Two bodies A and B having equal surface areas are maintained at temperatures `10^(@)C` . And `20^(@)C` . The thermal radiation emitted in a given time by A and B are in the ratio

Options

A. `1:1.15`
B. `1:02`
C. `1:04`
D. `1:16` .

Explanation

To find the ratio of thermal radiation emitted by two bodies A and B at different temperatures, we can use the Stefan-Boltzmann law, which states that the thermal radiation emitted by a body is proportional to the fourth power of its absolute temperature. Here’s how to solve the problem step by step:

### Step 1: Convert Temperatures to Kelvin
The temperatures of bodies A and B are given in Celsius. We need to convert these temperatures to Kelvin using the formula:
\[ T(K) = T(°C) + 273.15 \]

- For body A:
\[ T_A = 10°C + 273.15 = 283.15 K \]

- For body B:
\[ T_B = 20°C + 273.15 = 293.15 K \]

### Step 2: Apply the Stefan-Boltzmann Law
According to the Stefan-Boltzmann law, the thermal radiation emitted by a body is proportional to the fourth power of its absolute temperature. Therefore, we can write:
\[ \frac{E_A}{E_B} = \frac{T_A^4}{T_B^4} \]

### Step 3: Substitute the Temperatures
Now, we substitute the values of \( T_A \) and \( T_B \) into the equation:
\[ \frac{E_A}{E_B} = \frac{(283.15)^4}{(293.15)^4} \]

### Step 4: Calculate the Ratio
To find the numerical value of the ratio, we can calculate:
\[ \frac{E_A}{E_B} = \left(\frac{283.15}{293.15}\right)^4 \]

Calculating the fraction:
\[ \frac{283.15}{293.15} \approx 0.966 \]

Now raising this to the power of 4:
\[ (0.966)^4 \approx 0.866 \]

### Step 5: Final Ratio
Thus, the ratio of thermal radiation emitted by A and B is:
\[ \frac{E_A}{E_B} \approx 0.866 \]

This can be expressed as:
\[ E_A : E_B \approx 1 : 1.15 \]

### Conclusion
The thermal radiation emitted in a given time by bodies A and B are in the ratio of approximately 1:1.15.

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