The temperature of a perfect black body is `727^(@)C` and its area is `0.1 m^(2)`. If Stefan's constant is `5.67 xx 10^(-8) watt//m^(2)-s-K^(4)`, then heat radiated by it in 1 minute is:

Options

A. 8100 cal
B. 81000 cal
C. 810 cal
D. 81 cal

Explanation

To solve the problem of finding the heat radiated by a perfect black body in one minute, we will follow these steps:

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin**:
The temperature given is in Celsius. To convert it to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Substituting the given temperature:
\[ T = 727 + 273 = 1000 \, K \]

2. **Identify Given Values**:
We have the following values:
- Area \( A = 0.1 \, m^2 \)
- Stefan's constant \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)
- Time \( t = 1 \, \text{minute} = 60 \, \text{seconds} \)

3. **Apply Stefan-Boltzmann Law**:
The Stefan-Boltzmann law states that the power radiated per unit area of a black body is proportional to the fourth power of its temperature:
\[ \frac{dQ}{dt} = \sigma A T^4 \]
Substituting the values:
\[ \frac{dQ}{dt} = 5.67 \times 10^{-8} \times 0.1 \times (1000)^4 \]

4. **Calculate \( T^4 \)**:
First, calculate \( (1000)^4 \):
\[ (1000)^4 = 10^{12} \]

5. **Calculate Power Radiated**:
Now substitute \( T^4 \) into the equation:
\[ \frac{dQ}{dt} = 5.67 \times 10^{-8} \times 0.1 \times 10^{12} \]
\[ = 5.67 \times 10^{3} \, \text{W} = 5670 \, \text{W} \]

6. **Calculate Total Heat Radiated in 60 seconds**:
To find the total heat \( Q \) radiated in 60 seconds, we multiply the power by time:
\[ Q = \frac{dQ}{dt} \times t = 5670 \times 60 \]
\[ = 340200 \, \text{Joules} \]

7. **Convert Joules to Calories**:
To convert Joules to Calories, we use the conversion factor \( 1 \, \text{Calorie} = 4.2 \, \text{Joules} \):
\[ Q_{\text{calories}} = \frac{340200}{4.2} \approx 81000 \, \text{Calories} \]

### Final Answer:
The heat radiated by the perfect black body in one minute is approximately **81000 Calories**.

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