A lead 5 gram bullet hitting an obstacle completely melted. The temperature of the bullet before impact was 27 ° C.

A lead 5 gram bullet hitting an obstacle completely melted. The temperature of the bullet before impact was 27 ° C. How much heat was absorbed by the bullet before hitting?

Data: m (mass of the lead bullet taken) = 5 g (in SI m = 5 * 10 ^ -3 kg); the bullet has completely melted; t (bullet temperature before impact) = 27 ºС.

Reference values: Cc (specific heat of lead) = 140 J / (kg * K); tmelt (temperature of the beginning of melting) ≈ 327 ºС; λ (specific heat of fusion) = 0.25 * 10 ^ 5 J / kg.

The amount of heat that is sufficient to melt the bullet: Q = Q1 + Q2 = Cc * m * (tmelt – t) + λ * m.

Let’s perform the calculation: Q = 140 * 5 * 10 ^ -3 * (327 – 27) + 0.25 * 10 ^ 5 * 5 * 10 ^ -3 = 335 J.