The temperature at which a black body of unit area loses its energy at the rate of 1 joule/second is

Options

A. `65^(@)`C
B. `-65^(@)`C
C. 65 K
D. 72 K

Explanation

To solve the problem of finding the temperature at which a black body of unit area loses energy at the rate of 1 joule/second, we will use Stefan-Boltzmann Law. The law states that the power radiated per unit area of a black body is directly proportional to the fourth power of its absolute temperature (T).

### Step-by-Step Solution:

1. **Understanding Stefan-Boltzmann Law**:
The Stefan-Boltzmann Law is given by the equation:
\[ E = \sigma T^4 \]
where:
- \( E \) is the energy radiated per unit area (in watts, which is joules/second),
- \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \),
- \( T \) is the absolute temperature in Kelvin.

2. **Setting Up the Equation**:
Since we want to find the temperature at which the black body radiates 1 joule/second (1 watt), we set \( E = 1 \, \text{W} \):
\[ 1 = \sigma T^4 \]

3. **Substituting the Value of \(\sigma\)**:
Substitute the value of \(\sigma\) into the equation:
\[ 1 = 5.67 \times 10^{-8} T^4 \]

4. **Rearranging the Equation**:
Rearranging the equation to solve for \( T^4 \):
\[ T^4 = \frac{1}{5.67 \times 10^{-8}} \]

5. **Calculating \( T^4 \)**:
Now, calculate \( T^4 \):
\[ T^4 = \frac{1}{5.67 \times 10^{-8}} \approx 1.76 \times 10^7 \]

6. **Taking the Fourth Root**:
To find \( T \), take the fourth root of \( T^4 \):
\[ T = (1.76 \times 10^7)^{1/4} \]

7. **Calculating \( T \)**:
Using a calculator, we find:
\[ T \approx 65 \, \text{K} \]

### Final Answer:
The temperature at which a black body of unit area loses its energy at the rate of 1 joule/second is approximately **65 Kelvin**.