Question

A power drill of 400 W makes a hole in a lead cube of specific heat capacity 0.13 Jg⁻¹°C⁻¹ in 80 s. If the temperature of the lead rises from 27°C to 327°C, calculate the mass of the lead cube.

Answer 1

Energy supplied by the drill (E) = P × t

= 400 × 800

= 32000 J

Rise in temperature of lead (ΔT) = 327 − 27 = 300°C

Heat absorbed by lead (Q) = mcΔT

$$m=\frac{Q}{c\Delta T}$$

$$=\frac{32000}{0.13 \times 300}$$

$$=\frac{32000}{39}$$

= 820.5 g