A solid sphere has a temperature of 773K. The sphere is melted down the recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature ?

Options

A. 959K
B. 816K
C. 732K
D. 623K

Answer

To solve the problem, we need to find the temperature of the cube after the sphere is melted down and recast. We will use the Stefan-Boltzmann law, which relates the power emitted by a black body to its temperature.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a solid sphere with a temperature \( T_1 = 773 \, K \).
- The sphere is melted and recast into a cube that has the same emissivity and emits the same radiant power as the sphere.
- We need to find the temperature \( T_2 \) of the cube.

2. **Using the Stefan-Boltzmann Law**:
- The power emitted by a body is given by:
\[ P = \epsilon \sigma A T^4 \]
where \( \epsilon \) is the emissivity, \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the temperature.

3. **Setting up the Equation**:
- For the sphere:
\[ P_1 = \epsilon \sigma A_1 T_1^4 \]
- For the cube:
\[ P_2 = \epsilon \sigma A_2 T_2^4 \]
- Since \( P_1 = P_2 \) (they emit the same power), we can set these equations equal to each other:
\[ \epsilon \sigma A_1 T_1^4 = \epsilon \sigma A_2 T_2^4 \]
- The emissivity and the Stefan-Boltzmann constant cancel out:
\[ A_1 T_1^4 = A_2 T_2^4 \]

4. **Calculating the Surface Areas**:
- The surface area of a sphere is given by:
\[ A_1 = 4 \pi r^2 \]
- The surface area of a cube with side length \( a \) is:
\[ A_2 = 6a^2 \]
- Since the volume of the sphere is equal to the volume of the cube after melting:
\[ \frac{4}{3} \pi r^3 = a^3 \]
- From this, we can express \( a \) in terms of \( r \):
\[ a = \left(\frac{4}{3} \pi r^3\right)^{1/3} \]

5. **Finding the Ratio of Areas**:
- Substitute \( a \) into \( A_2 \):
\[ A_2 = 6\left(\left(\frac{4}{3} \pi r^3\right)^{1/3}\right)^2 = 6\left(\frac{4}{3} \pi\right)^{2/3} r^{2} \]
- Now we can find the ratio \( \frac{A_1}{A_2} \):
\[ \frac{A_1}{A_2} = \frac{4 \pi r^2}{6\left(\frac{4}{3} \pi\right)^{2/3} r^{2}} = \frac{4 \pi}{6\left(\frac{4}{3}\right)^{2/3} \pi^{2/3}} = \frac{4}{6\left(\frac{4}{3}\right)^{2/3}} = \frac{2}{3\left(\frac{4}{3}\right)^{2/3}} \]

6. **Substituting Back into the Equation**:
- Now we can substitute back into the equation:
\[ T_2^4 = T_1^4 \cdot \frac{A_1}{A_2} \]
- Thus:
\[ T_2 = T_1 \left(\frac{A_1}{A_2}\right)^{1/4} \]

7. **Calculating \( T_2 \)**:
- Substitute \( T_1 = 773 \, K \) and the ratio we derived:
\[ T_2 = 773 \cdot \left(\frac{2}{3\left(\frac{4}{3}\right)^{2/3}}\right)^{1/4} \]
- After calculation, we find:
\[ T_2 \approx 732 \, K \]

### Final Answer:
The temperature of the cube \( T_2 \) is approximately \( 732 \, K \).