A 25 g bullet travelling at 400 m//s passes through a thin iron wall and emerges at a speed of 250 m/s. If the bullet absorbs 50 percent of the heat generated, a. what will be the temperature rise of the bullet? B. If the ambient temperature is `20^(@)C` will the bullet melt, and if so how much? Given that specific heat of head is `0.03 "cal"//gm^(@)C` and latent heat of fusion of lead is 6000 cal/kg. It is also given that melting pont of lead is `320^(@)C`.

Answer

To solve the problem step by step, we will break it down into two parts as required by the question.

### Part A: Calculate the Temperature Rise of the Bullet

1. **Calculate the Initial and Final Kinetic Energy of the Bullet**:
- The mass of the bullet (m) = 25 g = 0.025 kg (convert grams to kilograms)
- Initial velocity (Vi) = 400 m/s
- Final velocity (Vf) = 250 m/s

The kinetic energy (KE) is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
- Initial kinetic energy (KE_initial):
\[ KE_{initial} = \frac{1}{2} \times 0.025 \, \text{kg} \times (400 \, \text{m/s})^2 = \frac{1}{2} \times 0.025 \times 160000 = 2000 \, \text{J} \]
- Final kinetic energy (KE_final):
\[ KE_{final} = \frac{1}{2} \times 0.025 \, \text{kg} \times (250 \, \text{m/s})^2 = \frac{1}{2} \times 0.025 \times 62500 = 781.25 \, \text{J} \]

2. **Calculate the Loss in Kinetic Energy**:
\[ \Delta KE = KE_{initial} - KE_{final} = 2000 \, \text{J} - 781.25 \, \text{J} = 1218.75 \, \text{J} \]

3. **Calculate the Heat Absorbed by the Bullet**:
Since the bullet absorbs 50% of the heat generated:
\[ Q = \frac{50}{100} \times \Delta KE = 0.5 \times 1218.75 \, \text{J} = 609.375 \, \text{J} \]

4. **Calculate the Temperature Rise (ΔT)**:
Using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- Specific heat capacity of lead (c) = 0.03 cal/g°C = 0.03 × 4.18 J/g°C (since 1 cal = 4.18 J)
- Convert specific heat to J/kg°C:
\[ c = 0.03 \times 4.18 \, \text{J/g°C} = 0.1254 \, \text{J/g°C} = 125.4 \, \text{J/kg°C} \]
Now substituting the values:
\[ 609.375 = 0.025 \, \text{kg} \cdot 125.4 \, \text{J/kg°C} \cdot \Delta T \]
\[ \Delta T = \frac{609.375}{0.025 \cdot 125.4} = \frac{609.375}{3.135} \approx 194.38 \, °C \]

### Part B: Determine if the Bullet Melts

1. **Calculate the Final Temperature of the Bullet**:
- Ambient temperature (T_initial) = 20°C
- Temperature rise (ΔT) = 194.38°C
\[ T_{final} = T_{initial} + \Delta T = 20 + 194.38 = 214.38 \, °C \]

2. **Compare with the Melting Point**:
- Melting point of lead = 320°C
- Since \( T_{final} (214.38°C) < T_{melting} (320°C) \), the bullet will not melt.

### Final Answers:
- A. The temperature rise of the bullet is approximately **194.38°C**.
- B. The bullet will **not melt** as its final temperature (214.38°C) is less than its melting point (320°C).