A 0.0500 kg lead bullet of volume `5.00 xx 10^(-6) m ^(3)` at `20.0^(@)C` hits a block that is made of an ideal thermal insulator and comes to rest at its center. At that time, the temperature of the bullet is `327^(@)C`. Use the following information for lead `:`
Coefficient of linear expansion `: alpha = 2.0 xx 10^(-5)// ""^(@)C`
Specific heat capacity `:` `c = 128 J // ( kg . ""^(@)C )`
Latent heat of fusion `:` `L_(1) = 23 300 J // kg`
Melting point `:` `T_("melt") = 327^(@)C`
What is the volume of the bullet when it comes to rest ?

Options

A. `5.00 xx 10^(-6) m^(3)`
B. `5.01 xx 10^(-6) m^(3)`
C. `5.09 xx 10^(-6) m^(3)`
D. `5.07 xx 10^(-6) m^(3)`

Answer

To solve the problem, we need to determine the volume of the lead bullet when it comes to rest at a temperature of 327°C. We will use the formula for volumetric expansion based on the coefficient of linear expansion.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the bullet, \( m = 0.0500 \, \text{kg} \)
- Initial volume, \( V_0 = 5.00 \times 10^{-6} \, \text{m}^3 \)
- Initial temperature, \( T_0 = 20.0 \, ^\circ C \)
- Final temperature, \( T_f = 327.0 \, ^\circ C \)
- Coefficient of linear expansion, \( \alpha = 2.0 \times 10^{-5} \, ^\circ C^{-1} \)

2. **Calculate the Temperature Change:**
\[ \Delta T = T_f - T_0 = 327.0 - 20.0 = 307.0 \, ^\circ C \]

3. **Calculate the Coefficient of Volume Expansion:**
The coefficient of volume expansion \( \gamma \) is related to the coefficient of linear expansion \( \alpha \) by the equation:
\[ \gamma = 3\alpha \]
Substituting the value of \( \alpha \):
\[ \gamma = 3 \times (2.0 \times 10^{-5}) = 6.0 \times 10^{-5} \, ^\circ C^{-1} \]

4. **Use the Volume Expansion Formula:**
The formula for the volume at the final temperature is given by:
\[ V = V_0 (1 + \gamma \Delta T) \]
Substituting the known values:
\[ V = 5.00 \times 10^{-6} \, \text{m}^3 \left(1 + (6.0 \times 10^{-5}) \times 307.0\right) \]

5. **Calculate \( \gamma \Delta T \):**
\[ \gamma \Delta T = (6.0 \times 10^{-5}) \times 307.0 = 0.01842 \]

6. **Calculate the Final Volume:**
\[ V = 5.00 \times 10^{-6} \, \text{m}^3 \times (1 + 0.01842) = 5.00 \times 10^{-6} \, \text{m}^3 \times 1.01842 \]
\[ V \approx 5.09 \times 10^{-6} \, \text{m}^3 \]

### Final Answer:
The volume of the bullet when it comes to rest is approximately \( 5.09 \times 10^{-6} \, \text{m}^3 \).

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