The rate of dissipation of heat by a black body at temperature `T` is `Q`. What will be the the rate of dissipation of heat by another body at temperature `2T` and emissivity `0.25`?

Options

A. 16 Q
B. 4Q
C. 8Q
D. 4.5Q

Explanation

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

\[ P = \varepsilon \sigma A T^4 \]

where:
- \( P \) is the power (rate of heat dissipation),
- \( \varepsilon \) is the emissivity of the body (1 for a black body),
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the body,
- \( T \) is the absolute temperature in Kelvin.

### Step-by-step Solution:

1. **Rate of Heat Dissipation for Black Body at Temperature T**:
For a black body at temperature \( T \), the emissivity \( \varepsilon = 1 \). Thus, the rate of heat dissipation is:
\[ P_1 = \sigma A T^4 \]
Given that this rate is \( Q \), we have:
\[ Q = \sigma A T^4 \quad \text{(1)} \]

2. **Rate of Heat Dissipation for Another Body at Temperature 2T**:
For the second body at temperature \( 2T \) with emissivity \( \varepsilon = 0.25 \), the rate of heat dissipation is:
\[ P_2 = \varepsilon \sigma A (2T)^4 \]
Substituting \( \varepsilon = 0.25 \):
\[ P_2 = 0.25 \sigma A (2T)^4 \]
Simplifying \( (2T)^4 \):
\[ (2T)^4 = 16T^4 \]
Therefore:
\[ P_2 = 0.25 \sigma A \cdot 16T^4 \]
\[ P_2 = 4 \sigma A T^4 \quad \text{(2)} \]

3. **Relating \( P_2 \) to \( Q \)**:
From equation (1), we know \( Q = \sigma A T^4 \). Now substituting this into equation (2):
\[ P_2 = 4 Q \]

4. **Conclusion**:
The rate of dissipation of heat by the body at temperature \( 2T \) and emissivity \( 0.25 \) is:
\[ P_2 = 4Q \]

Thus, the answer is \( 4Q \).