Question

A metal drill with a power of 500 W drills a hole in a lead cube of mass 0.25 kg in 6.5 s. If all the energy of the drill is transferred to the metal cube, calculate the rise in temperature.

[Specific heat capacity of lead is 130 J kg⁻¹K⁻¹]

Answer 1

Given:

Power rating (P) = 500 W (or 500 J s⁻¹)

Mass of lead cube (m) = 0.25 kg

Time (t) = 6.5 seconds

Specific heat capacity of lead (c) = 130 J kg⁻¹ K⁻¹

Rise in temperature = ΔT

Q = Power × Time

= 500 × 6.5

= 3250 J

Using Q = mcΔT

3250 = 0.25 × 130 × ΔT

3250 = 32.5 × ΔT

$$\Delta T = \frac{3250}{32.5}$$

ΔT = 100°C (or 100 K)

The rise in temperature of the lead cube is 100°C (or 100 K).