To solve the problem, we need to calculate the heat energy flowing through a small hole in an electric furnace that acts as a black body. The parameters given are:
- Area of the hole, \( A = 200 \, \text{mm}^2 \)
- Temperature of the metal, \( T = 727^\circ C \)
- Stefan-Boltzmann constant, \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)
### Step-by-Step Solution:
1. **Convert the Area from mm² to m²:**
\[
A = 200 \, \text{mm}^2 = 200 \times 10^{-6} \, \text{m}^2 = 0.0002 \, \text{m}^2
\]
**Hint:** Remember that \( 1 \, \text{mm}^2 = 10^{-6} \, \text{m}^2 \).
2. **Convert the Temperature from Celsius to Kelvin:**
\[
T = 727^\circ C + 273 = 1000 \, \text{K}
\]
**Hint:** To convert Celsius to Kelvin, add 273 to the Celsius temperature.
3. **Use the Stefan-Boltzmann Law to Calculate Heat Radiated per Second:**
The formula for the power radiated by a black body is given by:
\[
P = \sigma \cdot A \cdot E \cdot T^4
\]
Since the hole acts as a black body, the emissivity \( E = 1 \). Therefore, the equation simplifies to:
\[
P = \sigma \cdot A \cdot T^4
\]
4. **Substitute the Values into the Formula:**
\[
P = (5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4) \cdot (0.0002 \, \text{m}^2) \cdot (1000 \, \text{K})^4
\]
5. **Calculate \( T^4 \):**
\[
T^4 = (1000)^4 = 10^{12} \, \text{K}^4
\]
6. **Calculate the Power:**
\[
P = (5.67 \times 10^{-8}) \cdot (0.0002) \cdot (10^{12})
\]
\[
P = (5.67 \times 0.0002 \times 10^{4}) \, \text{W}
\]
\[
P = (5.67 \times 0.0002 \times 10000) \, \text{W}
\]
\[
P = (5.67 \times 2) \, \text{W} = 11.34 \, \text{W}
\]
7. **Final Result:**
The heat energy flowing through the hole per second is:
\[
P = 11.34 \, \text{J/s}
\]
### Conclusion:
The heat energy flowing through the hole per second is \( 11.34 \, \text{J} \).
**Final Answer:** \( 11.34 \, \text{J/s} \)
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