The volume of a solid aluminum ball with initial radius 20 cm increase by `347 cm^3` when the ball is heated . What is the temperature change ?

Answer

To solve the problem of finding the temperature change when the volume of a solid aluminum ball increases, we can follow these steps:

### Step 1: Gather Given Information
- The initial radius of the aluminum ball, \( r = 20 \, \text{cm} \)
- The change in volume, \( \Delta V = 347 \, \text{cm}^3 \)

### Step 2: Calculate the Initial Volume
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting the radius into the formula:
\[ V_i = \frac{4}{3} \pi (20)^3 \]
Calculating \( V_i \):
\[ V_i = \frac{4}{3} \times 3.14 \times 8000 \approx 3351.2 \, \text{cm}^3 \]

### Step 3: Determine the Volume Expansion Coefficient
The linear coefficient of thermal expansion for aluminum is given as:
\[ \alpha = 23 \times 10^{-6} \, \text{°C}^{-1} \]
The volume expansion coefficient \( \beta \) is three times the linear coefficient:
\[ \beta = 3 \alpha = 3 \times 23 \times 10^{-6} = 69 \times 10^{-6} \, \text{°C}^{-1} \]

### Step 4: Relate Volume Change to Temperature Change
The relationship between the change in volume and the change in temperature is given by:
\[ \Delta V = \beta V_i \Delta T \]
Rearranging to find \( \Delta T \):
\[ \Delta T = \frac{\Delta V}{\beta V_i} \]

### Step 5: Substitute Values into the Equation
Substituting the known values into the equation:
\[ \Delta T = \frac{347 \, \text{cm}^3}{(69 \times 10^{-6} \, \text{°C}^{-1}) \times (3351.2 \, \text{cm}^3)} \]

### Step 6: Calculate \( \Delta T \)
Calculating the denominator:
\[ \beta V_i = 69 \times 10^{-6} \times 3351.2 \approx 0.231 \, \text{cm}^3/\text{°C} \]
Now substituting back to find \( \Delta T \):
\[ \Delta T = \frac{347}{0.231} \approx 1507.76 \, \text{°C} \]

### Final Answer
The temperature change \( \Delta T \) is approximately:
\[ \Delta T \approx 1507.76 \, \text{°C} \]