The coefficient of volume expansion for gold is `4.20xx10^(-5)//""^@C` . The density of gold is 19 300 `kg//m^3` at `0.0^@C` . What is the density of gold at `1050^@C` ?

Options

A. `20 200 kg//m^3`
B. `18 500 kg//m^3`
C. `19 300 kg //m^3`
D. `18 800 kg//^3`

Answer

To find the density of gold at \( 1050^\circ C \), we can use the formula that relates the initial density (\( \rho_0 \)) to the density at a different temperature (\( \rho_\theta \)) considering the coefficient of volume expansion (\( \beta \)). The formula is given by:

\[ \rho_\theta = \frac{\rho_0}{1 + \beta \Delta T} \]

Where:
- \( \rho_0 \) is the initial density at \( 0^\circ C \)
- \( \beta \) is the coefficient of volume expansion
- \( \Delta T \) is the change in temperature

### Step-by-step Solution:

1. **Identify the given values:**
- Coefficient of volume expansion for gold (\( \beta \)) = \( 4.20 \times 10^{-5} \, \text{°C}^{-1} \)
- Initial density of gold (\( \rho_0 \)) = \( 19,300 \, \text{kg/m}^3 \)
- Initial temperature = \( 0^\circ C \)
- Final temperature = \( 1050^\circ C \)

2. **Calculate the change in temperature (\( \Delta T \)):**
\[ \Delta T = 1050^\circ C - 0^\circ C = 1050^\circ C \]

3. **Substitute the values into the formula:**
\[ \rho_\theta = \frac{19,300 \, \text{kg/m}^3}{1 + (4.20 \times 10^{-5} \, \text{°C}^{-1}) \times (1050 \, \text{°C})} \]

4. **Calculate the term in the denominator:**
\[ 1 + (4.20 \times 10^{-5}) \times 1050 = 1 + 0.0441 = 1.0441 \]

5. **Now calculate \( \rho_\theta \):**
\[ \rho_\theta = \frac{19,300 \, \text{kg/m}^3}{1.0441} \approx 18,485.5 \, \text{kg/m}^3 \]

6. **Final result:**
The density of gold at \( 1050^\circ C \) is approximately \( 18,485.5 \, \text{kg/m}^3 \).