Question

A lead bullet of 10 g travelling at 300 m/s strikes against a block of wood and comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in its temperature is (Specific heat of lead = 150J/kg, K)

• A. 100°C
• B. 125°C
• C. 150°C
• D. 200°C

Answer 1

Answer: (C)

150°C

Answer 2

Since specific heat of lead is given in Joules, hence use $W = Q$ instead of $W = JQ$.

⇒$\frac{1}{2} \times \left( \frac{1}{2}mv^{2} \right) = m.c.\Delta\theta$ ⇒$\Delta\theta = \frac{v^{2}}{4c} = \frac{(300)^{2}}{4 \times 150} = 150{^\circ}C$.