A gold sphere has a radius of 1.000 cm at `25.0^(@)C`. If 7650 J of heat is added to the sphere, what will the final volume of the sphere be ? Gold has a density of 19.300 kg `// m^(3)` at `25.0^(@)C`, a specific heat capacity of `129J // ( kg. ""^(@)C )`, and a coefficient of volume expansion of `42.0 xx 10^(-6//""^(@)C )`.

Options

A. `2.88 xx 10^(-6)m^(3)`
B. `3.01 xx 10^(-6) m^(3)`
C. `4.32 xx 10^(-6) m^(3)`
D. `3.91 xx 10^(-6) m^(3)`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the initial volume of the gold sphere.
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

Given:
- Radius \( r = 1.000 \, \text{cm} = 0.01 \, \text{m} \)

Now, substituting the radius into the volume formula:
\[ V_1 = \frac{4}{3} \pi (0.01)^3 \]
Calculating this gives:
\[ V_1 = \frac{4}{3} \pi (1 \times 10^{-6}) \approx 4.19 \times 10^{-6} \, \text{m}^3 \]

### Step 2: Calculate the temperature change (\( \Delta T \)).
The heat added to the sphere is given by:
\[ Q = m c \Delta T \]
where:
- \( Q = 7650 \, \text{J} \)
- \( c = 129 \, \text{J/(kg} \cdot \text{°C)} \)

To find the mass \( m \), we use the density \( \rho \):
\[ m = \rho V_1 \]
Given that the density of gold is \( \rho = 19,300 \, \text{kg/m}^3 \):
\[ m = 19,300 \times 4.19 \times 10^{-6} \approx 0.0809 \, \text{kg} \]

Now, substituting \( m \) and \( c \) into the heat equation to find \( \Delta T \):
\[ 7650 = 0.0809 \times 129 \times \Delta T \]
Solving for \( \Delta T \):
\[ \Delta T = \frac{7650}{0.0809 \times 129} \approx 733.54 \, \text{°C} \]

### Step 3: Calculate the final volume of the sphere.
The final volume \( V_f \) can be calculated using the coefficient of volume expansion \( \gamma \):
\[ V_f = V_1 (1 + \gamma \Delta T) \]
Given:
- \( \gamma = 42.0 \times 10^{-6} \, \text{°C}^{-1} \)

Substituting the values:
\[ V_f = 4.19 \times 10^{-6} (1 + 42.0 \times 10^{-6} \times 733.54) \]
Calculating the term inside the parentheses:
\[ 1 + 42.0 \times 10^{-6} \times 733.54 \approx 1 + 0.0308 \approx 1.0308 \]
Now substituting back:
\[ V_f \approx 4.19 \times 10^{-6} \times 1.0308 \approx 4.32 \times 10^{-6} \, \text{m}^3 \]

### Final Answer:
The final volume of the gold sphere is approximately:
\[ V_f \approx 4.32 \times 10^{-6} \, \text{m}^3 \]

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