Solar constant is 1370 `W//m^(2)` . `70%` of the light incident on the earth is absorbed by the earth and the earth’s average temperature is 288 K. The effective emissivity of the earth is :

Options

A. `0.2`
B. `0.3`
C. `0.6`
D. 1

Explanation

To find the effective emissivity of the Earth, we can follow these steps:

### Step 1: Understand the Given Data
We are given:
- Solar constant (intensity of sunlight) = 1370 W/m²
- Percentage of light absorbed by the Earth = 70%
- Average temperature of the Earth = 288 K

### Step 2: Calculate the Absorptive Power
The absorptive power (A) of the Earth can be calculated from the percentage of light absorbed:
\[ A = \frac{70}{100} = 0.7 \]

### Step 3: Use the Relation Between Absorptive Power and Emissivity
There is a fundamental relation in thermodynamics that states:
\[ A + E = 1 \]
where:
- \( A \) = absorptive power
- \( E \) = emissivity

### Step 4: Solve for Emissivity
We can rearrange the equation to find the emissivity (E):
\[ E = 1 - A \]
Substituting the value of A:
\[ E = 1 - 0.7 \]
\[ E = 0.3 \]

### Step 5: Conclusion
The effective emissivity of the Earth is:
\[ E = 0.3 \]