A black body having an area of `2xx10^(-4)m^(2)` for its radiating surface radiates energy of 16.42 J in 15 minutes. What is the temperature of the body ?

Explanation

To find the temperature of a black body given its radiating energy, we can use the Stefan-Boltzmann law. Here’s a step-by-step solution to the problem:

### Step 1: Identify the given values
- Area (A) = \(2 \times 10^{-4} \, m^2\)
- Energy radiated (Q) = 16.42 J
- Time (t) = 15 minutes

### Step 2: Convert time from minutes to seconds
Since the time is given in minutes, we need to convert it to seconds for consistency in SI units:
\[ t = 15 \, \text{minutes} \times 60 \, \text{seconds/minute} = 900 \, \text{seconds} \]

### Step 3: Use the Stefan-Boltzmann law
According to the Stefan-Boltzmann law, the power radiated by a black body is given by:
\[ P = \sigma A T^4 \]
where:
- \(P\) is the power (energy per unit time),
- \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \, W/m^2K^4\)),
- \(A\) is the area,
- \(T\) is the absolute temperature in Kelvin.

### Step 4: Calculate the power (P)
The power can be calculated as:
\[ P = \frac{Q}{t} = \frac{16.42 \, J}{900 \, s} \approx 0.01824 \, W \]

### Step 5: Rearrange the Stefan-Boltzmann equation to solve for temperature (T)
Rearranging the equation to find \(T\):
\[ T^4 = \frac{P}{\sigma A} \]
Substituting \(P\), \(\sigma\), and \(A\):
\[ T^4 = \frac{0.01824 \, W}{5.67 \times 10^{-8} \, W/m^2K^4 \times 2 \times 10^{-4} \, m^2} \]

### Step 6: Calculate the right-hand side
Calculating the denominator:
\[ \sigma A = 5.67 \times 10^{-8} \times 2 \times 10^{-4} = 1.134 \times 10^{-11} \, W \]
Now substituting back:
\[ T^4 = \frac{0.01824}{1.134 \times 10^{-11}} \approx 1.607 \times 10^{9} \]

### Step 7: Solve for T
Taking the fourth root:
\[ T = (1.607 \times 10^{9})^{1/4} \approx 200.0 \, K \]

### Final Answer
The temperature of the black body is approximately **200 K**.

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