Object A has an emissivity of 0.95, and its temperature is `25^(@)C` .At what temperature ( in degrees Celsius ) does object B, whose emissivity is 0.60, emit radiation at the same rate as object A if both objects have the same surface area ?

Options

A. `28^(@)C`
B. `40^(@)C`
C. `61^(@)C`
D. `73^(@)C`

Answer

To solve the problem, we need to 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. The formula for the power radiated by an object is given by:

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

where:
- \( P \) is the power radiated,
- \( \varepsilon \) is the emissivity of the object,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area,
- \( T \) is the absolute temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the given values**:
- Emissivity of object A, \( \varepsilon_A = 0.95 \)
- Temperature of object A, \( T_A = 25^\circ C \)
- Emissivity of object B, \( \varepsilon_B = 0.60 \)
- Both objects have the same surface area, \( A \).

2. **Convert the temperature of object A to Kelvin**:
\[ T_A(K) = T_A(C) + 273 = 25 + 273 = 298 \, K \]

3. **Write the power radiated by both objects**:
- For object A:
\[ P_A = \varepsilon_A \sigma A T_A^4 = 0.95 \sigma A (298)^4 \]
- For object B:
\[ P_B = \varepsilon_B \sigma A T_B^4 = 0.60 \sigma A T_B^4 \]

4. **Set the power radiated by both objects equal**:
Since both objects emit radiation at the same rate:
\[ P_A = P_B \]
Therefore,
\[ 0.95 \sigma A (298)^4 = 0.60 \sigma A T_B^4 \]

5. **Cancel out common terms**:
The \( \sigma \) and \( A \) terms cancel out:
\[ 0.95 (298)^4 = 0.60 T_B^4 \]

6. **Solve for \( T_B^4 \)**:
Rearranging gives:
\[ T_B^4 = \frac{0.95 (298)^4}{0.60} \]

7. **Calculate \( (298)^4 \)**:
\[ (298)^4 = 7.85 \times 10^9 \, K^4 \]
Thus,
\[ T_B^4 = \frac{0.95 \times 7.85 \times 10^9}{0.60} \approx 1.25 \times 10^{10} \, K^4 \]

8. **Calculate \( T_B \)**:
Taking the fourth root:
\[ T_B = (1.25 \times 10^{10})^{1/4} \approx 334 \, K \]

9. **Convert \( T_B \) back to Celsius**:
\[ T_B(C) = T_B(K) - 273 \approx 334 - 273 = 61^\circ C \]

### Final Answer:
The temperature of object B is approximately \( 61^\circ C \).