Question

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. Given Data: Latent heat of fusion of lead = \(2.5 \times 10^4 \, \text{J kg}^{-1}\) and specific heat capacity of lead = 125 J kg\(^{-1}\) K\(^{-1}\).

• A. 20
• B. 15
• C. 10
• D. 5

Hint:

To solve such problems, divide the total heat into parts corresponding to different processes (e.g., temperature change and phase change) and apply the formula for heat in each part. Then, solve for the unknown mass.

Answer 1

The Correct Option is C

The mass of the bullet undergoing melting upon heating is calculated using the provided data. The relevant parameters are:

• Initial temperature of the bullet, \(T_i = 300 \text{ K}\) 
• Melting temperature of the bullet, \(T_m = 600 \text{ K}\)
• Total heat absorbed, \(Q = 625 \text{ J}\)
• Specific heat capacity of lead, \(c = 125 \text{ J kg}^{-1} \text{ K}^{-1}\)
• Latent heat of fusion of lead, \(L = 2.5 \times 10^4 \text{ J kg}^{-1}\)

The total heat input encompasses two stages:

1. The thermal energy required to increase the bullet's temperature from \(300 \text{ K}\) to \(600 \text{ K}\).
2. The thermal energy required to facilitate the phase transition (melting) at \(600 \text{ K}\).

The total heat \(Q\) is the sum of the heat for temperature change ( \(Q_1\) ) and the heat for melting ( \(Q_2\) ):

• Heat for temperature increase: \(Q_1 = mc (T_m - T_i)\)
• Heat for melting: \(Q_2 = mL\)

The governing equation is:

\(Q = mc (T_m - T_i) + mL\)

Substituting the given values yields:

\(625 = m \times 125 \times (600 - 300) + m \times 2.5 \times 10^4\)

Simplification of the equation proceeds as follows:

\(625 = m \times (125 \times 300 + 2.5 \times 10^4)\)

\(625 = m \times (37500 + 25000)\)

\(625 = m \times 62500\)

Solving for the mass \(m\) gives:

\(m = \frac{625}{62500} = \frac{1}{100} \ \text{kg} = 0.01 \ \text{kg} = 10 \ \text{g}\)

The calculated mass of the bullet is 10 grams.