A lead bullet (specific heat=0.032cal/gm `.^(@)C`) is completely stopped when it strikes a target with a velocity of 300m/s. the heat generated is equally shared by the bullet and the target. The rise in temperature of bullet will be

Options

A. `16.7^(@)C`
B. `1.67^(@)C`
C. `167.4^(@)C`
D. `267.4^(@)C`

Explanation

To find the rise in temperature of the lead bullet after it strikes a target, we can follow these steps:

### 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. Given that \( v = 300 \, \text{m/s} \), we can express the kinetic energy as:
\[ KE = \frac{1}{2} m (300)^2 = 45000m \, \text{J} \]

2. **Heat Generated and Shared:**
The total heat generated when the bullet strikes the target is equal to the kinetic energy. Since the heat is equally shared between the bullet and the target, the heat absorbed by the bullet (\( Q_b \)) is:
\[ Q_b = \frac{1}{2} KE = \frac{1}{2} (45000m) = 22500m \, \text{J} \]

3. **Relate Heat to Temperature Change:**
The heat absorbed by the bullet can also be expressed in terms of its specific heat capacity and the rise in temperature:
\[ Q_b = m s \Delta T \]
where \( s \) is the specific heat capacity of lead (0.032 cal/g°C) and \( \Delta T \) is the rise in temperature. We need to convert the specific heat from cal/g°C to J/kg°C:
\[ 0.032 \, \text{cal/g°C} = 0.032 \times 4.2 \, \text{J/g°C} = 0.1344 \, \text{J/g°C} = 134.4 \, \text{J/kg°C} \]

4. **Set the Two Expressions for Heat Equal:**
Now, we can set the two expressions for heat equal to each other:
\[ 22500m = m \times 134.4 \times \Delta T \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 22500 = 134.4 \Delta T \]

5. **Solve for the Rise in Temperature (\( \Delta T \)):**
Rearranging the equation gives:
\[ \Delta T = \frac{22500}{134.4} \approx 167.4 \, \text{°C} \]

### Final Answer:
The rise in temperature of the bullet will be approximately **167.4 °C**.

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