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

To solve this problem, we need to calculate the mass of the bullet that melts upon heating. We are given the following data:

• Initial temperature of the bullet, \(T_i = 300 \text{ K}\) 
• Melting temperature of the bullet, \(T_m = 600 \text{ K}\)
• Total heat required, \(Q = 625 \text{ J}\)
• 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 problem involves two main heat processes:

1. Heating the bullet from \(300 \text{ K}\) to \(600 \text{ K}\).
2. Melting the bullet at \(600 \text{ K}\).

The total heat required can be divided into two parts:

• Heat required to raise the temperature: \(Q_1 = mc (T_m - T_i)\)
• Heat required to melt the bullet: \(Q_2 = mL\)

The total heat \(Q\) is given by the sum of \(Q_1\) and \(Q_2\):

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

Substituting the known values:

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

Simplifying this equation:

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

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

\(625 = m \times 62500\)

Solving for \(m\):

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

Thus, the mass of the bullet is 10 grams.

Answer 2

The total heat required is the sum of the heat needed to raise the temperature of the bullet from 300 K to 600 K and the heat required to melt the bullet at the melting point.
Using the formula for heat \( Q = ms \Delta T \), where:
\( m \) is the mass,
\( s \) is the specific heat,
\( \Delta T \) is the change in temperature.
The first part of the heat is required to raise the temperature: \[ Q_1 = ms\Delta T = m \times 125 \times (600 - 300) = m \times 125 \times 300 \] The second part is required to melt the bullet at 600 K: \[ Q_2 = mL = m \times (2.5 \times 10^4) \] The total heat is given as 625 J: \[ 625 = ms\Delta T + mL \] Substitute the values: \[ 625 = m \times 125 \times 300 + m \times 2.5 \times 10^4 \] Simplifying: \[ 625 = m \times 37500 + m \times 25000 \] \[ 625 = m \times 62500 \] Solving for \( m \): \[ m = \frac{625}{62500} = \frac{1}{100} \, \text{kg} \] Since \( 1 \, \text{kg} = 1000 \, \text{grams}, \) we have: \[ m = 10 \, \text{grams} \] Thus, the mass of the bullet is 10 grams.