An aluminium baseball bat has a length of 0.86 m at a temperature of `17^@C` . When the temperature of the bat is raised , the bat lengthens by 0.000 16 m . Determine the final temperature of the bat.

Options

A. `19^@C`
B. `25^@C`
C. `34^@C`
D. `29^@C`

Answer

To determine the final temperature of the aluminum baseball bat, we can follow these steps:

### Step 1: Identify the given values
- Original length of the bat, \( L_1 = 0.86 \, \text{m} \)
- Initial temperature, \( T_i = 17 \, \text{°C} \)
- Change in length, \( \Delta L = 0.00016 \, \text{m} \)
- Coefficient of linear expansion for aluminum, \( \alpha = 23 \times 10^{-6} \, \text{°C}^{-1} \)

### Step 2: Use the formula for linear expansion
The formula for linear expansion is given by:
\[ \Delta L = L_1 \cdot \alpha \cdot \Delta T \]
Where:
- \( \Delta L \) is the change in length
- \( L_1 \) is the original length
- \( \alpha \) is the coefficient of linear expansion
- \( \Delta T \) is the change in temperature

### Step 3: Rearrange the formula to find \( \Delta T \)
From the formula, we can rearrange it to find the change in temperature:
\[ \Delta T = \frac{\Delta L}{L_1 \cdot \alpha} \]

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:
\[ \Delta T = \frac{0.00016 \, \text{m}}{0.86 \, \text{m} \cdot (23 \times 10^{-6} \, \text{°C}^{-1})} \]

### Step 5: Calculate \( \Delta T \)
Calculating the denominator:
\[ 0.86 \cdot (23 \times 10^{-6}) = 0.00001978 \, \text{m/°C} \]
Now substituting this back into the equation for \( \Delta T \):
\[ \Delta T = \frac{0.00016}{0.00001978} \approx 8.09 \, \text{°C} \]

### Step 6: Determine the final temperature
Now, we can find the final temperature \( T_f \):
\[ T_f = T_i + \Delta T = 17 \, \text{°C} + 8.09 \, \text{°C} \approx 25.09 \, \text{°C} \]

### Final Answer
Rounding to the nearest whole number, the final temperature of the bat is approximately:
\[ T_f \approx 25 \, \text{°C} \]