted from kg to grams to match the specific heat unit. Solving for ΔT yields the final temperature rise.
Solution Steps
Step 1: Convert units. Power P = 10,000 W; Mass m = 8000 g; Time t = 150 s.
Step 2: Calculate useful power. P_useful = 50% of 10 kW = 5000 W.
Step 3: Calculate heat energy supplied. Q = P_useful × t = = 750,000 J.
Step 4: Apply specific heat formula Q = msΔT.
Step 5: Solve for ΔT. ΔT = 750,000 / () 103 °C.