A gun fires a lead bullet of temperature 300 K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J , then the mass of the bullet is _______ grams. (Latent heat of fusion of lead =2.5 × 10⁴ JKg⁻¹ and specific heat capacity of lead =125 JKg⁻¹ K⁻¹)

Solution

Physics sir · 4 steps

The 625 J does two jobs: it first heats the lead up to 600 K, then melts it. Add both and solve for m.

1. Two stages. The bullet is first heated from 300 K to its melting point of 600 K, so Δ T = 300 K. At 600 K it then melts completely, at a constant temperature.

   Q = mc Δ T + mL = m(c Δ T + L)

2. Heat per kilogram. Heating needs c Δ T = 125 × 300 = 37500 J kg⁻¹. Melting needs L = 25000 J kg⁻¹. Together that is 62500 J kg⁻¹.
3. Solve for mass. m = 625/62500 = 0.01 kg, which is 10 g.
4. So. The mass of the bullet is 10 g, which is option (d).

The idea to keep. When a solid is heated and then melted, the total heat is mc Δ T + mL. The temperature rise stops at the melting point, and the latent heat is added on top.

Answer from the NTA final answer key. Solution by PrepWiser.