What is the volume of a lead ball at `20.00^@C` if the ball's volume at `60.00^@C` is `33.58 cm^3` ?

Answer

To find the volume of a lead ball at \(20.00^\circ C\) given its volume at \(60.00^\circ C\) is \(33.58 \, cm^3\), we can use the formula for volume expansion due to temperature change. The formula is:

\[ V_f = V_i \left(1 + \beta \Delta T\right) \]

Where:
- \(V_f\) is the final volume (volume at \(60.00^\circ C\))
- \(V_i\) is the initial volume (volume at \(20.00^\circ C\))
- \(\beta\) is the coefficient of volume expansion for lead
- \(\Delta T\) is the change in temperature

### Step 1: Identify the known values
- \(V_f = 33.58 \, cm^3\)
- \(\beta = 87 \times 10^{-6} \, K^{-1}\)
- Initial temperature \(T_i = 20.00^\circ C\)
- Final temperature \(T_f = 60.00^\circ C\)

### Step 2: Calculate the change in temperature
\[ \Delta T = T_f - T_i = 60.00^\circ C - 20.00^\circ C = 40.00 \, K \]

### Step 3: Rearrange the volume expansion formula to solve for \(V_i\)
\[ V_i = \frac{V_f}{1 + \beta \Delta T} \]

### Step 4: Substitute the known values into the equation
\[ V_i = \frac{33.58 \, cm^3}{1 + (87 \times 10^{-6} \, K^{-1}) \times 40.00 \, K} \]

### Step 5: Calculate the denominator
\[ 1 + (87 \times 10^{-6}) \times 40 = 1 + 0.00348 = 1.00348 \]

### Step 6: Calculate \(V_i\)
\[ V_i = \frac{33.58 \, cm^3}{1.00348} \approx 33.46 \, cm^3 \]

### Conclusion
The volume of the lead ball at \(20.00^\circ C\) is approximately \(33.46 \, cm^3\).

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