If the absolute temperature of a black body is doubled the percentage increase in the rate of loss of heat by radiation is .

Options

A. `15%`
B. `16%`
C. `1600%`
D. `1500%`

Explanation

To solve the problem of finding the percentage increase in the rate of loss of heat by radiation when the absolute temperature of a black body is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Stefan-Boltzmann Law**:
The power radiated by a black body is given by the formula:
\[ P = E \sigma A T^4 \]
where:
- \( P \) is the power (rate of heat loss),
- \( E \) is the emissivity (for a black body, \( E = 1 \)),
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area,
- \( T \) is the absolute temperature in Kelvin.

2. **Define Initial and Final Temperatures**:
Let the initial absolute temperature be \( T_1 \) and the final temperature be \( T_2 \). According to the problem, if the absolute temperature is doubled:
\[ T_2 = 2T_1 \]

3. **Calculate Initial and Final Power**:
- The initial power \( P_1 \) at temperature \( T_1 \):
\[ P_1 = E \sigma A T_1^4 \]
- The final power \( P_2 \) at temperature \( T_2 \):
\[ P_2 = E \sigma A (2T_1)^4 = E \sigma A \cdot 16 T_1^4 \]

4. **Finding the Ratio of Powers**:
To find the ratio of the final power to the initial power:
\[ \frac{P_2}{P_1} = \frac{E \sigma A \cdot 16 T_1^4}{E \sigma A T_1^4} = 16 \]

5. **Calculate the Increase in Power**:
The increase in power can be calculated as:
\[ P_2 - P_1 = P_2 - P_1 = 16P_1 - P_1 = 15P_1 \]

6. **Percentage Increase in Power**:
To find the percentage increase:
\[ \text{Percentage Increase} = \left(\frac{P_2 - P_1}{P_1}\right) \times 100 = \left(\frac{15P_1}{P_1}\right) \times 100 = 1500\% \]

### Final Answer:
The percentage increase in the rate of loss of heat by radiation when the absolute temperature of a black body is doubled is **1500%**.

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