The temperature of a piece of iron is `27^(@)C` and it is radiating energy at the rate of `Q kWm^(-2)` . If its temperature is raised to `151^(@)C` , the rate of radiation of energy will become approximately

Options

A. `2 QkWm^(-2)`
B. `4Q kWm^(-2)`
C. `6Q kWm^(-2)`
D. `8Q kWm^(-2)`

Answer

To solve the problem, we will use the Stefan-Boltzmann law, which states that the power radiated per unit area of a black body is proportional to the fourth power of its absolute temperature. The formula is given by:

\[ P = \sigma T^4 \]

where:
- \( P \) is the power radiated per unit area,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( T \) is the absolute temperature in Kelvin.

### Step 1: Convert the temperatures from Celsius to Kelvin
The absolute temperature in Kelvin can be calculated by adding 273 to the Celsius temperature.

1. For the initial temperature of 27°C:
\[ T_1 = 27 + 273 = 300 \, K \]

2. For the final temperature of 151°C:
\[ T_2 = 151 + 273 = 424 \, K \]

### Step 2: Write the expression for the initial and final power radiated
Using the Stefan-Boltzmann law, we can express the initial and final power radiated per unit area:

1. Initial power radiated (\( P_1 \)):
\[ P_1 = \sigma T_1^4 = \sigma (300)^4 \]

2. Final power radiated (\( P_2 \)):
\[ P_2 = \sigma T_2^4 = \sigma (424)^4 \]

### Step 3: Find the ratio of the final power to the initial power
To find how the power changes when the temperature increases, we can take the ratio of \( P_2 \) to \( P_1 \):

\[ \frac{P_2}{P_1} = \frac{\sigma (424)^4}{\sigma (300)^4} = \left(\frac{424}{300}\right)^4 \]

### Step 4: Calculate the numerical value
Now we will calculate the numerical value of the ratio:

1. Calculate \( \frac{424}{300} \):
\[ \frac{424}{300} \approx 1.4133 \]

2. Raise this value to the power of 4:
\[ \left(1.4133\right)^4 \approx 5.737 \]

### Step 5: Calculate the new rate of radiation
If the initial rate of radiation is \( Q \) kW/m², then the new rate of radiation \( Q' \) can be calculated as:

\[ Q' = Q \times 5.737 \]

Thus, the new rate of radiation will be approximately \( 5.737Q \) kW/m².

### Final Answer
The rate of radiation of energy when the temperature is raised to 151°C will be approximately \( 5.737Q \) kW/m².

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