A man, the surface area of whose skin is `2m^(2)` , is sitting in a room where air temperature is `20^(@)C` if his skin temperature is `28^(@)C` and emissivity of his skin equals 0.97, find the rate at which his body loses heat.

Explanation

To solve the problem of calculating the rate at which a man's body loses heat, we will use Stefan-Boltzmann's law. The formula for the rate of heat loss (dQ/dt) is given by:

\[ \frac{dQ}{dt} = \sigma \epsilon A (T^4 - T_0^4) \]

Where:
- \(\sigma\) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4\)
- \(\epsilon\) is the emissivity of the skin
- \(A\) is the surface area
- \(T\) is the absolute temperature of the skin in Kelvin
- \(T_0\) is the absolute temperature of the surrounding air in Kelvin

### Step-by-step Solution:

1. **Identify the given values:**
- Surface area \(A = 2 \, \text{m}^2\)
- Air temperature \(T_0 = 20^\circ C\)
- Skin temperature \(T = 28^\circ C\)
- Emissivity \(\epsilon = 0.97\)

2. **Convert temperatures from Celsius to Kelvin:**
- \(T_0 = 20 + 273 = 293 \, \text{K}\)
- \(T = 28 + 273 = 301 \, \text{K}\)

3. **Substitute the values into the formula:**
\[ \frac{dQ}{dt} = \sigma \epsilon A (T^4 - T_0^4) \]
Substituting the known values:
\[ \frac{dQ}{dt} = (5.67 \times 10^{-8}) \times (0.97) \times (2) \times ((301)^4 - (293)^4) \]

4. **Calculate \(T^4\) and \(T_0^4\):**
- \(301^4 = 8.2501 \times 10^9\)
- \(293^4 = 7.6881 \times 10^9\)

5. **Calculate the difference \(T^4 - T_0^4\):**
\[ T^4 - T_0^4 = 8.2501 \times 10^9 - 7.6881 \times 10^9 = 0.5620 \times 10^9 \]

6. **Substitute back into the equation:**
\[ \frac{dQ}{dt} = (5.67 \times 10^{-8}) \times (0.97) \times (2) \times (0.5620 \times 10^9) \]

7. **Calculate the final value:**
\[ \frac{dQ}{dt} = 5.67 \times 10^{-8} \times 0.97 \times 2 \times 0.5620 \times 10^9 \]
\[ \frac{dQ}{dt} \approx 92.2 \, \text{W} \]

### Final Answer:
The rate at which the man's body loses heat is approximately **92.2 Watts**.