If the temperature of a black body incresese from `7^(@)C` to `287^(@)C` then the rate of energy radiation increases by

Options

A. a.`((287)/(7))^(4)`
B. b.16
C. c.4
D. d.2

Explanation

To solve the problem of how the rate of energy radiation increases when the temperature of a black body increases from \(7^\circ C\) to \(287^\circ C\), we can follow these steps:

### Step 1: Convert Celsius to Kelvin
First, we need to convert the temperatures from Celsius to Kelvin because the Stefan-Boltzmann Law applies to absolute temperatures.

- Initial temperature \(T_1\):
\[ T_1 = 7^\circ C = 273 + 7 = 280 \, K \]

- Final temperature \(T_2\):
\[ T_2 = 287^\circ C = 273 + 287 = 560 \, K \]

### Step 2: Use the Stefan-Boltzmann Law
The rate of energy radiation \(E\) from a black body is given by the Stefan-Boltzmann Law, which states that:
\[ E \propto T^4 \]
This means that the energy radiation is proportional to the fourth power of the absolute temperature.

### Step 3: Calculate the ratio of energy radiation
We can express the ratio of the energy radiated at the two temperatures as:
\[ \frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4 \]

### Step 4: Substitute the values
Now we substitute the values of \(T_1\) and \(T_2\) into the equation:
\[ \frac{E_2}{E_1} = \left(\frac{560}{280}\right)^4 \]

### Step 5: Simplify the ratio
Calculating the ratio:
\[ \frac{560}{280} = 2 \]
Thus,
\[ \frac{E_2}{E_1} = 2^4 = 16 \]

### Step 6: Conclusion
This means that the rate of energy radiation increases by a factor of 16 when the temperature of the black body increases from \(7^\circ C\) to \(287^\circ C\).

### Final Answer
The rate of energy radiation increases by **16 times**.

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