Question

Two bodies A and B at temperatures ' $\text{T}_1$ ' $\text{K}$ and ' $\text{T}_2$ ' $\text{K}$ respectively have the same dimensions. Their emissivities are in the ratio $16 : 1$. At $\text{T}_1 = x\text{T}_2$, they radiate the same amount of heat per unit area per unit time. The value of $x$ is

• A. 8
• B. 4
• C. 2
• D. 0.5

Hint:

If emissivity increases, temperature must decrease to maintain the same power output.

Answer 1

The Correct Option is D

Step 1: Concept
Stefan-Boltzmann Law: $E = e\sigma T^4$.
Step 2: Analysis
Given $E_A = E_B$. So, $e_A \sigma T_1^4 = e_B \sigma T_2^4 \implies \frac{e_A}{e_B} = \left( \frac{T_2}{T_1} \right)^4$.
Step 3: Calculation
Given $e_A/e_B = 16$.
$16 = \left( \frac{T_2}{T_1} \right)^4 \implies 2^4 = \left( \frac{T_2}{T_1} \right)^4 \implies \frac{T_2}{T_1} = 2$.
Thus, $T_1 = \frac{1}{2} T_2 = 0.5 T_2$. Comparing with $T_1 = xT_2$, we get $x = 0.5$.
Step 4: Conclusion
Hence, the value of $x$ is 0.5.
Final Answer: (D)