A lead bullet strikes a target with velocity of 480 m/s. if the bullet falls dead, then the rise in temperature of bullet is, (Assuming that heat produced is equally shared between the bullet and target).
`(J=4.2xx10^(3)J//kcal,C=0.03kcal//kgK`)

Options

A. `557^(@)C`
B. `457^(@)C`
C. `857^(@)C`
D. `754^(@)C`

Answer

To solve the problem, we need to calculate the rise in temperature of a lead bullet after it strikes a target and comes to rest. The energy lost by the bullet will be converted into heat energy, which will raise the temperature of both the bullet and the target. We will assume that the heat produced is equally shared between the bullet and the target.

### Step-by-Step Solution:

1. **Calculate the Kinetic Energy (KE) of the Bullet:**
The kinetic energy of the bullet can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the bullet and \( v \) is its velocity.

2. **Substituting the Given Values:**
Given that the velocity \( v = 480 \, \text{m/s} \), we can express the kinetic energy as:
\[ KE = \frac{1}{2} m (480)^2 = \frac{1}{2} m \times 230400 = 115200 m \, \text{J} \]

3. **Convert Specific Heat Capacity:**
The specific heat capacity \( C \) of the bullet is given as \( 0.03 \, \text{kcal/kgK} \). We need to convert this to joules:
\[ 1 \, \text{kcal} = 4200 \, \text{J} \quad \Rightarrow \quad C = 0.03 \times 4200 = 126 \, \text{J/kgK} \]

4. **Heat Energy Shared:**
Since the heat produced is equally shared between the bullet and the target, the heat energy available for the bullet is:
\[ Q = \frac{KE}{2} = \frac{115200 m}{2} = 57600 m \, \text{J} \]

5. **Relating Heat Energy to Temperature Rise:**
The heat energy absorbed by the bullet can also be expressed as:
\[ Q = mc\Delta T \]
where \( \Delta T \) is the rise in temperature of the bullet.

6. **Setting the Equations Equal:**
Setting the two expressions for heat energy equal gives:
\[ 57600 m = m \times 126 \times \Delta T \]

7. **Canceling Mass \( m \):**
Since \( m \) is common on both sides, we can cancel it out (assuming \( m \neq 0 \)):
\[ 57600 = 126 \Delta T \]

8. **Solving for \( \Delta T \):**
Now, we can solve for \( \Delta T \):
\[ \Delta T = \frac{57600}{126} \approx 457.14 \, \text{K} \]

9. **Final Result:**
The rise in temperature of the bullet is approximately \( 457.14 \, \text{K} \) or \( 457.14 \, \text{°C} \).