If the rates of loss of energy by unit area of an iron ball are `E_(1) , E_(2) ` and `E_(3)` when it cools from `75^(@)` to `70^(@)C` , `70^(@)C` to` 65^(@) C` and `65^(@)`C to `60^(@)C` respectively then -

Options

A. `E_(1) = E_(2) = E_(3)`
B. `E_(1) gt E_(2) gt E_(3)`
C. `E_(1) lt E_(2) lt E_(3)`
D. `E_(1) lt E_(3) lt E_(2)`

Answer

To solve the problem, we need to analyze the rates of energy loss by the iron ball as it cools down from different temperature intervals. We'll use Stefan's Law of radiation, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The iron ball is cooling from 75°C to 70°C, then from 70°C to 65°C, and finally from 65°C to 60°C.
- We denote the rates of energy loss per unit area during these intervals as \( E_1 \), \( E_2 \), and \( E_3 \) respectively.

2. **Applying Stefan's Law**:
- According to Stefan's Law, the rate of energy loss per unit area \( E \) can be expressed as:
\[ E = \epsilon \sigma A (T_b^4 - T_s^4) \]
where \( \epsilon \) is the emissivity of the material (constant for iron), \( \sigma \) is the Stefan-Boltzmann constant (also constant), \( A \) is the area (constant for a sphere), \( T_b \) is the temperature of the body in Kelvin, and \( T_s \) is the temperature of the surroundings in Kelvin.

3. **Converting Celsius to Kelvin**:
- Convert the temperatures from Celsius to Kelvin:
- \( T_b(75°C) = 75 + 273 = 348 K \)
- \( T_b(70°C) = 70 + 273 = 343 K \)
- \( T_b(65°C) = 65 + 273 = 338 K \)
- \( T_b(60°C) = 60 + 273 = 333 K \)

4. **Calculating Energy Loss for Each Interval**:
- For the interval from 75°C to 70°C:
\[ E_1 = \epsilon \sigma (348^4 - T_s^4) \]
- For the interval from 70°C to 65°C:
\[ E_2 = \epsilon \sigma (343^4 - T_s^4) \]
- For the interval from 65°C to 60°C:
\[ E_3 = \epsilon \sigma (338^4 - T_s^4) \]

5. **Comparing the Rates of Energy Loss**:
- Since \( T_s \) (surrounding temperature) remains constant, we can focus on the differences in \( T_b \):
- The values of \( E_1 \), \( E_2 \), and \( E_3 \) depend on the fourth power of the body temperatures:
- \( E_1 \) will be the highest since \( T_b \) is highest (348 K).
- \( E_2 \) will be less than \( E_1 \) since \( T_b \) is lower (343 K).
- \( E_3 \) will be the least since \( T_b \) is the lowest (338 K).
- Thus, we conclude:
\[ E_1 > E_2 > E_3 \]

### Final Conclusion:
The rates of energy loss per unit area are ordered as follows:
\[ E_1 > E_2 > E_3 \]