Two black bodies A and B have equal surface areas and are maintained at temperatures `27^@ C` and `177^@ C` respectively. What will be the ratio of the thermal energy radiated per second by A to that by B?

Options

A. `4 :9`
B. `2 :3`
C. `16 :81`
D. `27 :177`

Explanation

To solve the problem, we need to find the ratio of the thermal energy radiated per second by two black bodies A and B, which are at different temperatures. We will use the Stefan-Boltzmann law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature.

**Step-by-step solution:**

1. **Convert the temperatures from Celsius to Kelvin:**
- For body A:
\[ T_A = 27^\circ C + 273 = 300 \, K \]
- For body B:
\[ T_B = 177^\circ C + 273 = 450 \, K \]

2. **Use the Stefan-Boltzmann law to express the thermal energy radiated per second:**
- The formula for the power radiated by a black body is given by:
\[ Q = \sigma A T^4 \]
- Where \( Q \) is the thermal energy radiated per second, \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the absolute temperature in Kelvin.

3. **Set up the ratio of the thermal energy radiated per second by bodies A and B:**
- Since both bodies have equal surface areas and are black bodies, we can write:
\[ \frac{Q_A}{Q_B} = \frac{\sigma A T_A^4}{\sigma A T_B^4} = \frac{T_A^4}{T_B^4} \]

4. **Substitute the temperatures into the ratio:**
- Now substituting the temperatures we found:
\[ \frac{Q_A}{Q_B} = \frac{(300)^4}{(450)^4} \]

5. **Simplify the ratio:**
- This can be simplified as:
\[ \frac{Q_A}{Q_B} = \left(\frac{300}{450}\right)^4 = \left(\frac{30}{45}\right)^4 = \left(\frac{2}{3}\right)^4 \]

6. **Calculate the final ratio:**
- Now calculating \(\left(\frac{2}{3}\right)^4\):
\[ \left(\frac{2}{3}\right)^4 = \frac{2^4}{3^4} = \frac{16}{81} \]

7. **Final answer:**
- Therefore, the ratio of the thermal energy radiated per second by A to that by B is:
\[ Q_A : Q_B = 16 : 81 \]