There is a small hole in a container. At what temperature should it be maintained in order that it emits one calorie of energy per second per `"meter"^(2)` :-

Options

A. `10K`
B. `500K`
C. `200K`
D. `100K`

Explanation

To solve the problem of determining the temperature at which a small hole in a container emits one calorie of energy per second per square meter, we can use Stefan's Law of radiation. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
- We need to emit 1 calorie of energy per second per square meter.
- Convert calories to joules:
\[ 1 \text{ calorie} = 4.2 \text{ joules} \]
- Therefore, the energy emitted per second (ΔQ) is:
\[ ΔQ = 4.2 \text{ joules} \]
- The area (A) is given as 1 m².

### Step 2: Apply Stefan's Law
Stefan's Law states that the power (energy per unit time) emitted by a black body is given by:
\[ P = \sigma A T^4 \]
Where:
- \( P \) is the power (in watts),
- \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \),
- \( A \) is the area (in m²),
- \( T \) is the absolute temperature (in Kelvin).

### Step 3: Substitute Known Values
Since we have:
- \( P = 4.2 \text{ joules/second} = 4.2 \text{ watts} \)
- \( A = 1 \text{ m}^2 \)

We can substitute these values into the equation:
\[ 4.2 = 5.67 \times 10^{-8} \times 1 \times T^4 \]

### Step 4: Solve for \( T^4 \)
Rearranging the equation gives:
\[ T^4 = \frac{4.2}{5.67 \times 10^{-8}} \]

### Step 5: Calculate \( T^4 \)
Calculating the right side:
\[ T^4 = \frac{4.2}{5.67 \times 10^{-8}} \approx 74000000.0 \]

### Step 6: Find \( T \)
Now, take the fourth root to find \( T \):
\[ T = (74000000.0)^{1/4} \]

### Step 7: Approximate \( T \)
Calculating the fourth root:
\[ T \approx 100 \text{ K} \]

### Conclusion
Thus, the temperature at which the container should be maintained to emit one calorie of energy per second per square meter is approximately:
\[ \boxed{100 \text{ K}} \]