Emissive power of an ideal black body at `127^(@)C` is E. the temperature at which it increases to 102% is

Options

A. 400 K
B. 100 K
C. 402 K
D. 502 K

Answer

To solve the problem, we need to find the temperature at which the emissive power of an ideal black body increases to 102% of its initial value at 127°C. Let's break this down step by step.

### Step 1: Convert the initial temperature to Kelvin
The initial temperature \( T_1 \) is given as \( 127^\circ C \). To convert this to Kelvin, we use the formula:
\[ T_1 = 127 + 273 = 400 \, K \]

### Step 2: Write the expression for emissive power
The emissive power \( E \) of an ideal black body is given by Stefan-Boltzmann Law:
\[ E = \sigma A \epsilon T^4 \]
For an ideal black body, the emissivity \( \epsilon = 1 \). Thus, we can simplify the equation to:
\[ E = \sigma A T^4 \]

### Step 3: Write the initial emissive power
Substituting \( T_1 \) into the emissive power equation, we have:
\[ E = \sigma A (400)^4 \]

### Step 4: Calculate the new emissive power at 102%
The new emissive power \( E' \) when it increases to 102% of the initial emissive power \( E \) is:
\[ E' = 1.02 E \]

### Step 5: Write the expression for the new emissive power
Using the emissive power formula again for the new temperature \( T_2 \):
\[ E' = \sigma A (T_2)^4 \]

### Step 6: Set up the equation
Now we can set up the equation:
\[ 1.02 E = \sigma A (T_2)^4 \]
Substituting \( E \) from Step 3:
\[ 1.02 (\sigma A (400)^4) = \sigma A (T_2)^4 \]

### Step 7: Cancel out common terms
Since \( \sigma A \) is common on both sides, we can cancel it out:
\[ 1.02 (400)^4 = (T_2)^4 \]

### Step 8: Solve for \( T_2 \)
Taking the fourth root of both sides, we get:
\[ T_2 = 400 \times (1.02)^{1/4} \]

### Step 9: Calculate \( (1.02)^{1/4} \)
Using a calculator or approximation, we find:
\[ (1.02)^{1/4} \approx 1.005 \]
Thus,
\[ T_2 \approx 400 \times 1.005 = 402 \, K \]

### Step 10: Convert back to Celsius (if needed)
To convert \( T_2 \) back to Celsius:
\[ T_2 = 402 - 273 \approx 129^\circ C \]

### Final Answer
The temperature at which the emissive power increases to 102% is approximately \( 402 \, K \) or \( 129^\circ C \).

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