A bullet of mass 5 g travelling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is `0.030 (cal)/(g - ^@C) (1 cal = 4.2 xx 10^7" ergs")` close to :

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert the mass of the bullet to kilograms
Given:
- Mass of the bullet, \( m = 5 \, \text{g} = \frac{5}{1000} \, \text{kg} = 0.005 \, \text{kg} \)

### Step 2: Calculate the initial kinetic energy of the bullet
The kinetic energy (KE) of the bullet before striking the target is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 0.005 \, \text{kg} \times (210 \, \text{m/s})^2 \]
Calculating this:
\[ KE = \frac{1}{2} \times 0.005 \times 44100 = 110.25 \, \text{J} \]

### Step 3: Determine the heat absorbed by the bullet
According to the problem, half of the kinetic energy is converted into heat in the bullet:
\[ q_{\text{bullet}} = \frac{1}{2} KE = \frac{1}{2} \times 110.25 = 55.125 \, \text{J} \]

### Step 4: Convert specific heat from calories to joules
Given:
- Specific heat of the bullet, \( C = 0.030 \, \frac{\text{cal}}{\text{g} \cdot \degree C} \)
We need to convert this to SI units (J/kg·°C):
\[ C = 0.030 \, \text{cal/g°C} \times \frac{1000 \, \text{g}}{1 \, \text{kg}} \times 4.2 \, \text{J/cal} = 0.030 \times 1000 \times 4.2 \, \text{J/kg°C} = 126 \, \text{J/kg°C} \]

### Step 5: Calculate the rise in temperature of the bullet
Using the formula:
\[ q = mC\Delta T \]
We can rearrange this to find \( \Delta T \):
\[ \Delta T = \frac{q_{\text{bullet}}}{mC} \]
Substituting the values:
\[ \Delta T = \frac{55.125 \, \text{J}}{0.005 \, \text{kg} \times 126 \, \text{J/kg°C}} = \frac{55.125}{0.63} \approx 87.5 \, \degree C \]

### Final Answer
The rise in temperature of the bullet is approximately \( 87.5 \, \degree C \).

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