A hot water radiator at 310 K temperature radiates thermal radiation like a black body. Its total surface area is `1.6m^(2).` Find the thermal power radiated by it.

Answer

To find the thermal power radiated by the hot water radiator, we can use the Stefan-Boltzmann law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature and its surface area. The formula is given by:

\[ P = \varepsilon \sigma A T^4 \]

Where:
- \( P \) is the thermal power (in watts),
- \( \varepsilon \) is the emissivity of the surface (for a black body, \( \varepsilon = 1 \)),
- \( \sigma \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)),
- \( A \) is the surface area (in square meters),
- \( T \) is the absolute temperature (in Kelvin).

### Step-by-step Solution:

1. **Identify the given values**:
- Temperature, \( T = 310 \, \text{K} \)
- Surface area, \( A = 1.6 \, \text{m}^2 \)
- Emissivity for a black body, \( \varepsilon = 1 \)
- Stefan-Boltzmann constant, \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)

2. **Calculate \( T^4 \)**:
\[ T^4 = (310)^4 \]
First, calculate \( 310^2 \):
\[ 310^2 = 96100 \]
Then, calculate \( 310^4 \):
\[ T^4 = 96100^2 = 9236060100 \, \text{K}^4 \]

3. **Plug the values into the power formula**:
\[ P = \varepsilon \sigma A T^4 \]
Substituting the values:
\[ P = 1 \times (5.67 \times 10^{-8}) \times (1.6) \times (9236060100) \]

4. **Calculate the power**:
First, calculate \( 5.67 \times 10^{-8} \times 1.6 \):
\[ 5.67 \times 10^{-8} \times 1.6 = 9.072 \times 10^{-8} \]
Now multiply by \( T^4 \):
\[ P = 9.072 \times 10^{-8} \times 9236060100 \]
Performing this multiplication gives:
\[ P \approx 837.3 \, \text{W} \]

### Final Answer:
The thermal power radiated by the radiator is approximately \( 837.3 \, \text{W} \).