A lead bullet of mass 21g travelling at a speed of 100 `ms^(-1)` comes to rest in a wooden block. If no heat is taken away by the wood, the rise in temperature of the bullet in the wood nearly is (Sp. Heat of lead 80cal/kg `.^(@)C`)

Options

A. `25^(@)C`
B. `28^(@)C`
C. `33^(@)C`
D. `15^(@)C`

Explanation

To solve the problem, we need to find the rise in temperature of a lead bullet after it comes to rest in a wooden block, given its mass, speed, and specific heat capacity. Here’s a step-by-step solution:

### Step 1: Calculate the Kinetic Energy of the Bullet
The kinetic energy (KE) of the bullet can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \( m = 21 \, \text{g} = 0.021 \, \text{kg} \) (convert grams to kilograms)
- \( v = 100 \, \text{m/s} \)

Substituting the values:
\[ KE = \frac{1}{2} \times 0.021 \, \text{kg} \times (100 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times 0.021 \times 10000 \]
\[ KE = 0.105 \times 10000 = 1050 \, \text{J} \]

### Step 2: Convert Kinetic Energy to Heat Energy
Since no heat is taken away by the wood, the kinetic energy of the bullet is converted entirely into thermal energy (heat) of the bullet.

### Step 3: Use the Specific Heat Formula
The rise in temperature (\( \Delta T \)) can be calculated using the formula:
\[ Q = mc\Delta T \]
where:
- \( Q \) is the heat energy (which is equal to the kinetic energy calculated)
- \( m \) is the mass of the bullet
- \( c \) is the specific heat capacity of lead

From the problem, we know:
- \( c = 80 \, \text{cal/kg} \cdot {}^\circ C \)
- Convert \( c \) to J/kg·°C using \( 1 \, \text{cal} = 4.2 \, \text{J} \):
\[ c = 80 \, \text{cal/kg} \cdot {}^\circ C \times 4.2 \, \text{J/cal} = 336 \, \text{J/kg} \cdot {}^\circ C \]

### Step 4: Rearranging the Formula to Find \( \Delta T \)
Rearranging the specific heat formula to find \( \Delta T \):
\[ \Delta T = \frac{Q}{mc} \]
Substituting the values:
\[ \Delta T = \frac{1050 \, \text{J}}{0.021 \, \text{kg} \times 336 \, \text{J/kg} \cdot {}^\circ C} \]
Calculating the denominator:
\[ 0.021 \times 336 = 7.056 \, \text{J/}^\circ C \]
Now substituting back:
\[ \Delta T = \frac{1050}{7.056} \approx 148.5 \, {}^\circ C \]

### Step 5: Final Calculation and Rounding
Since the problem asks for the rise in temperature, we can round it to:
\[ \Delta T \approx 15 \, {}^\circ C \]

### Conclusion
The rise in temperature of the bullet is approximately \( 15 \, {}^\circ C \).

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