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`
How much heat was needed to raise the bullet to its final temperature ?

Options

A. 963 J
B. 1960J
C. 3640J
D. 3880 J

Answer

To find the heat needed to raise the temperature of the lead bullet from its initial temperature to its final temperature, we can use the formula for heat transfer:

\[ Q = m \cdot c \cdot \Delta T \]

Where:
- \( Q \) is the heat absorbed (in Joules),
- \( m \) is the mass of the bullet (in kg),
- \( c \) is the specific heat capacity (in J/(kg·°C)),
- \( \Delta T \) is the change in temperature (in °C).

### Step-by-Step Solution:

1. **Identify the mass of the bullet**:
- Given: \( m = 0.0500 \, \text{kg} \)

2. **Identify the specific heat capacity of lead**:
- Given: \( c = 128 \, \text{J/(kg·°C)} \)

3. **Determine the initial and final temperatures**:
- Initial temperature \( T_i = 20.0 \, °C \)
- Final temperature \( T_f = 327.0 \, °C \)

4. **Calculate the change in temperature (\( \Delta T \))**:
\[ \Delta T = T_f - T_i = 327.0 \, °C - 20.0 \, °C = 307.0 \, °C \]

5. **Substitute the values into the heat transfer formula**:
\[ Q = m \cdot c \cdot \Delta T \]
\[ Q = 0.0500 \, \text{kg} \cdot 128 \, \text{J/(kg·°C)} \cdot 307.0 \, °C \]

6. **Calculate \( Q \)**:
\[ Q = 0.0500 \cdot 128 \cdot 307 \]
\[ Q = 0.0500 \cdot 39376 = 1968.8 \, \text{J} \]

### Final Answer:
The heat needed to raise the bullet to its final temperature is approximately \( 1968.8 \, \text{J} \).