A body of area `0.8xx10^-2m^2` and mass `5xx10^-4kg` directly faces the sun on a clear day. The body has an emissivity of 0.8 and specific heat of `0.8 cal//kg` K. The surroundings are at `27^@C`. (solar constant`=1.4 kW//m^(2)`).
The temperature that the body would reach if it lost all its heat by radiation is

Options

A. 396 K
B. `296^@C`
C. `85^@C`
D. `85K`

Explanation

To solve the problem, we need to find the temperature that the body would reach if it lost all its heat by radiation. We will use the concept of thermal equilibrium, where the rate of heat absorbed equals the rate of heat radiated.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Area of the body, \( A = 0.8 \times 10^{-2} \, \text{m}^2 \)
- Mass of the body, \( m = 5 \times 10^{-4} \, \text{kg} \)
- Emissivity, \( \epsilon = 0.8 \)
- Specific heat, \( c = 0.8 \, \text{cal/kg} \, \text{K} \) (Convert to Joules: \( 1 \, \text{cal} = 4.184 \, \text{J} \))
- Surrounding temperature, \( T_{\text{surroundings}} = 27^\circ C = 300 \, \text{K} \)
- Solar constant, \( S = 1.4 \, \text{kW/m}^2 = 1400 \, \text{W/m}^2 \)

2. **Convert Specific Heat to Joules:**
\[ c = 0.8 \, \text{cal/kg} \, \text{K} \times 4.184 \, \text{J/cal} = 3.3472 \, \text{J/kg} \, \text{K} \]

3. **Calculate the Power Absorbed by the Body:**
The power absorbed by the body from the sun is given by:
\[ P_{\text{absorbed}} = S \times A \]
Substituting the values:
\[ P_{\text{absorbed}} = 1400 \, \text{W/m}^2 \times 0.8 \times 10^{-2} \, \text{m}^2 = 11.2 \, \text{W} \]

4. **Set Up the Equation for Power Radiated:**
The power radiated by the body can be expressed using the Stefan-Boltzmann law:
\[ P_{\text{radiated}} = \epsilon \sigma A (T^4 - T_{\text{surroundings}}^4) \]
Where \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2 \, \text{K}^4 \).

5. **Equate Power Absorbed and Power Radiated:**
At thermal equilibrium:
\[ P_{\text{absorbed}} = P_{\text{radiated}} \]
Thus:
\[ 11.2 = 0.8 \times 5.67 \times 10^{-8} \times 0.8 \times 10^{-2} (T^4 - 300^4) \]

6. **Calculate the Right Side:**
First, calculate \( 0.8 \times 5.67 \times 10^{-8} \times 0.8 \times 10^{-2} \):
\[ = 0.8 \times 5.67 \times 10^{-8} \times 0.008 = 4.54 \times 10^{-10} \]

7. **Rearranging the Equation:**
\[ 11.2 = 4.54 \times 10^{-10} (T^4 - 300^4) \]
Rearranging gives:
\[ T^4 - 300^4 = \frac{11.2}{4.54 \times 10^{-10}} \]

8. **Calculate \( 300^4 \):**
\[ 300^4 = 8.1 \times 10^8 \, \text{K}^4 \]

9. **Calculate the Right Hand Side:**
\[ \frac{11.2}{4.54 \times 10^{-10}} \approx 2.46 \times 10^{10} \]

10. **Combine and Solve for \( T^4 \):**
\[ T^4 = 2.46 \times 10^{10} + 8.1 \times 10^8 \]
Approximating gives:
\[ T^4 \approx 2.46 \times 10^{10} \]

11. **Taking the Fourth Root:**
\[ T \approx (2.46 \times 10^{10})^{1/4} \approx 85 \, \text{°C} \]

### Final Answer:
The temperature that the body would reach if it lost all its heat by radiation is approximately \( 85 \, \text{°C} \).