If the emission rate of a blackbody at `0 ^@C` is R , then the rate of emission at `273^@C` is

Options

A. 2R
B. 4R
C. 8R
D. 16R

Answer

To solve the problem of finding the rate of emission of a blackbody at 273°C given that the rate of emission at 0°C is R, we can use Stefan-Boltzmann's law, which states that the power radiated per unit area of a black body is proportional to the fourth power of its absolute temperature (in Kelvin).

### Step-by-Step Solution:

1. **Convert temperatures to Kelvin:**
- The temperature at 0°C is:
\[ T_1 = 0 + 273 = 273 \, \text{K} \]
- The temperature at 273°C is:
\[ T_2 = 273 + 273 = 546 \, \text{K} \]

2. **Apply Stefan-Boltzmann Law:**
- According to Stefan-Boltzmann law, the rate of emission \(E\) is given by:
\[ E = \sigma T^4 \]
- For the first temperature (0°C), we have:
\[ R = \sigma T_1^4 = \sigma (273)^4 \]
- For the second temperature (273°C), we have:
\[ R' = \sigma T_2^4 = \sigma (546)^4 \]

3. **Establish the relationship between \(R'\) and \(R\):**
- From the above equations, we can write:
\[ \frac{R'}{R} = \frac{\sigma (546)^4}{\sigma (273)^4} \]
- The \(\sigma\) cancels out:
\[ \frac{R'}{R} = \frac{(546)^4}{(273)^4} \]

4. **Simplify the ratio:**
- This can be simplified as:
\[ R' = R \left(\frac{546}{273}\right)^4 \]
- Since \(546 = 2 \times 273\), we can substitute:
\[ R' = R \left(2\right)^4 \]
- Therefore:
\[ R' = R \times 16 \]

5. **Final Answer:**
- The rate of emission at 273°C is:
\[ R' = 16R \]